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Two charges 2μC and –2μC are placed at points A and B, 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of electric field at every point on this surface?
Given two charges placed at points A and B.
A spherical conductor of radius 12 cm has a charge of 1.6 x 10–7 C distributed uniformly on its surface. What is the electric fields
(a) inside the sphere,
(b) just outside the sphere,
(c) at a point 18 cm from the centre of the sphere?
Given,
(a) Since, charge resides on the outer surface of the sphere, magnitude of electric field is 0 inside.
(b) Electric field just outside the sphere is given by
(c) At a point 0.18 m from the centre of the sphere
Three capacitors each of capacitance 9 pF are connected in series:
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Given, three capacitors of capacitances
(a) Since the capacitors are connected in parallel we have,
(b) Given,
q1,q2 and q3 represent the charge on each capacitor respectively.
Explain what would happen if in the capacitor given in Question 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
(a) while the voltage supply remained connected.
(b) after the supply was disconnected.
(a) Given,
Dielectric constant of mica sheet, k = 6
Thickness ofmica sheet, t =3 mm = m
Using the formula Q=CV
So, while voltage supply remained connected we have
(b) After the supply was disconnected, the charge remains same
i.e.,
and as
The charge remains constant i.e., Q = 1.08 x 10–8 C after the supply was disconnected and the voltage will come down to 16.6 V.
Given,
Electrostatic energy stored in the capacitor is given as
Given,
Charge, q =8 mC=8 x 10–3 C is located at origin and the small charge q0 =–2 x 10–9 C is taken from point P (0,0,3 cm) to a point Q (0,4 cm,0) through point R (0,6 cm,9 cm) as shown in the figure below.
Initial separation between q0 and q is rp = 3 cm = 0.03 m
Final separation between q0 and q is rQ = 4 cm = 0.04 m
Work done in taking the charge q0 from point P to Q does not dependent on the path followed and depends only upon rp and rQ i.e., initial and final positions.
Note: Point R is irrelevant to the answer.
Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this mid-point in a plane normal to the line and passing through the mid-point.
A ‘+ q’ charge which is placed at the centre of the shell will induce a ‘–q’ charge on the inner surface of the shell.
As a result of induction of '-q' on the inner surface, the charge on the outer surface will increase by '+q'.
Therefore, there will be total (Q + q) charge on the outer surface of the shell and, '–q' charge on the inner surface of the shell.
i) Now,
Thus,
Charge density on the outer surface
and
Charge density on the inner surface
ii) As charge resides on the outer surface so, the net charge on the inner surface of the cavity is zero as per the Gaussian theorem. Although, the net charge is zero yet the electric field may not be zero if the cavity is not spherical because, the surface may not have equal number of positive and negative charges.
We assume a loop for this reason, some portion of which is inside the cavity and rest of its part is inside the conductor. Now, consider that there is some electric field inside the cavity. Since inside the conductor total electric field is zero and net work done by the field in bringing a test charge over this loop will not be zero. But this is not possible for an electrostatic field. Therefore, we must conclude that there is no electric field inside the cavity irrespective of its shape.
where, E1' and E2' are the tangential components of E1 and E2 respectively.
Hence, E’1 = E’2
Therefore, the tangential component of the electrostatic field is continuous across the surface.
Given,
a) Potential energy =
Potential energy is zero at infinite separation.
Hence, the potential energy of the system is (–27.17–0) or 27.17 eV if zero of potential energy is taken at infinite separation.
b) Kinetic Energy of electron is given by
Total energy of electron = -27.2 + 13.6
= - 13.6 eV
Amount of work required to free the electron = Increase in energy of electron
= 0 – (–13.6)
= 13.6 eV
c) If the zero of Potential Energy is taken at 1.06 Å (1.06 x 10–10 m) separation, then the potential energy of system
The amount of work done to free the electron in this case is,
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Therefore, the ratio of the electric field of first sphere to that of the second sphere is b:a.
The surface charge densities of the two spheres are given as
This implies,
The surface charge densities are inversely proportional to the radii of the sphere.
The surface charge density on the sharp and pointed ends of a conductor is higher than on its flatter portion since a flat portion may be taken as a spherical surface of large radius and a pointed portion as that of small radius.
where p = q.(2a) is the dipole moment.
Electrostatic potential at P(0,0,z) when closer to charge -q is given by [see fig. below] :
So potential at point P (0, 0, ± z) is
Now, Electrostatic potential at point (x, y, 0 ) is
Therefore, Electrostatic potential at point (x,y,0) is equal to 0.
Given,
Two charges -q and +q are located at (0,0,-a) and (0,0,a) respectively.
To find: the amount of work done to move a small test charge from point (5,0,0) to (-7,0,0) along x-axis.
In moving small test charge q from the point (5, 0, 0) to (–7, 0, 0) the work done W = q(V1– V2) = q x 0 = 0.
The work done by the electrostatic field between two points is not dependant on path connecting two points. Therefore, work done by any charge along X axis, Y axis or on equitorial plane is zero as the potential does not change.
Here we have,
Total required capacitance, C = 2
Potential difference applied across the circuit, V =1 kV= 1000 V
Capacitance of each capacitor, C1 = 1
Maximum Potential difference applied across each capacitor, V1 = 400 V
Let, the possible arrangement of circuit be such that n capacitors of 1 each, be connected in series in a row and m such rows be connected in parallel.
Therefore, total number of capacitors-N = m x n
Since the potential in each row is 1000 V, the number of capacitors in each row of series arrangement is
Since, the number of capacitors cannot be in fraction, therefore, we take n=3.
Now, the capacitance of n capacitors in series, in m such rows is
So, the resultant capacitance of all capacitors is equal to
Hence, 1 capacitor each, should be connected in 6 parallel rows, where, each row contains 3 capacitors each in series.
The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.
Given,
Area of the parallel plate capacitor, A=90 cm2=
Distance between the parallel plates, d=2.5 mm =
Potential applied across the capacitor, V= 400 V
(a) Electrostatic energy stored in the capacitor, U
(b) Volume of the medium in between the plates
Energy stored per unit volume, u =
So, relation between magnitude of electric field and u
[]
Given, a 4 μF capacitor is charged by a 200 V supply.
Then,
Charge on the capacitor-Q= CV
= 4 x 10–6 x 200
= 8 x 10–4 C
Also given, it is then disconnected and, connected to another uncharged 2μF () capacitor.
Thus, total capacitance-C'= (4 + 2)μF =6 μF
Until both the capacitor acquire a common potential, charge on the first capacitor is shared between them.
After the combination, the common potential-V' =Q/C
Electrostatic energy of the first capacitor, before the combination,
Electrostatic energy of the system, after the combination,
Now, electrostatic energy lost by the first capacitor in the form of heat and electromagnetic radiation is
A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 μC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the sphere?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
The dependance of solid angle on radius 'r' is given by
.
Solid angle is that angle an arbitary area makes at a point P.
Gauss's law which happens to be an equivalent of Coulomb's law will not hold true if, Coulomb's law involved 1/r3 dependance instead of 1/r2 dependance.
Given, a cylindrical capacitor has two co-axial cylinders, of which, the outer cylinder is earthed.
Length of cylinder-l= 15 cm =
Radius of inner cylinder-a = 1.4 cm=
Radius of outer cylinder-b = 1.5 cm=
Charge on the inner cylinder-q =
Capacitance of the sysyem=
Since, the outer cylinder is earthed the potential of inner cylinder will be equal to the potential difference between inner and outer cylinder.
Hence, potential of inner cylinder
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Given,
Voltage rating for the design of parallel plate capacitor-V= 1 KV=1000 V
Dielectric constant of material-K== 3
dielectric strength of material = 107 V/m
For safety, electric field at the most should be 10% of dielectric strength.
E=10% of 107
= V/m
E= 106 V/m
Area of the plates-A = ?
Capacitance of plates-C=
Now, using the formula
∴
Therefore,
Hence, the minimum area of plates required is .
Given,
Potential difference across the capacitor-V = 200 V
charge stored in the capacitor-q = 0.1 C
Energy released on discharging = energy stored on charging =
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Electrical energy of the atmosphere is dissipated as
(i) light energy, involved in lightning.
(ii) heat and sound energy, in accompanying thunder.
Given,
The charges are placed on x-axis.
Electrostatic potential energy of the system of charge is,
A parallel plate capacitor, each with plate area A and separation d, is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab of thickness d and dielectric constant k is now placed between the plates. What change, if any, will take place in
(i) charge on the plates,
(ii) electric field intensity between the plates,
(iii) capacitance of the capacitor.
Justify your answer in each case.
Capacitance of the capacitor before a dielectric slab is placed in between is given by,
Potential difference = V
Initial charge on capacitor -qo=
i) When, the battery is disconnected, charge on the capacitor remains unchanged.
i.e, q= qo= .
(ii) Initial electric field between the plates-
when, dielectric is introduced in between the plates, the permittivity of medium becomes
Now, electric field in between the plates-
Thus, electric field is reduced by a factor of .
(iii) The capacitance increases due to the decrease in potential difference and for any dielectric, K>1
The given graph shows the variation of charge q versus potential difference for two capacitors C1 and C2. The two capacitors have same plate separation, but the plate area of C2 is double that of C1. Which of the lines in the graph corresponds to C1 and C2 and why?
Given,
length of the elctric dipole = 4cm=
angle made with the elctric field-
Torque experienced by the dipole- =
charge on the dipole-q=
i) magnitude of Electric field-E
(ii) magnitude of potential energy of the dipole
In which of the following two cases, more work will be done in increasing the separation between the plates of a charged capacitor and why?
(i) The charging battery remains connected to the capacitor.
(ii) The battery is removed after charging the capacitor.
In the same question, if battery is removed after charging the condenser and dielectric slab introduced how are all the five parameters affected?
A surface that has the same electric potential at every point is called an equipotential surface.
Given,
Electric field E, directed along PQ= 300 NC-1
(i) Work done to take a positive charge from point A to B is 0 because, the charge is moving perpendicular to the direction of the electric field.
Potential difference between A and B = 0
Therefore, potential difference between B and C
In a parallel plate capacitor, how is the capacity affected, when without changing the charge:
(i) the distance between the plates is doubled;
(ii) the area of the plate is halved.
Capacitance is given by
where,
'A' is the area of plates and 'd' is the distance between plates.
(i) when distance between the plates is doubled, capacity is halved.
(ii) Since, C ∝ A; when area of plates is halved, capacity is also halved.
A parallel plate capacitor with air as dielectric is charged by a d.c. source to a potential V. Without disconnecting the capacitor from the source, air is replaced by another dielectric medium of dielectric constant K. State with reason, how does
(i) potential difference
(ii) electric field between the plates
(iii) capacity
(iv) charge and
(v) energy stored in the capacitor, change.
Given, a parallel plate capacitor with air as dielectric is charged by d.c source. Later on, without disconnecting the source, air is replaced by another dielectric medium K.
Initial capacitance when, air is the dielectric medium, Co=
Capacitance when, dielectric medium of dielectric constant K is introduced, C=
(i) Potential difference will remain constant as the capacitor remains connected to the battery.
(ii) Electric field is given by .
Since, neither the potential difference nor the separation between the plates is changing therefore, the electric field remains unchanged.
(iii) When, dielectric medium is introduced, capacity would increase by a factor of K.
So,
(iv) Since, capacitance is increased by a factor of K and potential remains unchanged therefore charge is increased by a factor of K. Additional charge flows from the battery to the plates.
(v) Energy stored in the capacitor is,
.
Therefore, the energy is increased by a factor of K.
Electric potential at a point is defined as the amount of work done, in moving a unit positive charge with zero acceleration from infinity to that point.
Potential energy at a position is equal to the amount of work done to carry the total charge from infinity to that position, against the electrostatic forces.
Thus, potential energy = potential x charge.
Potential energy, is the energy possessed by the charge by virtue of its particular position.
Given,
Charge, q1=
Charge, q2=
Distance betwen the charges, d=9-(-9)=18 cm=
i) Electrostatic potential energy of the system is,
(ii) Work done to seperate the two charges infinitely away from each other
= 0.7 J.
(c) Given,
When, the system is placed in an external electric field, electrostatic energy is,
[ E=V/r]
Given,
Thickness of the dielectric slab, t =1 cm=10-2 m
Dielectric constant, εr =K = 5
Area of the plates of the capacitor, A= 0.01 m2= 10-2 m2
Distance between parallel plates of the capacitor, d =2 cm = 2 x 10–2 m
Therefore,
Capacity with air in between the plates
Capacity with dielectric slab in between the plates
Capacity with conducting slab in between the plates
Increase in capacity on introduction of dielectric
C – C0 = 7.375 x 10–12 – 4.425 x 10–12
= 2.95 x 10–12 farad
Increase in capacity on introduction of conducting slab
C’ – C0 = 8.85 x 10-12 – 4.425 x 10–12
= 4.425 x 10–12 farad.
A parallel plate capacitor with air as dielectric is charged by a d.c. source to a potential ‘V’. Without disconnecting the capacitor from the source, air is replaced by another dielectric medium of dielectric constant 10. State with reason, how does (i) electric field between the plates and (ii) energy stored in the capacitor change.
(ii) Surface A is given a constant potential of 10V. Therefore, the potential difference between X and Y is given as
Work done is W=
= 0
Therefore, no work is done in moving a charge from X and Y.
Given, two point charges q1 and q2 are brought from infinity to points P1 and P2 in the presence of an external field.
Let be the required external field.
The potential energy of the system is equl to the total amount of work done in assembling the configuration.
Therefore,
Work done on q1 against the external field = q1.V( )
Work done on q2 against the external electric field
and,
Work done on q2 against the field due to q1= Work done on q1 against the field due to q2 which is given by,
where, r12 is the distance between q1 and q2.
Using the superposition principle for fields, we add up the work done on q2 against the two fields.
Therefore, work done in carrying q2 to r2 is
Hence, Potential energy of the system is given by,
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Given,
Charge, Q = 12 μC
Potential applied across the capacitor, V = 1200 V
Dielectric strength of air = 3 x 107 V/m
We have to find the minimum area of the capacitor, A.
Since, dielectric strength is the maximum electric field, a material can tolerate without break down. Therfore, for safety let the electric field never exceed ~ 10% of dielectric strength.
That is,
minimum area of capacitor is given by,
Given,
Capacitance on capacitor 1, C1 = 1.0 μF
Maximum voltage applied across capacitor 1, V1 = 6 kV = 6 x 103 V
Capacitance on capacitor 2, C2 = 2 μF
Maximum voltage applied across capacitor 2, V2 =4 kV = 4 x 103 V
Now,
Charge on first capacitor, q1 = C1 V1
= 1.0 x 6 x 103 μC
= 6000 μC
Charge on second capacitor, q2 = C2V2
= 2.0 x 4 x 103 μC
= 8000 μC
In series combination, the magnitude of charge on each capacitor must be the same.
As maximum charge on C1 is 6000 μC, therefore, maximum charge on C2 must also be 6000 μC.
Hence, maximum voltage for the combination is,
If a piece of metal has a charge + 0.1 μC and is placed inside a hollow metal sphere of radius 20 cm (without touching it), what is the potential of the sphere? What will the potential of the sphere become, if
(a) the sphere is temporarily earthed and then left insulated,
(b) the metal subsequently touched the inside of the sphere?
(i)When the sphere is temporarily earthed and then left insulated, the potential of the sphere is momentarily reduced to zero. The positive charge on the outer surface disappears but the induced negative charge inside remains. Therefore, the potential of the sphere is zero.
(ii) The induced negative charge is neutralized when the metal touches the sphere and no charge remains on the metal or sphere. Both are at the same potential that is, the sphere has zero potential.
Calculate the potential at P due to the charge configuration as shown in the following figure.
If r >> a, then how will you modify the result?
Let, C be the capacitance of the infinite ladder.
As the ladder is infinite, addition of one more elements of two capacitors (1 μF and 2 μF) across the points X and Y should not change the total capacitance. Therefore, total capacity of the arrangement shown in figure must remain C only. The resultant circuit diagram for the circuit is re-drawn as shown in the fig. below.
As seen, capacitor of 2 μF is in series with capacitance C.
using the formula, for series combination,
Combined capacity =
This combination is in parallel with 1 μF capacitor.
Therefore,
The equivalent capacity of the arrangement is
i.e.,
Since, the value of capacitance cannot be negative.
Given,
Charge on the particle, q= 2 μC
Mass of the particle, m= 10 mg
Velocity with which the particle is moving, v= 1000 ms–1
Strength of uniform electric field, E= 102 NC–1
Therfore,
Acceleration experienced by the particle after 10 sec,
And, velocity attained by the particle after 10 sec, perpendicular to the direction of motion is,
vy = uy + ayt
= 0 + 20 x 10
= 200 ms–1
Hence, there is no force acting along the direction of motion.
∴ velocity along the direction of motion remains same i.e., vn = 1000 ms–1
∴ Net velocity after 10 s
Given, dipole moment of an electric dipole, p= 3 x 10–3
Electric field intensity of dipole, E= 104 NC–1
Initially, dipole is in a stable equilibrium.
This implies,
and,
Therefore,
Work done in rotating the dipole from an angle of θ1 to θ2, W = pE (cos θ1 – cos θ2)
W = 3 x 10–3 x 104 [1 – (– 1)]
= 60 J.
D.
depends on the ratio of the material of the sphere
B.
the potential difference across the plates of A is 6 V and across the plates of B is 4 VA surface with constant value of potential at all points on the surface is termed as an equipotential surface.
Two charges 2 μC and – 2 μC are placed at points A and B 6 cm apart.
(i) Identify an equipotential surface of the system.
(ii) What is the direction of the electric field at every point on this surface?
We are given, two charges 2 μC and – 2 μC placed 6 cm apart at points A and B.
i) The potential is equivalent when, the plane is normal to AB and is passing through the middle point of AB.
ii) The electric field is normal to the plane directing towards AB.
Consider, two charges q and -q seperated by distance 2a such that, it's total charge is 0.
Potential due to charge q =
Potential due to charge -q =
Potential due to dipole, is the sum of potential due to charges q and -q.
Therefore,
where, r1 and r2 are distances of charges q and -q from point P.
Now,
Considering r much greater than a, r>>a.
Therefore,
Putting these values in the above equation of potential we have,
where,
For potential at any point on axis,
V=
Potential is positive when and potential is negative when .
Electrical potential falls off at large distance, as and not as characteristic of the potential due to a single charge.
Given, an electric charge 10-3 μC is placed at origin of (x,-y) coordinate system.
Point A is at
Point B is at (2,0).
Let, rA be the distance from point A to origin and,
rB be the distance from point B to (0,0).
Therefore,
rA=
rB=
and, charge is, q =
Now,
Potential difference between points A and B is given by,
VA-VB=
Therfore, the potential difference is 0 between points A and B.
Given that, the charges are placed on the vertices of the square.
Let, the distance from the centre of the square to the vertex be 'a'.
Potential before and after interchanging the charges is 0 .
Direction of electric field before and after interchainging the charges is shown below. we can see that, after interchanging the charges A and B with C and D, the direction of electric field has reversed.
Therefore, electric field changes and potential remains constant.
While charging the capacitor, a certain amount of work is done on the system. And, when the dielectric is removed, an equal amount of work is done by the system. Since, the work done on the system and work done by the system is equal, net amount of work done is zero.
Consider, E as the emf of the battery.
Then, charge gained by the battery is q=CE
and,
Work done by the battery is given by, W=qE
Now, energy stored in the capacitor-U=
Therefore, the ratio of energy stored in the capacitor and work done by the battery is,
Consider, a capacitor consisting of two thin conducting plates 1 and 2, each of area 'A' held parallel to each other, at a distance 'd' apart.
One of the plates is insulated and the other plate is earth connected.
When a charge +Q is given to the insulated plate , then a charge of -Q is induced on the nearer face of the plate 2 and +Q is induced on the farther face of plate 2. As the plate 2 is earthed charge +Q being free, flows to the earth.
Surface charge density of plate 1 ,
Surface charge density of plate 2 = -
Electric field intensity in between the plates, E =
Taking this field localised between the plates as uniform throughout,
Potential difference between the plates, V =
Capacity of a parallel plate capacitor is given by, C = = =
Now, when the plates of the capacitor are seperated by a dielctric medium of relative permittivity then,
Capacity, Cm =
i.e., Cm = K Co
Therefore, the capacity becomes K times the initial capacity when dielectric medium is inserted in between the plates.
Given, a parallel plate capacitor is charged to a potential difference V. If the distance between the plates is doubled then,
i) Electric field in between the plates is given by
Since, E does not depend upon distance between the capacitor plates, electric field remains unaffected.
ii) Capacitance is given by C=
When distance between the plates is doubled C'=
That is, capacitance becomes half of it's initial value.
iii) Energy is given by U=
There is no change in Q but, capacitance changes. Since Capacitance becomes half, stored energy of the capacitor becomes twice of it's initial value. [Capacitance and energy have inverse dependance].
Given,
Charge, q= 0.5 μC
Distance between point X and point charge, r=10 cm
i) Electric potential at X, V is given by
ii) Amount of work done in bringing a charge of 3 from infinioty to X.
Work done = charge potential at point X
= 3 10-9 4.5 104
= 13.5 10-5 J
A metal wire is bent into a circle.
Radius of the circle so formed, r= 10 cm
Charge distributed on circle, q =200 μC=.
The potential at the centre of circle is simply given by
Given, a hollow metallic sphere.
Charge given to the sphere, q= 12 μC
radius of the sphere, r= 0.1 m
i) Potential at the surface of sphere, V=
ii) Potential at the centre is same as that of the potential at the surface. The charge will get accumulated at the centre of the hollow sphere.
Let, the point charges be given by qA and qB.
Charge, qA=
Charge, qB=
Distance betwen charges, d=20 cm
Let, the point where the potential is 0 be O.
Assume, the distance of O from point charge A be x.
then, distance of O from point charge B will be (.2-x)
Since the electric potential is zero at point O therefore,
Potential at point O due to qA=Potential at point O due to qB
Therefore, the point O is at a distance of 0.08 m from charge placed at A.
Given, a parallel plate capacitor.
Capacitance of plates, C= 2F
Seperation between the plates, d=0.5 cm=
Capacitance of a parallel plate capacitor is given by the formula C=
Therefore, area of the plates, A =
Given,
Charge, q1 = 8 nC = 8 10-9 C
Charge, q2 = -3 nC = C
Electrostatic potential energy is given by,
i) Capacitance of a capacitor is defined as the ratio of the electric charge on the capacitor to the electric potential of capacitor due to it's charge.
ii) Dielectric strength of a dielectric is defined as the maximum value of electric field that can be applied to the dielectric without it's electric breakdown.
The unit of dielectric strength is Volt per metre.
iii) When a dielectric slab is introduced in between the plates of capacitor, the electric field gets reduced.
Consider a parallel plate capacitor with vaccum in between it's plates. The capacitor is charged up with battery such that electric field is set up between it's plates.
Then,
where, is the surface charge density of the plates.
Now, as soon as the dielectric is introduced in between the plates each molecule of the dielectric get's polarised. Charges are induced on the surface of the dielectric and, these induced charges set up an electric field EP inside the dielectric.
Therefore, the resultant electric field gets reduced and is given as
E=Eo- EP
iv) When there is vaccum in between the plates, capacitance is given by C= .
When, dielctric is inserted in between the plates, Capacitance increases by a factor of K.
where, K is the dielectric constant.
Capacitance becomes C=.
Given, a right circular cylinder of length l cm and radius r cm along the x- axis.
E = Exi N/C for x > 0 and,
E = Ex i N/C for x < 0
E and are parallel for the perimeter area of cylinder.
Therefore, the outer flux is given by,
Along the curved surface are of the cylinder E and is perpendicular to each other.
Therefore, the outer flux is 0.
Hence, the net outward flux is given by,
The net charge within the cylinder as per gauss law is given by q .
C
The diagonally opposite points lie at equal distance from the centre of the sphere.
Let,
The distance from the centre of the square to opposite diagonal points be 'r'.
Opposite diagonal points be A and B
Therefore,
Since, VA=VB there is no potential difference across A and B.
Hence, the work done in moving a charge is 0.
Let, the charges on the vertices of triangle be qA, qB and qC.
Given, charge q is 1.6 x 10–10 C.
The work done to dissosciate the system would be the amount of potential energy required to bring a charge from one point to another.
Potential energy is given by
Therefore, work done to dissosciate the system is W=-U
W= 2.3 10-8 J
Given, two point charges.
Charge 1, q1 = 10 x 10–8 C
Charge 2, q2 = – 2 x 10–8 C
Distance between the charge, d= 60 cm =0.6 m
i) Let, the required point where the electric potential is 0 be at a distance'x' from the charge 1 and (0.6-x) distance from charge 2.
Therefore, V1+ V2=0
Hence, Potential is zero at a point 0.5 m away from the first charge.
ii) Electrostatic potential energy of the system,U=
Given, two point charges 4Q and Q seperated by 1 m in air.
Let, the point where the electric field intensity is 0 be at a distance x from charge Q.
Therefore,
Hence, at a distance 0.3 m from the first charge, electric field intensity is 0.
A parallel plate capacitor us charged to a potential V. After some time, spacing between the plates is halved and dielectric medium of K=10 is introduced without disconnecting the d.c source.
When the d.c source remains connected, potential across the plates remain same.
i) Initial capacitance, C=
When the spacing between the plates is halved d=d/2
A dielectric slab of K=10 is inserted.
New capacitance, C'= K==20 C
New capacitance becomes 20times that of the initial capacitance.
ii) Electric field is given by
When spacing is halved.
New electric field becomes E'=
Therefore, electric field becomes twice that of the initial field.
Two parallel plate capacitors, X and Y have the same area of plates and same separation between them. X has air between the plates while Y contained a dielectric medium of εr = 4. (0 Calculate capacitance of each capacitor if equivalent capacitance of combination is 4 μF.
(ii) Calculate the potential difference between the plates of X and Y.
(iii) What is the ratio of electrostatic energy stored in X and Y?
Given,
Using the relation
Putting values,
Given,
Since,
Therefore,
Power consumed = Current x Voltage
=
= 2.2 x 200
= 484 watt.
The peak voltage of an ac supply is 300 V. What is the rms voltage?
Given,
Now,
Using the relation,
The rms value of current in an ac circuit is 10 A. What is the peak current?
Given,
Hence, using the formula,
which is the required value of peak current.
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Given,
Inductance, L = 44 mH =
Now,
Given,
Rms value of current,
Now,
On putting the values, we get
Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?
Here,
Inductance, L = 2.0 H
Capacitance,
Resistance, R = 10 Ohm
Inductive reactance,
Suppose the initial charge on the capacitor in Question 7.7 is 6 mC. What is the total energy stored in the cirucit initially? What is the total energy at later time?
Given, initial charge in capacitor = 6 mC = 6
Since,
Energy of capacitor =
Since, energy is not absorbed by capacitor, there is no loss of energy. Therefore, the total energy remains the same.
A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?
[Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.]
Frequency of radio, f1 = 800 kHz = 800 × 103 Hz,
Frequency tuned to, f2 = 1200 kHz = 1200 × 103 Hz
Inductance of LC circuit, L = 200 μH = 200 × 10–6 H
We know that the resonant frequency is given by
Therefore,
Similarly,
Hence, the range of the variable condenser is from 87.88 pF to 197.73 pF.
Here, we are given a LCR circuit.
Inductance, L = 5.0 H
Resistance, R = 40
Capacitance,
Effective voltage,
Peak voltage,
(a) Resonance angular frequency is given by,
(b) Impedance of the circuit,
At resonance,
Therefore,
Amplitude of current at resonating frequency
(c) Potential drop across L
Potential drop across R
Potential drop across C
Therefore,
Potential drop across LC circuit
Thus, the potential drop across the LC combination is zero at the resonating frequency.
An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0.
(a) What is the total energy stored initially? Is it conserved during LC oscillations?
(b) What is the natural frequency of the circuit?
(c) At what time is the energy stored
(i) completely electrical (i.e., stored in the capacitor)?
(ii) completely magnetic (i.e., stored in the inductor)?
(d) At what times is the total energy shared equally between the inductor and the capacitor?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?
Given,
Inductance, L = 0.50 H
Resistance, R = 100
We can calculate,
Angular frequency,
Peak voltage,
(a) Maximum amount of current in the coil is given by,
(b) In LR circuit,
If ,
Now,
At
At
∴ Time lag between voltage maximum and current maximum
Using the below given relation, we can find the value of .
Therefore,
Thus,
is the required time lag between the maximum voltage and maximum current.
A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.
(a) What is the maximum current in the circuit?
(b) What is the time lav between current maximum and voltage maximum?
Here given,
Capacitance,
Resistance, R = 40 Ω
a.) In RC circuit, as
Therefore,
b.) In RC circuit, voltage lags behind the current by phase angle ϕ,
where ϕ is given by,
Hence,
i.e., we can see that, ϕ is nearly zero at high frequency.
It is clear from here that at high frequency, capacitor acts like a conductor.
For a D.C. circuit, after steady state has been reached, ω = 0.
Hence,
Therefore, capacitor C amounts to an open circuit.
Given,
Instantaneous current, i = 10 sin 314 t A
Instantaneous voltage, V = 50 sin (314 t + /2) V
Since,
Therefore, from the above equations we get
Peak current, peak voltage and as,
Power dissipation in the circuit is given by
Therefore,
The number of turns in secondary coil of a transformer is 100 times the number of turns in the primary coil. What is the transformation ratio?
What do you mean by power factor? On what factors does it depend?
The power factor is defined as the cosine of the phase angle between alternating e.m.f. and current in a.c. current.
Power factor of an a.c. circuit is given by
Power factor depends upon the frequency of the a.c. source.
In a transformer with transformation ratio 0.1, 220 volt a.c. is fed to primary. What voltage is obtained across the secondary?
Given,
Transformation ratio = 0.1
Voltage of the primary coil, EP = 220 V
Now, using the formula,
"
which is the required voltage across secondary.
When a lamp is connected to an alternating voltage supply, if lights with the same brightness as when connected to a 12 V DC battery. What is the peak value of alternating voltage source?
Given,
Rms value of voltage,
Effective voltage,
So,
Peak voltage,
The number of turns in the secondary coil of a transformer is 500 times that in primary. What power is obtained from the secondary when power fed to the primary is 10 W?
Given, the power fed to the primary is 10 W.
Therefore,
If there is no loss of energy, then the output power will be 10 W.
What is the power dissipated in an a.c. circuit in which voltage and current are given by
V = 230 sin (ωt + /2) and I = 10 sin ωt?
What is the power dissipated in an a.c. circuit in which voltage and current are given by
V = 230 sin and I = 10 sin ?
Phase different between V and I =
A small dc motor operates at 110 V dc. What is back emf when its efficiency is maximum?
In a series LCR circuit, the voltage across an inductor, capacitor and resistor are 20 V, 20 V and 40 V respectively. What is the phase difference between the applied voltage and the current in the circuit?
Find the capacitance of the capacitor that have a reactance of 100 Ω when used with an a.c source of frequency 5/ kHz.
The numerical formula for power factor is given by,
As the value of capacitor (C) is changed, the value of mpedence (Z) also changes, hence power factor can be imporved with the help of appropriate capacitance in the circuit.
A bulb and a capacitor are connected in series to an a.c. source of variable frequency. How will the brightness of the bulb change on increasing the frequency of the a.c. source?
The power factor of an a.c. cirucit is 0.5. What will be the phase difference between voltage and current in this circuit?
Given,
Power factor, cos = 0.5 = .
Therefore,
Phase difference between voltage and current is 60° or radian.
Peak value of emf of an a.c. source is E0. What is its rms value?
The peak value of an a.c. circuit supply is 300 V. What is the r.m.s. voltage?
Given, peak voltage = 300 V
Hence,
Rms
What is iron loss in a transformer?
What happens to the power dissipation if the value of electric current passing through a conductor of constant resistance is doubled?
(i) State the law that gives the polarity of the induced emf.
(ii) A 15.0 μF capacitor is connected to 220 V, 50 Hz source. Find the capacitive reactance and the rms current.
(i) The law that gives the polarity of induced emf is given by, Faraday's law of electromagnetic inductionwhich states that :
→ Induced e.m.f. is produced whenever magnetic flux linked with a circuit changes.
→ As long as the change in the magnetic flux continues, the induced e.m.f. lasts.
→ The magnitude of the induced e.m.f. is directly proportional to the rate of change of the magnetic flux linked with the circuit.
(ii) Given,
Capacitance, C = 15.0 μF = 15 × 10–6
Capacitive reactance,
Rms value of current is ,
An a.c. voltage, V = Vm sin ωt, is applied across a
(i) series RC circuit in which the capacitative impedance is 'a' times the resistance in the circuit.
(ii) series RL circuit in which the inductive impedance is 'b' times the resistance in the circuit. Calculate the value of the power factor of the circuit in each case.
Given, A.C voltage, V = Vm sin ωt
Capacitive reactance is 'a' times the resistance.
Inductive impedence is 'b' times the resistance.
(i) Power factor in RC circuit is
Here
(ii) Power factor in RL circuit is
Here,
Given,
Inductor, L = 200 μH
Capacitor, C = 500 μF
Resistor, R = 10 Ω
Effective voltage, V = 100 V
(i) Power factor,
So,
(ii) The current amplitude at this frequency,
(iii) The Q-factor,
State the condition under which the phenomenon of resonance occurs in a series LCR circuit. Plot a graph showing variation of current with frequency of a.c. source in a series LCR circuit.
An alternating voltage of frequency f is applied across a series LCR circuit. Let fr be the resonance frequency for the circuit. Will the current in the circuit lag, lead or remain in phase with the applied voltage when (i) f > fr, (ii) f < fr? Explain your answer in each case.
In the following circuit, calculate,
(i) the capacitance 'c' of the capacitor if the power factor of the circuit is unity, and
(ii) also calculate the Q-factor of the circuit. 
(i) Power factor is given by,
[For power factor unity ]
(ii) Quality factor of the circuit is given by,
A capacitor and a resistor are connected in series with an a.c. source. If the potential difference across C, R are 120 V, 90 V respectively and if the r.m.s. current of the circuit is 3A, calculate the (i) impedance, (ii) power factor of the circuit.
Given,
Rms current,
Effective volatge,
(i) Impedance of the circuit,
(ii) Power factor,
If the voltage in a.c. cirucit is represented by the equation.
Calculate (a) peak and rms value of the voltage, (b) average voltage, (c) frequency of a.c.
(a) For a.c. voltage, V = V0 sin (ωt – ϕ)
The peak voltage value is ,
The rms value of voltage
(b) Average voltage in full cycle is zero.
Average voltage in half cycle is
(c) As,
i.e.,
which, is the required frequency of AC .
Effective value of a.c.: The value of direct current which produces the same heating effect in a given resistor as is produced by the given alternating current when passed for the same time is termed as effective value of a.c.
Peak value of a.c.: The maximum value attained by an alternating current in either of its half cycle is called its peak value.
Given,
Comparing with we get
(i) Rms value/ effective value of current
(ii) Frequency of the source,
Therefore,
Given, E = 50 sin
Resistance , R = 10
(i) Rms value of the potential,
(ii) Frequency of a.c. supply,
(iii) Initial phase ( from the given equation)
(iv) Rms value of current,
An a.c. voltage E = E0 sin ωt is applied across an inductor L. Obtain an expression for current I.
A resistor of resistance R, an inductor inductance L and a capacitor of capacitance C all are connected in series with an a.c. supply. The resistance of R is 16 ohm and for a given frequency, the inductive reactance of L is 24 ohm and capacitive reactance of C is 12 ohm. If the current in the circuit is 5 amp., find
(a) the potential difference across R, L and C
(b) the impedance of the circuit
(c) the voltage of a.c. supply
(d) phase angle
Given, an LCR circuit where all the components are connected in series with an a.c. supply.
Resistance, R = 16
Inductive reactance, = 24 ohm
Capacitive reactance,
Current flowing in the circuit, I = 5 A
(a) Potential difference across resistance,
VR = iR
= 5 × 16
= 80 volt
Potential difference across inductance,
VL = i × (ωL)
= 5 × 24
= 120 volt
Potential difference across condenser,
(b) The impedance of the circuit is given as
(c) The voltage of a.c. supply is given by
V = iz
= 5 × 20
= 100 volt
(d) Phase angle
A series circuit consists of a resistance of 15 ohms, an inductance of 0.08 henry and a condenser of capacity 30 microfarad. The applied voltage has a fequency of 500 radians. Does the current lead or lag the applied voltage and by what angle.
Impedence, 
An LCR series circuit with 100 Ω resistance is connected to an a.c. source of 200 V and angular frequency 300 radians per second. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. Calculate the current and power dissipated in LCR circuit.
Given, an LCR series circuit.
Resistance, R = 100 Ω
Rms voltage, V = 200 V
Angular frequency = 300 radians per second.
Current lags behind the voltage by 60o
Using the formula,
or,
Impedance of circuit,
Current in the circuit,
Average power,
But,
Now,
Average power in LCR circuit:
Let, the alternating e.m.f. applied to an LCR circuit is
E = E0 sin ωt ...(i)
If alternating current developed lags behind the applied e.m.f. by a phase angle ϕ then,
I = I0 sin (ωt – ϕ)
Total work done over a complete cycle is
Thus,
∴ Average power in LCR circuit over a complete cycle is
An a.c. source of voltage V = Vm sin ωt is connected, one-by-one, to three circuit elements X, Y and Z. It is observed that the current flowing in them,
(i) is in phase with applied voltage for element X.
(ii) lags the applied voltage, in phase, by /2 for element Y.
(iii) leads the applied voltage, in phase, by /2 for element Z. Identify the three circuit elements.
Find an expression for the (a) current flowing in the circuit, (b) net impedance of the circuit, when the same a.c. source is connected across a series combination of the elements X, Y and Z. (c) If the frequency of the applied voltage is varied, set up the condition of frequency when the current amplitude in the circuit is maximum. Write the expression for this current amplitude.
A 20 volts 5 watt lamp is used in a.c. main of 220 volts 50 c.p.s. calculate the
(i) capacitance of capacitor.
(ii) inductance of inductor, to be put in series to run the lamp.
(iii) what pure resistance should be included in place of the above device so that the lamp can run on its voltage?
(iv) which of the above arrangements will be more economical and why?
An alternating emf is applied across a capacitor. Show mathematically that current in it leads the applied emf by a phase angle of /2. What is its capacitive reactance ? Draw a graph showing the variation of capacitive reactance with the frequency of the a.c. source.
What is meant by root mean square value of alternating current? Derive an expression for r.m.s. value of alternating current.
Root mean square (r.m.s.) or virtual value of a.c.: It is that steady current, which when passed through a resistance for a given time will produce the same amount of heat as the alternating current does in the same resistance and in the same time. It is denoted
Irms or Iv.
Derivation of r.m.s. value of current:
The instantaneous value of a.c. passing through a resistance R is given by
I = I0 sin ωt
The alternating current changes continuously with time.
Suppose, that the current through the resistance remains constant for an infinitesimally small time dt.
Then, small amount of heat produced the resistance R in time dt is given by
dH = I2 R dt
= (10 sin ωt)2 R dt
= I02 R sin2 ωt dt.
The amount of heat produced in the resistance in time T/2 is
...(i)
If Irms be the r.m.s. value of a.c., then by definition,
...(ii)
From equtions (i) and (ii), we have
A 12 ohm resistance and an inductance of 0.05/ henry with negligible resistance are connected in series. Across the end of this circuit is connected a 130 volt alternating voltage of frequency 50 cycles/second. Calculate the alternating current in the circuit and potential difference across the resistance and that across the inductance.
A coil of self -inductance 0.16 H is connected to a condenser of capacity 0.81 F. What should be the frequency of alternating current that should be applied so that there is resonance in the circuit? Resistance of circuit is negligible.
Give expression for average value of a.c. voltage V = V0 sin ωt over time interval t = 0 to
A.C. voltage, V= V0 sin ωt
As,
therefore, first half cycle (T/2).
Hence, average value of AC voltage,
An a.c., source of angular frequency ω is fed across a resistor R and a capacitor C in series. The current registered is i. If now the frequency of the source is changed to ω/3 (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance of resistance at the original frequency.
(i)Out of the two arrangements given below, for winding of primary and secondary coils in a transformer, which arrangement do you think will have higher efficiency and why?
(ii) Show that, in ideal transformer, when the voltage is stepped up by a certain factor, the current gets stepped down by the same factor.
(iii) State any two causes of energy loss in transformer.
A.
Any angle between 0 andB.
less resistanceas large as possible
equal to natural frequency of LCR system
B.
equal to natural frequency of LCR system
A choke coil is an inductor used to block high frequency alternating current in an electrical circuit , while passing lower frequency current or direct current.
A choke coil is preferred over resistances in a.c. circuit because a choke coil has large value of self inductance and hence, the power dissipation is 0 for choke coil.
Power factor is 0 for an inductance.
In fig. 1 there will be no effect on the flow of current as resistance is independent of frequency.
In fig. 2 since , when frequency is doubled impedence increases and thus, flow of current decreases.
In fig. 3 the flow of current will increase as the capacitive reactance decreases when the frequency is doubled.
[ ]
For circuit 1 we have,
Therefore,
Power factor,
For circuit 2 we have,
Therefore,
Power factor,
Thus, the ratio of the power factor is given by,
On comparing the two equations we can see that, phase difference between current and voltage is .
Therefore, the power consumption in the circuit is 0. [ because, cos
Given,
Inductance of the coil = 10 H
Resistance = 5
Voltage of battery = 5 V
Time , t = 2 sec
Maximum current , I =
During growth of current ,
which is the required amount of current flowing through the circuit.
Given,
Efficiency of the transformer, = 80 %
Power, P = 4 kW
Voltage across primary , EP = 100 V
Voltage across seconadry , Es = 200 V
EP IP = 4 kW = 4000 W
Now,
Current across the primary coil, IP =
Using the formula for the efficiency of transformer,
which is the current across the secondary coil of the transformer.
Given,
Instantaneous value of voltage, e = 80 sin 100 t.
Impedence = 20
Peak value of voltage, Eo = 80.
Therefore,
Current through the circuit, I =
Given,
Power dissipated in an a.c circuit, P = VrmsIrms cos
where, cos.
For an ideal capacitor, R = 0
Therefore,
Power dissipated in circuit , P = 0
Let, the circuit contain a resistor of resistance R and inductor of inductance L in series.
Voltage across R is VR
Voltage across L is VL
Current flowing through the circuit is I.
Applied voltage, V = Vo sin
VR and I are in phase whereas, VL leads the current by an angle by /2.
Applied voltage is the resultant of VR and VL
i.e.,
But, VR = Ri and,
VL =
where, is the inductive reactance.
V =
and
Impedence, Z =
which is the required expression for impedence.
Given, an LCR circuit. L, C and R are arranged in parallel and the source frequency is kept equal to the resonating frequency.
Then,
Using the formula for resonant frequency,
Since elements are in parallel, reactance X of L and C in parallel is given by
Impedance of R and X in parallel is given by
which is less than resistance R.
At resonant frequency,
and
Then, impedance Z = R and will be maximum.
Hence, current will be minimum at resonant frequency in the parallel LCR circuit.
From Ex. 11:
Inductance, L = 5H
Capacitance, C = 80 × 10–6 F
Resistnace, R = 40 Ω.
Therfore,
Current through L and C will be in opposite phase. Hence, current in circuit will be only 5.75 A as, circuit impedance will be equal to R only.
A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.
(a) Obtain the current amplitude and rms value.
(b) Obtain the rms values of potential drops across each element.
(c) What is the average power transferred to the inductor?
(d) What is the average power transferred to the capacitor?
(e) What is the total average power absorbed by the circuit? ['Average' implies 'averaged over one cycle'.]
Here,
Inductance, L = 80 mH = 80 x 10-3H
Capacitance,
Resistance, R = 0
RMS voltage, Ev = 230 V
Frequency of Ac supply,
(a) We have to find
Therefore,
and,
Negative sign appears as
e.m.f lags behind the current by
(b) Rms value of potential drop across L,
Rms value of potential drop across C,
As voltages across L and C are 180° out of phase, therefore, they get subtracted.
That is why applied r.m.s. voltage = 436.84 – 206.74
= 230.1 volt.
(c) Average power transferred over a complete cycle by the source to inductor is always zero because of phase difference of between voltage and current through the inductor.
(d) Average power transferred over a complete cycle by the source to the capacitor is also zero because of phase difference of between voltage and current through capacitor.
(e) Total average power absorbed by the circuit is also, therefore zero.
Here,
Reistance,
Impedance,
Average power transferred to inductor,
Average power transferred to capacitor=
Average power transferred to R =
A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
(c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
(d) What is the Q-factor of the given circuit?
(a) Here,
Inductance, L = 0.12 H
I0 would be maximum, when
(b) Average power absorbed by the circuit is maximum, when I = I0
(c) The two angular frequencies for which the power transferred to the circuit is half the power at the resonant frequency,
When,
∴ angular frequencies at which power transferred is half = ωr ± ∆ω
= 4166.7 ± 95.83 = 4262.3 and 4070.87 rad s–1
Current amplitude at these frequencies is
(d) Q-factor =
Given,
Inductance, L = 3.0 H
Capacitance, C = 27
Resistance, R = 7.4
Therefore,
Resonant frequency,
Q-factor of the circuit,
For improvement in sharpness of resonance by a factor of 2, Q should be doubled. To double Q with changing ωr, R should be reduced to half, i.e., to 3.7 Ω.
A choke coil is needed in the use of fluorescent tubes to reduce alternating current without loss of power. If we use an ordinary resistor, a.c. will reduce, but loss of power due to heating will be there.
∴ Power dissipated = Ev Iv cos ϕ
In a resistor, ϕ = 0°
∴ Power dissipated = Ev Iv cos 0°
= Ev Iv
which, is the maximum value.
In a choke coil, ϕ = 90°
∴ Power dissipated = Ev Iv cos 90° = zero.
Here given,
Height of water pressure head, h = 300 m
Volume of the water flowing per second, V = 100 m3
Mass of water flowing per second, m = 100 × 103 kg
= 105 kg
acceleration due to gravity, g = 9.8 ms–2
Potential energy of water fall during one second
Therefore, input power =
A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterise the step-up transformer at the plant.
a) Power required, P = 800 kW = 800 × 103 W
Total resistance of two wire lines
R = 2 × 15 × 0.5 = 15 Ω.
Since supply is through 4000 – 230 V transformer
As,
where, IV is the rms value of the current.
Line power loss in the form of heat is,
= (IP)2 Resistance of line
= (200)2 0.5 15 2
= 60 104 W
= 600 kW
(b) If there is no power loss due to leakage,
then the essential plant power supply is
= 800 + 600
= 1400 kW
(c) Voltage drop on the line =
Voltage from transmission = 3000 + 4000 = 7000 V.
Since, the power is generated at 440 volt, the step-up transformer needed at the plant is 440 V - 7000 V.
(b) Power supplied by the plant = 800 + 6
= 806 kW
(c) Voltage drop on the line = IP × R
= 20 × 15
= 300 V
Voltage output of the step-up transformer at the plant = 40000 + 300= 40300 V
∴ The step-up transformer at the plant need a voltage of 440 V – 40300 V.
Power loss at lower voltage =
Power loss at higher voltage =
Hence, it is evident that power loss is minimal when voltage is high. Therefore, high voltage transmission is produced.
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Given, capacitance, C1 = 8 pF
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
Capacitance having a dielectric of thickness ‘t’ is given by,
When the thickness of the plates is reduced to half, t = d/2 then,
Capacitance becomes, 
A capacitor, made of two parallel plates each of plate area A and separation d, is being charged by an external ac source. Show that the displacement current inside the capacitor is the same as the current charging the capacitor.
The charge on the plates is because of the conduction current flowing in the wires.
According to Maxwell’s equation, displacement current between the plated is given by,
where, 
is the lectric flux.
Now, using Gauss theorem, 
So, 
Therefore, both the displacement current and conduction currents are equal.
(a) For a given a.c.,
show that the average power dissipated in a resistor R over a complete cycle is 
(b) A light bulb is rated at 100 W for a 220 V a.c. supply. Calculate the resistance of the bulb.
a) Average power consumed in resistor R is given by, 
b) In case of ac, we have
(a) Draw a schematic sketch of a cyclotron. Explain clearly the role of crossed electric and magnetic field in accelerating the charge. Hence derive the expression for the kinetic energy acquired by the particles.
(b) An –particle and a proton are released from the centre of the cyclotron and made to accelerate.
(i) Can both be accelerated at the same cyclotron frequency? Give reason to justify your answer.
(ii) When they are accelerated in turn, which of the two will have higher velocity at the exit slit of the dees?
a)
The schematic sketch of cyclotron is as shown below:
Electric field: It helps in accelerating the charged particle passing through the gap with the help of electric oscillator. Electric oscillator imparts the energy to charged particle till it comes out from the exit slit.
Magnetic field: The magnetic force exerts a centripetal force when the accelerated charge particle enters normally to the uniform magnetic field. Centripetal force makes the particle move in a semicircular path of increasing radii in each Dee.
Kinetic energy acquired by the particle is given by,
b)
i) Now, using equation (1), we have
Here we can see that, cyclotron frequency depends upon (q/m) ratio.
So, 
That is, 
ii) Kinetic energy is given by, 
That is, 
That is, proton acquires higher velocity.
(a) Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.
(b) Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero.
Electric dipole moment is the product of either charges or the distance between two equal and opposite charges.
It is a vector quantity.
Electric dipole moment at a point on the equatorial plane:
Consider a point P on broad side on the position of dipole formed of charges + q and - q at separation 2l. The distance of point P from mid-point O of electric dipole is r. 
Let E1 and E2 be the electric field strength due to charges +q and –q of electric dipole.
From the fig. we have
Now, inorder to find the resultant electric field, we resolve the components along and perpendicular to AB.
The components perpendicular to AB are sin components and they being equal and opposite to each other cancel each other.
Therefore,
Resultant electric field is given by,
E1 = E1 
But, 
From the fig. we can see that,
If dipole is infinitesimal and point P is far away, then l2 can be neglected as compared to r2.
Therefore, 
b) Equipotential surfaces due to an electric dipole is given by,
Electric potential is zero at all points in the plane passing through the dipole equator.
The resistance offered by the capacitor when connected to an electrical circuit is known as capacitive reactance.
It is given by,
where, 
is the angular frequency of the source, and
C, is the capacitance of the capacitor.
SI Unit is ohm.
Two capacitors of unknown capacitances C1 and C2 are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of C1 and C2. Also, calculate the charge on each capacitor in parallel combination.
When the capacitors are connected in parallel,
Equivalent resistance, 
Energy stored in capacitors, EP = 
When the capacitors are connected in series,
Putting this in equation (2), we get
C2 = 50 – 11.2 = 38.2 μF
The charge on capacitor is given by, Q = CV
Charge on capacitor 1, C1 = 
Charge on capacitor 2, C2 = 
An inductor L of inductance XL is connected in series with a bulb B and an ac source.
How would brightness of the bulb change when (i) number of turn in the inductor is reduced, (ii) an iron rod is inserted in the inductor and (iii) a capacitor of reactance XC = XL is inserted in series in the circuit. Justify your answer in each case.i) Net resistance in the circuit is given by, 
Inductance is given by, 
As number of turns decreases, L decreases.
Inductance is given by, 
Therefore, 
will also decrease thereby, reducing the net resistance in the circuit. Thus, current increases and brightness of the bulb is increased.
ii) When a soft iron rod is inserted in the circuit, L increases. Therefore, inductive reactance also increases. Net resistance increases and flow of current in the circuit decreases. Thus, the brightness of the bulb will decrease.
iii) When a capacitor of reactance XL = XC is connected in series with the circuit net resistance becomes,
; where R is the resistance of the bulb.
Here, we will have Z= R which is the condition of resonance.
At resonance, maximum current will flow through the circuit. Therefore, the brightness of the bulb will increase.
A group of students while coming from the school noticed a box marked “Danger H.T. 2200 V” at a substation in the main street. They did not understand the utility of such a high voltage, while they argued; the supply was only 220 V. They asked their teacher this question the next day. The teacher thought it to be an important question and therefore explained to the whole class.
Answer the following questions:
(i) What device is used to bring the high voltage down to low voltage of a.c. current and what is the principle of its working?
(ii) Is it possible to use this device for bringing down the high dc voltage to the low voltage? Explain.
iii) Write the values displayed by the students and the teacher.
i) The device which is used to bring the high voltage down to low voltage is step down transformer. The working principle of step down transformer is mutual induction. Whenever there is change in current in one circuit, emf is induced in the neighboring circuit.
ii) Transformer cannot be used to bring down the high d.c voltage to low voltage because its working is based on electromagnetic induction, which is associated with varying magnetic flux. But, current is constant for a d.c source. So, magnetic flux will not vary. Hence, output cannot be obtained from the transformer.
iii) There is a curiosity among the students to gain knowledge. And the teachers are able to make use of their knowledge and impart it to students to explain the practical application.
(a) Explain, using suitable diagrams, the difference in the behaviour of a (i) conductor and (ii) dielectric in the presence of external electric field. Define the terms polarization of a dielectric and write its relation with susceptibility.
(b) A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge Q/2 is placed at its centre C and another charge +2Q is placed outside the shell at a distance x from the centre as shown in the figure. Find (i) the force on the charge at the centre of shell and at the point A, (ii) the electric flux through the shell.
When a conductor like a metal is subjected to external electric field, the electrons experience a force in the opposite direction and they accumulate on the left hand side. A positive charge is therefore induced on the right hand side. This creates on opposite electric field (
) that balances out (E0). The net electric field inside the conductor becomes zero. 
Here, E0 is the external field and Ein is the internal field created by the re-distribution of electrons inside the metal.
Dielctric: When external electric field is applied, dipoles are created for non-polar dielectrics and in case of polar dielectrics dipoles are aligned. An internal electric field is created such that it reduces the external electric field. The placement of dipoles is as shown in the given figure.
The dipole moment per unit volume of the polarized dielectric is known as the Polarization of a dielectric. It is denoted by P.
The formula for polarization is given by, 
b) i) The electric field inside a spherical shell is zero. Therefore, force on the charge placed at the centre of the shell is zero. For a charge placed at A, shell will behave as if the entire charge Q is placed at the centre of the shell.
Therefore, total charge is, 
Distance from 2Q is ‘x’. 
ii) Total charge enclosed by the shell is 
So, according to Gauss’s law,
Total flux, 
The carrier wave is given by C (t) = 2sin (8πt) Volt.
The modulating signal is a square wave as shown. Find modulation index. 
The generalized equation of a carrier wave is given by, 
The generalized equation of a modulating signal is given by,
On comparing the given equation of carrier wave with the generalized equation, we get,
Amplitude of modulating signal, Am = 1 V
Amplitude of carrier wave, Ac = 2 V
Modulation index is the ratio of the amplitude of modulating signal to the amplitude of carrier wave 
.
It is denoted by,
So, 
A capacitor 'C', a variable resistor 'R' and a bulb 'B' are connected in series to the ac mains in circuit as shown. The bulb glows with some brightness. How will the glow of the bulb change if (i) a dielectric slab is introduced between the plates of the capacitor, keeping resistance R to be the same; (ii) the resistance R is increased keeping the same capacitance? 
(i) When a dielectric slab is introduced between the plates of the capacitor, its capacitance will increase. Hence, the potential drop across the capacitor will decrease (V= Q/C ) thereby decreasing the potential drop across the bulb because, both the bulb and capacitor is connected in series. So, brightness of the bulb will increase.
(ii) When resistance (R) is increased keeping the capacitance same, the potential drop across the resistor will increases. Therefore, the potential drop across the bulb will decrease because both are connected in series. So, brightness of the bulb will decrease.
An electric dipole of length 4 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of 4 Nm. Calculate the potential energy of the dipole, if it has charge ±8 n C.
Given,
Length of the dipole, l = 4 cm
Torque, 
= 4
N m
Using the formula for torque, we have
Torque, τ = PEsinθ = (Ql) Esinθ ...(1)
Potential energy, U= −PEcosθ= −(Ql)Ecosθ ...(2)
Dividing equation (2) by (1),
(a) Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor.
(b) The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving a charge q over a closed loop a b c d a.
(a) Consider a parallel-plate capacitor of plate area A.
Let us say, charge Q is given to the capacitor. Now, in order to increase the separation between the plates, plate b is slowly pulled away from plate a.
Distance between the two capacitor plates = d
Force on plate b due to plate a is given by, 
Work done inorder to displace the plate from its fixed position is, W=Fd
; where 
is the capacitance of the capacitor.
Work done is equal to increase in energy of the system.
Electric field is created in a volume which is given by, V = Ad
So, Energy stored per unit volume is given by, 
where, E is the intensity of the electric field.
(b) Work done, W= F.d ;
where, F is the force exerted on electrical charge and d is the displacement.
Since, the charge is moved along a closed path, net displacement is zero.
Therefore, work done= 0
When electrons drift in a metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction?
No. When electric field is applied, the net drift of the electrons is from lower to higher potential. But, locally electrons collide with ions and may change its direction during the course of their motion.
When an ideal capacitor is charged by a dc battery, no current flows. However, when an ac source is used, the current flows continuously. How does one explain this, based on the concept of displacement current?
When an ideal capacitor is charged by dc battery, charge flows momentarily till the capacitor gets fully charged. Ideal capacitor means infinite resistance for dc.
When an ac source is connected to the capacitor, conduction current, 
continuously flows in the connecting wire. Due to changing current, charge deposited on the plates of the capacitor changes with time. This causes change in electric field between the plates of the capacitor which causes the electric flux to change and gives rise to a displacement current in the region between the plates of the capacitor.
Displacement current is given by, 
Define self-inductance of a coil. Show that magnetic energy required to build up the current I in a coil of self-inductance L is given by 
Self-inductance of a coil is numerically equal to the magnetic flux linked with the coil when the current through coil is one Ampere.
Mathematically, it is given by,
where, L is the constant of proportionality and is called the self-inductance.
Energy stored in an inductor:
Consider a source of emf connected to an inductor L.
With increase in current, the opposing induced emf is given by, 
If the source of emf sends a current i through the inductor for a small time dt, then the amount of work done by the source, 
Hence, the total amount of work done by source of emf when the current increases from its initial values (i = 0) to its final value (I) is given by,
This work done gets stored in the inductor in the form of energy.
Explain the principle of a device that can build up high voltages of the order of a few million volts.
Draw a schematic diagram and explain the working of this device.
Is there any restriction on the upper limit of the high voltages set up in this machine? Explain.
Van de Graaff generator is capable of producing very high potential of the order of 10 MV.
The underlying principle of Van de graaf generator is that a charge given to a hollow conductor is transferred to outer surface and is distributed uniformly over it.
Construction: It consists of a large hollow metallic sphere S mounted on two insulating columns A and B and an endless belt of rubber or silk is made to run on two pulleys P1 and P2 by the means of an electric motor. C1 and C2 are two sharp metallic spikes in the form of combs. The lower comb C1 is connected to the positive terminal of a very high voltage source and the upper comb C2 is connected to the inner surface of metallic sphere S.
Working: When comb C1 is given very high potential, then it produces ions in its vicinity, due to action of sharp points. The positive ions, so produced, get sprayed on the belt due to the repulsion between positive ions and comb C1. These positive ions are carried upward by the moving belt. The pointed end of C2 just touches the belt. The comb C2 collects positive charge from the belt which immediately moves to the outer surface of sphere S. As the belt goes on revolving, it continues to take (+) charge upward, which is collected by comb C2 and transferred to outer surface of sphere S. Thus the outer surface of metallic sphere S gains positive charge continuously and its potential rises to a very high value.
When the potential of a metallic sphere gains very high value, the dielectric strength of surrounding air breaks down and its charge begins to leak, to the surrounding air. The maximum potential is reached when the rate of leakage of charge becomes equal to the rate of charge transferred to the sphere. To prevent leakage of charge from the sphere, the generator is completely enclosed in an earthed connected steel tank which is filled with air under high pressure.
Van de Graff generator is used to accelerate stream of charged particles to very high velocities. But, there is a restriction on the upper limit of high voltage set up in the machine. The high voltages can be built up to the breakdown field of the surrounding medium.
A point charge Q is placed at point O as shown in the figure. Is the potential difference VA –VB positive, negative, or zero, if Q is (i) positive (ii) negative? 
With increase in distance, potential due to point charge decreases. So,
i) VA - VB is positive when charge Q is positive.
ii) VA - VB is negative.
Define the term 'wattless current'.
Current flowing in an ac circuit without any net dissipation of power is called wattless current.
Two uniformly large parallel thin plates having charge densities +
and –
are kept in the X-Z plane at a distance 'd' apart. Sketch an equipotential surface due to electric field between the plates. If a particle of mass m and charge '-q' remains stationary between the plates, what is the magnitude and direction of this field?
Equipotential surface between the plates due to the electric field is given by the figure shown below. We can see that, the equipotential surface is at a distance d/2 from either plate in X-Z plane.
Given, a particle of mass ‘m’ and charge ‘-q’ remains stationary in between the plates. For the negative charge,
i) weight mg acts, vertically downward
ii) electric force qE acts vertically upward.
So, mg = qE
, is acting vertically downwards (along –Y axis).
Figure shows two identical capacitors, C1 and C2, each of 1 μF capacitance connected to a battery of 6 V. Initially switch 'S' is closed. After sometime 'S' is left open and dielectric slabs of dielectric constant K = 3 are inserted to fill completely the space between the plates of the two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?
When switch S is closed, potential difference across capacitor is 6V
Potential across the battery, V1 = V2 = 6V
Capacitance of the capacitor, C1 = C2 = 1 
Charge on each capacitor, 
When switch S is opened, the potential difference across C1 remains 6 V, while the charge on capacitor C2 remains 6 
. After insertion of dielectric between the plates of each capacitor, the new capacitance of each capacitor becomes
i)
Now, charge on capacitor C1 ,q1’ = C1’V1 = 
Charge on capacitor C2 remains the same, i.e., 6 
ii)
Potential difference across C1 remains the same.
Potential difference across C2 is, 
Derive an expression for the impedance of a series LCR circuit connected to an AC supply of variable frequency.
Plot a graph showing variation of current with the frequency of the applied voltage.
Explain briefly how the phenomenon of resonance in the circuit can be used in the
Tuning mechanism of a radio or a TV set.a)
Expression for impedance in LCR series circuit:
Suppose a resistance R, inductance L and capacitance C are connected to series and an alternating voltage V = 
is applied across it. 
Since L, C and R are connected in series, current flowing through them is the same. The voltage across R is VR, inductance across L is VL and across capacitance is VC.
The voltage VR and current i are in the same phase, the voltage VL will lead the current by angle 90° while the voltage VC will lag behind the current by 90°.
Clearly VC and VL are in opposite directions, therefore their resultant potential difference =VC -VL (if VC >VC).
Thus, VR and (VC -VL) are mutually perpendicular and the phase difference between them is 90°. As seen in the fig, we can say that, as applied voltage across the circuit is V, the resultant of VR and (VC -VL) will also be V.
So, 
Therefore, Impedance of the circuit is given by, 
This is the impedence of the LCR series circuit.
b)
The below graph shows the variation of current with the frequency of the applied voltage. 
c)
A radio or a TV set has a L-C circuit with capacitor of variable capacitance C. The circuit remains connected with an aerial coil through the phenomenon of mutual inductance. Suppose a radio or TV station is transmitted a programme at frequency f, then waves produce alternating voltage of frequency f, in area, due to which an emf of same frequency is induced in LC circuit, When capacitor C is in circuit is varied then for a particular value of capacitance, 
, the resonance occurs and maximum current flows in the circuit; so the radio or TV gets tuned.
What is the geometrical shape of equipotential surfaces due to a single isolated charge?
The equipotential surface for an isolated charge is concentric circles. The distance between the shell decreases with increase in electric field.
A capacitor has been charged by a dc source. What are the magnitudes of conduction and displacement currents, when it is fully charged?
When capacitor is charged by a dc source,
Conduction current is equal to the displacement current.
i.e., 
Because, 
is maximum when, capacitor is fully charged.
A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 C . When potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 C .
Calculate:
(i) The potential V and the unknown capacitance C.
(ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120V?
OR
A hollow cylindrical box of length 1m and area of cross-section 25cm2 is placed in a three dimensional coordinate system as shown in the figure. The electric field in the region is given by, where E is in NC-1 and x is in metres. Find:
(i) Net flux through the cylinder.
(ii) Charge enclosed by the cylinder.
The charge on an unknown capacitor is given by,
Q = CV
CV = 360 C ... (1)On reducing the potential by 120 V, charge on the capacitor is reduced which is given by,
Q’ = C(V-120)
C(V-120) = 120 C … (2)
On solving equation (1) and (2), we have
Unknown capacitance from equation (1),
Q = CV
ii) Charge on the capacitor, if voltage is increased by 120V
Q = C (V+120)
= 2 (180+120)
= 600 C
OR
i) Electric flux through a surface, 
Flux through the left surface, 
Since x = 1m,
ii) Using Gauss Theorem, we can calculate the charge inside the cylinder.
In a series LCR circuit connected to an ac source of variable frequency and voltage 
draw a plot showing the variation of current (I) with angular frequency (
) for two different values of resistance R1 and R2 (R1>R2). Write the condition under which the phenomenon of resonance occurs. For which values of the resistance out of the two curves, a sharper resonance is produced? Define Q-factor of the circuit and give its significance.
At certain frequency 
the flow of current through the series combination,
Condition of resonance – If system (LCR) of natural frequency w0 is driven by an energy source at a frequency w, the amplitude of the current flow increases, however the amplitude of the current rises to its maximum value, if frequency of the energy source becomes exactly equal to the natural frequency.
For resistance R2 < R1, series LCR shows a sharp resonance.
Q-factor - The ratio of reactance (either inductive or capacitive) at natural frequency to the resistance of the current is called Q - factor.
Significance of Q factor is given by,
(i) If resistance R is low or inductance L is large then Q – factor is large and the circuit is more selective.
(ii) If resonance is less sharp, tuning of the circuit will not be good.
Draw a graph to show variation of capacitive-reactance with frequency in an a.c. circuit.
Here, XC is the capacitive-reactance and is the frequency in an a.c circuit.
A series LCR circuit is connected across an a.c. source of variable angular frequency ‘
’. Plot a graph showing variation of current 'i' as a function of ‘’ for two resistances R1 and R2 (R1 > R2).
Answer the following questions using this graph:
(a) In which case is the resonance sharper and why?
(b) In which case in the power dissipation more and why?a) The variation of current with angular frequency for the two resistances R1 and R2 is shown in the graph below:
We can see that the resonance for the resistance R2 is sharper than for R1 because resistance R2 is less than resistance R1. Therefore, at resonance, the value of peak current will rise more abruptly for a lower value of resistance.
b) Power associated with resistance is given by, 
From the above graph, we can say that virtual current in case of R2 is more than the virtual current in case of R1. Therefore, the power dissipation for circuit with R2 is more than that with R1.
Find the equivalent capacitance of the network shown in the figure, when each capacitor is of 1 
F. When the ends X and Y are connected to a 6 V battery, find out
(i) the charge and
(ii) the energy stored in the network
Equivalent circuit can be drawn as, 
In the above circuit, in one branch there are two capacitors in series.
Therefore, resultant capacitance is, 
The circuit can be further re-arranged as, 
We can see the two capacitors are in parallel.
Therefore, their resultant capacitance is,
Cres = 2 + 2 = 4 
Voltage of the battery = 6V
i) Charge across the capacitor, q = CV = 
ii) Energy stored in the capacitor, E is given by,
Draw a necessary arrangement for winding of primary and secondary coils in a step-up transformer. State its underlying principle and derive the relation between the primary and secondary voltages in terms of number of primary and secondary turns. Mention the two basic assumptions used in obtaining the above relation.
State any two causes of energy loss in actual transformers.
Underlying principle of a step-up transformer: A transformer which increases the ac voltage is known as a step up transformer.
Working of step-up transformer is based on the principle of mutual inductance and it converts the alternating low voltage to alternating high voltage. The number of turns in the secondary coil is greater than the number of turns in the primary coil.
i.e., NS > NP
Working: When an A.C source is connected to the ends of the primary coil, the current changes continuously in the primary coil. Hence, the magnetic flux which is linked with the secondary coil changes continuously. So, the emf which is developed across the secondary coil is same as that in the primary coil. The emf is induced in the coil as per Faraday’s law.
Let, EP be the alternating emf applied to primary coil and np be the number of turns in it.
Consider 
as the electric flux associated with it.
Then, 
Assumption: We assume that there is no leakage of flux so that, the flux linked with each turn of primary coil is same as flux linked with secondary coil.
Two sources of energy loss in the transformer:
Joule Heating – Energy is lost in resistance of primary and secondary windings in the form of heat.
H = I2Rt
Flux leakage - Energy is lost due to coupling of primary and secondary coils not being perfect, i.e., whole of magnetic flux generated in primary coil is not linked with the secondary coil.
iii)
Conservation of law of energy is not violated in step-up transformer. When output voltage increases, the output current automatically decreases. Thus, there is no loss of energy.
Why is the use of A.C. voltage preferred over D.C. voltage? Give two reasons.
The reason why AC voltage is preferred over DC is because of the following reasons:
i) AC can be transmitted with much lower energy losses as compared to DC.
ii) Alternating current can be generated easily.
A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.
Energy stored in a capacitor is given by, 
When capacitor is connected in parallel combination, equivalent capacitance = 2C
Here, total charge Q remains the same.
Initial energy is given by, 
Final energy is given by, 
So, the ratio of energy is given by, 
A voltage V = V0 sin ωt is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle.
Under what condition (i) no power is dissipated even though the current flows through the circuit, (ii) maximum power is dissipated in the circuit?Voltage V = Vo sin 
t is applied to an LCR series circuit.
Current is lagging with respect to the voltage applied.
Therefore,
Current in the circuit is given by,
is the phase lag of the current w.r.to voltage applied.Instantaneous Power dissipation is given by, 
Draw a labelled diagram of Van de Graff generator. State its working principle to show how by introducing a small charged sphere into a larger sphere, a large amount of charge can be transferred to the outer sphere. State the use of this machine and also point out its limitations.
1. Used to accelerate charged particles such as electrons, protons, ions etc, used for nuclear disintegration.
Limitations:
1. Generator allows only one route for the movement of charge as it is a series combination.
2. It can accelerate only charged particles and not the uncharged particles.
Mention the two characteristic properties of the material suitable for making core of a transformer.
Two characteristic properties are:
i) Low hysteresis loss
ii) Low coercivity.
A test charge ‘q’ is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in the figure. (i) Calculate the potential difference between A and C. (ii) At which point (of the two) is the electric potential more and why? 
i) Potential difference between A and C
ii) Direction of electric field is in higher to lower potential. So, Vc > VA
State the underlying principle of a transformer. How is the large scale transmission of electric energy over long distances done with the use of transformers?
The principle of transformer is based upon the principle of mutual induction which states that due to continuous change in the current in the primary coil an emf gets induced across the secondary coil. At the power generating station, the step up transformers step up the output voltage which reduces the current through the cables and hence reduce resistive power loss. Then, at the consumer end, a step down transformer steps down the voltage. Therefore, the large scale transmission of electric energy over long distances is done with the uses of transformers.
Deduce the expression for the electrostatic energy stored in a capacitor of capacitance ‘C’ and having charge ‘Q’.
How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant ‘K’?When a capacitor is charged by a battery, work is done by the charging battery at the expense of its chemical energy. This work is stored in the capacitor in the form of electrostatic potential energy.
Consider a capacitor of capacitance C. Initial charge on capacitor is zero. Initial potential difference between capacitor plates =0. Let a charge Q be given to it in small steps. When charge is given to capacitor, the potential difference between its plates increases.
Potential difference between its plates, V= q/C
Work done to give an infinitesimal charge dq to the capacitor is given by, 
If V is the final potential difference between capacitor plates, then Q = CV
Work is stored in the form of electrostatic potential energy.
Electrostatic potential energy, U = 
When battery is disconnected,
i) Energy stored will decrease.
Energy becomes, U = 
So, energy is reduced to 1/K times its initial energy.
i) In the presence of dielectric, electric field becomes, E = 
A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation.
Suppose a resistance R, inductance L and capacitance C connected in series.
An alternating source of voltage V = Vo sin 
t is applied across it. Since all the components are connected in series, the current flowing through all is same.
Voltage across resistance R is VR, voltage across inductance L is VL and voltage across capacitance C is VC.
VR and (VC -VL) are mutually perpendicular and the phase difference between them is 90°.
From the figure above, we have 
and
The phase difference between current and voltage is given by,
From the graph, we can see that with increase in frequency, current first increases and then decreases. At resonant frequency, current amplitude is maximum.
A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. What is the potential at the centre of the sphere?
Potential at centre of sphere is equal to 10 V.
How are radio waves produced?
Radio waves are produced by the accelerated motion of charges in conducting wires.
Two bar magnets are quickly moved towards a metallic loop connected across a capacitor ‘C’ as shown in the figure. Predict the polarity of the capacitor. 
Left face of the coil will act as North Pole and right face as South Pole so as to oppose the current induced in the coil. As seen from left, the direction of the current in the coil will be anti-clockwise. 
That is, plate of the capacitor A will be positive and plate B will be negative.
Upper plate is positive and lower plate is negative.
Net capacitance of three identical capacitors in series is 1 
F. What will be their net capacitance if connected in parallel?
Let,C be the capacitance of each capacitor,
In a series combination, 
When these capacitors are connected in parallel, net capacitance, CP = 3 C = 33 = 9F
Ration of energy stored is given by,
Draw a labelled diagram of a full wave rectifier circuit. State its working principle. Show the input-output waveforms.
The circuit of Full wave rectifier is as shown below:
Working principle: The underlying working principle of full wave rectifier is that the p-n junction conducts when it is forward biased and does not conduct when it is reverse biased.
The input and output waveform is shown below:
Two heating elements of resistance R1 and R2 when operated at a constant supply of voltage, V, consume powers P1 and P2 respectively. Deduce the expressions for the power of their combination when they are, in turn, connected in (i) series and (ii) parallel across the same voltage supply.
i) When they are in series combination,
Net resistance, R = R1 + R2 … (1)
Heating elements are operated at same voltage V, we have
So, from equation (1), we have
ii) When the resistors are connected in parallel combination,
Net resistance is given by,
A point charge +Q is placed at point O as shown in the figure. Is the potential
difference VA – VB is positive, negative or zero?
Electric potential at a distance r from the point charge +Q is given by,
Potential at point A is, 
Similarly, potential at point B is given by, 
Since rA < rB 
Quality factor relates the maximum or peak energy stored in the circuit (the reactance) to the energy dissipated (the resistance) during each cycle of oscillation.
Quality factor is expressed as, 
Q-factor is a dimensionless quantity.
i)
Capacitance of a parallel plate capacitor is given by, 
ii)
Potential difference between plates X and Y can be calculated as,
Q = CV 
iii)
The ratio of electrostatic energies can be calculated as: 
Voltage of the source is given by,
V = Vo sin ωt
Let current of the source be I = Io sin ωt
Maximum voltage across R is VR = Vo R, represented along OX
Maximum voltage across L = VL = IO XL, represented along OY and is 90o ahead of Io.
Maximum voltage across C = VC = Io XC, represented along OC and is lagging behind Io by 90o
Hence, reactive voltage is VL - VC, represented by OB'
the vector sum of VR, VL and VC is resultant of OA and OB', represented along OK.
OK = Vo = 
i.e., Vo = 
Impedance, Z = 
When, XL = XC , the voltage and current are in the same phase.
In such a situation, the circuit is known as non-inductive circuit.
ii)
Given,
Power factor, P1 = R/Z
Thus, 
i) A transformer is a device that changes a low alternating voltage into a high alternating voltage or vice versa.
Transformer works on the principle of mutual induction.
A changing alternate current in primary coil produces a changing magnetic field, which induces a changing alternating current in secondary coil. 
Energy losses in transformer:
Flux leakage due to poor structure of the core and air gaps in the core.
Loss of energy due to heat produced by the resistance of the windings.
Eddy currents due to alternating magnetic flux in the iron core, which leads to loss due to heating.
Hysteresis, frequent and periodic magnetisation and demagnetization of the core, leading to loss of energy due to heat.
ii)
a) Number of turns in secondary coil is given by, 
b) Current in primary is given by, 
Define an equipotential surface. Draw equipotential surfaces:
(i) in the case of a single point charge and in a constant electric field in Z-direction.
(ii) Why the equipotential surfaces about a single charge are not equidistant?
(iii) Can electric field exist tangential to an equipotential surface? Give reason.
i) Equipotential surface for a single point charge is:
(i) When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.
(ii) A lamp is connected in series with an inductor and an AC source. What happens to the brightness of the lamp when the key is plugged in and an iron rod is inserted inside the inductor ? Explain.
Average power supplied by the source over a complete cycle is,
P = VI cos 
;
cos 
is called as the power factor.
For a pure inductive circuit,
Phase difference between current and voltage = 
And, cos 
Therefore,
Average power dissipated = 0
ii) The brightness of the lamp decreases because when an iron rod is inserted in the inductor, the value of inductance L increases. Therefore, the current flowing across the bulb will decrease, thus decreasing the brightness of the bulb.
(i) Identify the part of the electromagnetic spectrum which is:
(a) suitable for radar system used in aircraft navigation,
(b) produced by bombarding a metal target by high-speed electrons.
ii) Why does a galvanometer show a momentary deflection at the time of charging or discharging a capacitor ? Write the necessary expression to explain this observation.
i) a) Microwaves are suitable for radar system used in aircraft navigation.
b) X-rays are produced by bombarding a metal target with high-speed electrons.
ii) A capacitor is connected to the battery. So, electrons start moving towards the plate connected to the negative terminal of the battery and electrons leave from the plate connected to the positive terminal of the battery.
Transfer of electrons takes place until the potential of the capacitor becomes equal to that of the battery. The whole process happens very quickly, and the charging current produces deflection.
The reverse process is repeated at the time of discharging a capacitor and again galvanometer shows a momentary deflection.
Galvanometer acts as a resistance and the circuit behaves like a R-C circuit, having time constant equal to RC.
Therefore, the required expression is given by,
(i) Draw a labelled diagram of a step-down transformer. State the principle of its working.
(ii) Express the turn ratio in terms of voltages.
(iii) Find the ratio of primary and secondary currents in terms of turn ratio in an ideal transformer.
(iv) How much current is drawn by the primary of a transformer connected to 220 V supply when it delivers power to a 110 V- 550 W refrigerator?
How does Ampere-Maxwell law explain the flow of current through a capacitor when it is being charged by a battery? Write the expression for the displacement current in terms of the rate of change of electric flux.
During charging, electric flux between the plates of capacitor keeps on
changing; this results in the production of a displacement current
between the plates.
Expression for the displacement current
(i) Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase: current or voltage?
(ii) Without making any other change, find the value of the additional capacitor C, to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity.
Given V = Vosin (1000 +Φ)
ω = 1000s-1
L = 100 mH
C =2μF
R = 400Ω
Phase difference 
XL = ωL = 1000 x100 x10-3 = 100Ω
=
(ii) For power factor to be unity, R = Z
or XL = XC
For two capacitance in parallel, resultant capacitance C'=C + C1
10-5 = 0.2 x 10-5 + C1
⇒ C1 = 8μF
Construction
Main parts of an ac generator:
Armature − Rectangular coil ABCD
Filed Magnets − Two pole pieces of a strong electromagnet
Slip Rings − The ends of coil ABCD are connected to two hollow metallic rings R1 and R2.
Brushes − B1 and B2 are two flexible metal plates or carbon rods. They are fixed and are kept in tight contact with R1 and R2 respectively.
Theory and Working − As the armature coil is rotated in the magnetic field, angle θ between the field and normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes. An emf is induced in the coil. According to Fleming’s right-hand rule, current induced in AB is from A to B and it is from C to D in CD. In the external circuit, current flows from B2 to B1.
To calculate the magnitude of emf induced:
Suppose
A → Area of each turn of the coil
N → Number of turns in the coil
→ Strength of magnetic field
θ → Angle which normal to the coil makes with 
at any instant t
∴ Magnetic flux linked with the coil in this position:
= NBA cosθ= NBA cosωt …(i)
Where, ‘ω’ is angular velocity of the coil
As the coil rotates, angle θchanges. Therefore, magnetic flux Φ linked with the coil changes and hence, an emf is induced in the coil. At this instant t, if e is the emf induced in the coil, then
A circular coil of cross-sectional area 200 cm2 and 20 turns is rotated about the vertical diameter with angular speed of 50 rad s−1 in a uniform magnetic field of magnitude 3.0 × 10−2T. Calculate the maximum value of the current in the coil.
Here, A = 200 cm2
N=20
ω = 50 rad/s
B = 3.0× 10−2 T
for maximum current emf induced should be maximum.
So, for maximum emf sinωt in e = NAB ω sinωt should be 1.
Hence, e = NABω
e = 20 × 200/10000 × 50 × 3 × 10−2 = 0.6 V
Since resistance of the circular loop is not given, let us consider its resistance to be R.
Therefore, current I in the coil is
(a) Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of number of turn and currents in the two coils.
(b) A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 300 turns. Find the number of turns in the secondary to get the power output at 220 V.
A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 µF and 9 µF capacitors), at a point distant 30 m from it, would equal:
240N/C
360N/C
420N/C
480N/C
C.
420N/C
Resultant circuit,
As, charge on 3μF = 3μF x 8V = 24μC
Charge on 3μF = 3μF x 2V = 18 μC
charge on 4μF +Charge on 9μF
= (24 + 18)μC = 42μC
therefore,
Electric field at a point distant 30 m
A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ∞) on its surface. For this sphere, the equipotential surfaces with potentials 3Vo/2, 5Vo/4, 3V/4 and Vo/4 have radius R1, R2,R3 and R4 respectively. Then
R1 = 0 and R2>(R4-R3)
R1≠0 and (R2-R1)>(R4-R3)
R1 = 0 and R2<(R4-R3)
2R<R4
C.
R1 = 0 and R2<(R4-R3)
The potential at the surface of the charged sphere.
As potential decreases for outside points. Thus, according to the question, we can write
Similarly,
In the given circuit, charge Q2 on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF. Q2 as a function of ‘C’ is given properly by : (figures are drawn schematically and are not to scale)
A.
From the given circuit,
Qc= Q1+Q2
or
C (E-V) = 1 x V + 2 XV
Or V (C+3) = CE
As, C1 varied from 1μF to 3μF, charge increases with decreasing slope.
A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is:
1.73 V/m
2.45 V/m
5.48 V/m
7.75 V/m
B.
2.45 V/m
Consider the LED as a point source of light.
Let power of the LED is P
Intensity at r from the source
... (i)
As we know that,
.... (ii)
From eqs. (i) and (ii) we can write
An LCR circuit is equivalent to a damped pendulum. In an LCR circuit, the capacitor is charged to Q0 and then connected to the L and R as shown.
If a student plots graphs of the square of maximum charge ( QMax2 ) on the capacitor with time (t) for two different values L1 and L2 (L1 > L2) of L then which of the following represents this graph correctly? (Plots are schematic and not drawn to scale)
D.
Inductance is inertia of circuit. It means inductance oppose the flow of charge, more inductance means decay of charge is slow.
In option (d), in a given time to 
.
So, L1> L2
Assume that an electric field 
exists in space. Then the potential difference VA – VO, where VO is the potential at the origin and VA the potential at x = 2 m is:
120 J
-120 J
-80 J
80 J
C.
-80 J
As we know, potential difference VA -Vo is
dV = - Edx
A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close to
6 x 10-7 C/m2
3 x 10-7 C/m2
3 x 104 C/m2
6x 104 C/m2
A.
6 x 10-7 C/m2
When free space between parallel plate capacitor, 
When dielectric is introduced between parallel plates of capacitor, 
Electric field inside dielectric 
where, K = dielectric constant of medium = -2.2
εo = permitivity of free space = 8.85 x 10-12
σ = 2.2 x 8.85 x 10-12 x 3 x 104
= 6.6 x 8.85 x 10-8 = 5.841 x10-7
= 6 x 10-7 C/m2
In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be
8 A
10 A
12 A
14 A
C.
12 A
Total power (P) consumed
= (15 x 40) + ( 5x 100) + ( 5x 80) + (1 x 1000) = 2500
Power, P = VI
Minimum capacity should be 12 A.
A conductor lies along the z-axis at -1.5 ≤ z< 1.5 m and carries a fixed current of 10.An in-az direction (see figure).
B = 3.0 x 10-4 e-0.2x ay T, find the power required to move the conductor at constant speed to x = 2.0 , y = 0 in 5 x10-3 s. Assume parallel motion along the x axis.
1.57 W
2.97 W
04.85 W
29.7 W
B.
2.97 W
When force exerted on a current carrying conductor
Fext = BIL
Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then
5C1 = 3C2
3C1 = 5C2
3C1 = 5C2 = 0
9C1 = 4C2
B.
3C1 = 5C2
q1 = q2
C1V1 = C2V2
120C1 = 200C2
⇒ 3C1 = 5C2
In an LCR circuit as shown below both switches are open initially. Now switch S1 is closed, S2 kept open. (q is charge on the capacitor and τ = RC is capacitive time constant). Which of the following statement is correct?
Work done by the battery is half of the energy dissipated in the resistor
At t = τ, q = CV/2
At t = 2τ, q = CV (1-e-2)
At t = τ/2, q = CV (1-e-1)
C.
At t = 2τ, q = CV (1-e-2)
For charging of capacitor
q = CV (1-e1/τ)
At t = 2τ, q =CV (1-e-2)
The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant τ of this circuit lies between
150 sec and 200 sec
0 and 50 sec
50 sec and 100 sec
100 sec and 150 sec
D.
100 sec and 150 sec
Time constant τ is the duration when the value of potential drops by 63% of its initial maximum value (i.e, Vo/e)
Here, 37% of 25 V = 9.25 V which lies between 100s to 150 s in the graphs.
A fully charged capacitor C with initial charge q0 is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is
A.
As initially, charge is maximum
∴ q = qocos ωt
Current, 
In a series LCR circuit, R = 200 Ω and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30º. On taking out the inductor from the circuit the current leads the voltage by 30º. The power dissipated in the LCR circuit is
305 W
210 W
zero
242 W
D.
242 W
The given circuit is under resonance as XL = XC. Hence, power dissipated in the circuit is
P = V2/R = 242 W
A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is :
24
32
2
16
B.
32
To hold 1 KV potential difference minimum four capacitors are required in series
⇒ C1 = 1/4
for one series.
So for Ceq to be 2μF, 8 parallel combinations are required.
⇒ Minimum no. of capacitors = 8 × 4 = 32
This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement – 1: For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q.
Statement-2: The net work done by a conservative force on an object moving along a closed loop is zero.
Statement-1 is true, Statement-2 is false
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
Statement-1 is false, Statement-2 is true.
B.
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
Work done by conservative force does not depend on the path. The electrostatic force is a conservative force.
A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other tow corners. If the net electrical force on Q is zero, then Q/q equals
1
-1
A.
Three forces F41, F42 and f43 acting on Q are shown Resultant of F41 + F43
Resultant on Q becomes zero only when ‘q’ charges are of negative nature.
An electric dipole is placed at an angle of 30° to a non-uniform electric field. The dipole will experience
a torque only
a translational force only in the direction of the field
a translational force only in a direction normal to the direction of the field
a torque as well as a translational force
D.
a torque as well as a translational force
A fully charged capacitor has a capacitance ‘C’ it is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity ‘s’ and mass ‘m’. If the temperature of the block is raised by ‘∆T’. The potential difference V across the capacitance
D.
Dimensionally only 4th option is correct.
A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C then the resultant capacitance is
(n − 1)C
(n + 1)C
C
nC
A.
(n − 1)C
Ceq=(n−1) C (Q all capacitors are in parallel)
A.
6 x 10-7 C/m2 When free space between parallel plate capacitor,
Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +σ, −σ and +σ respectively. The potential of shell B is :
C.
Outside the sphere (P lies outside the sphere)
where,
σ-surface charge density
At the surface of Sphere,
V= R
Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is:
A.
Using parallel axes theorem, moment of inertia about 'O'
Again, moment of inertia about point P,
Ip = Io +md2
A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K= 5/3 is inserted between the plates, the magnitude of the induced charge will be:
0.9 n C
1.2 n C
0.3 n C
2.4 n C
B.
1.2 n C
Charge on Capacitor, Q1 =CV
After inserting dielectric of dielectric constant = K
Qf = (kC)V
Induced charges on the dielectric
Qind =Qf-Qi = KCV-CV
= (K-1)CV =
The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B2. The ratio B1/B2 is:
2
D.
Dipole moment of circular loop is m = IA
m1 = I.A = I.πR2 { R = Radius of the loop}
If moment is doubled (keeping current constant)
R becomes
A capacitor of 2
is charged as shown in the figure. When the switch s is turned to position 2, the percentage of its stored energy dissipated is,
20%
75%
80%
0%
C.
80%
Consider the figure given above.
When switch S is connected to point 1, then initial energy stored in the capacitor is given as,
When the switch S is connected to point 2, energy dissipated on connection across 8
will be,
Therefore, per centage loss of energy = 
A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of a dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. which of the following is incorrect?
The potential difference between the plates decreases K times
The energy stored in the capacitor decreases K times
The change in energy stored is 
The charge on the capacitor is not conserved
C.
The change in energy stored is
A potentiometer wire has length 4 m and resistance 8Ω. The resistance that must be connected in series with the wire and an accumulator of emf 2V, so as to get a potential gradient 1mV per cm on the wire is
32Ω
40Ω
44Ω
48Ω
A.
32Ω
Given, l=4m,
R= potentiometer wire resistance = 8Ω
Across a metallic conductor of non-uniform cross -section, a constant potential difference is applied. The quantity which remains constant along the conductor is
Current density
Current
drift velocity
electric field
B.
Current
As the cross-sectional area of the conductor is non-uniform so current density will be different.
As, I=JA
It is clear from Eq. (i) when an area increases the current density decreases so the number of the flow of electrons will be same and thus the current will be constant.
A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance such that the impedance of the circuit becomes 'Z' the power drawn will be
P
A.
when a resistor is connected to an AC source. The power drawn will be
P=Vrms.Irms
A small signal voltage V(t) = Vo sin 
is applied across an ideal capacitor C:
over a full cycle the capacitor C does not consume any energy from the voltage source
current I(t) is in phase with voltage V(t)
Current I(t) leads voltage V(t) by 180o
Current I(t) lags voltage V(t) by 90o
A.
over a full cycle the capacitor C does not consume any energy from the voltage source
For an ac circuit containing capacitor only, the phase difference between current and voltage will be 
(i.e., 90o).
In this case, the current is ahead of voltage by 
Therefore, power is given by,
P = VI cos 
where, 
is the phase difference between voltage and current.
P = VI cos 90o = 0
An inductor of 20 mH, a capacitor of 50 
and a resistor of 40 ohm are connected in series across a source of emf V = 10 sin 340 t. the power loss in the ac circuit is,
0.67 W
0.76 W
0.89 W
0.51 W
D.
0.51 W
Given,
Inductance, L = 20 mH
Capacitance, C = 50 
Resistance, R = 40 ohm
emf, V = 10 sin 340 t
Power loss in AC circuit will be given as,
Pav = IV2 R = 
= 
A parallel plate capacitor has a uniform electrinc field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy storedf in the capacitor is df and area of each plate is A, the energy stored in the capacitor is
C.
Energy density for a parallel plate capacitor
Two metallic spheres of radii 1 cm and 3 cm are given charges of (-1 x 10-2 C) and 5 x 10-2 C , respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
2 x 10-2 C
3 x 10-2 C
4 x 10-2 C
1 x 10-2 C
B.
3 x 10-2 C
Charge flows from high potential to low potential
The potential energy of a particle in a force field is,
where A and B are positive constants are r is the distance of a particle from the centre of the field. For stable equilibrium, the distance of a particle is
B/2A
2A/B
A/B
B/A
B.
2A/B
Given, the potential energy of particle in a force field
If voltage across a bulb rated 220 V -100W drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is
20%
2.5%
5%
10%
C.
5%
Power P= V2/R
For small variation
ΔP/P x 100% 
Therefore, power would decrease by 5%
Four point charge -Q,-q,2q and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the centre of the square is zero is
Q= -q
Q=-1/q
Q=q
Q =1/q
A.
Q= -q
A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are:
300 V, 15 A
450 V, 15 A
450 V, 13.5 A
600 V, 15 A
B.
450 V, 15 A
Initial power = 3000 W
As the efficiency is 90%, then final power = 
= 2700 W
That is,
... (i)
So, 
A, B and C are three points in a uniform electric field. The electric potential is
EMaximum at A
Maximum at B
Maximum at C
same at all the three points A, B and C
B.
Maximum at B
The electric field maximum at B, because electric field is directed along decreasing potential VB>VC>VA.
In the given figure, a diode D is connected to an external resistance R = 100 Ω and an e. m. f of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be 
30 mA
40 mA
20 mA
35 mA
A.
30 mA
Given, external resistance R = 100 Ω and an emf is 3.5 V.
Potential barrier across the diode is 0.5 V
Potential difference on R = 3.5 V - 0.5 V = 3.0 V
Current in circuit, I = V/R = 3/100
= 0.03 A = 30 mA
A series R -C circuit is connected to an alternating voltage source. Consider tow situations:
1. When the capacitor is air filled.
2. When the capacitor is mica filled.
Current through resistor is i and voltage across capacitor is V then
Va < Vb
Va > Vb
ia >ib
Va = Vb
B.
Va > Vb
Net reactive capacitance,
in 
when mica is introduced capacitance will increase hence, voltage across capacitor get decrease.
The electric potential V at any point (x, y,z), all in meters in space is given by V= 4x2 volt. The electric field at the point (1,0,2) in volt/meter is
8 along positive X-axis
16 along negative x- axis
16 along positive X- axis
8 along negative X - axis
D.
8 along negative X - axis
We know that
Three charges, each +q, are placed at the corner of an isosceles triangle ABC of sides BC and AC, 2a. D and E are the midpoints of BC and CA. The work done in taking a charge Q from D to E is.
zero
C.
zero
Here, AC = BC
VD = VE = V
W = Q [ VE-VD]
W = Q [V-V]
W = 0
A 220 V input is supplied to a transformer. The output circuit draws a current 2.0 A at 440 V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is
3.6 A
2.8 A
2.5 A
5.0 A
D.
5.0 A
Efficiency is defined as the ratio of output power and input power
i.e. 
Two positive ions, each carrying a charge q, are separated by a distance d. If F is the force of repulsion between the ions, the number of electrons missing from each ion will be (e being the charge on an electron)
C.
Two positive ions each carrying a charge q are kept at a distance d, then it is found that force of repulsion between them is
A series combination of n1 capacitors, each of value of C1, is charged by a source of potential difference 4V. When another parallel combination of n2 capacitors, each of value C2, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of C2, in terms of C1, is then
D.
Case I . When the capacitors are joined in series
Case II. When the capacitors are joined in parallel.
Two parallel metal plates having charges +Q and -Q faces each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will
become zero
increase
decrease
remain same
B.
increase
The electric field in a vacuum
When the plates are dipped in kerosene oil tank, then the electric field between the plate will
so, the electric field between the plates will decrease.
Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be
C/3, V/3
3C, V/3
C/3, 3V
3C ,3V
C.
C/3, 3V
In series arrangement charge on each plate of each capacitor has the same magnitude. The potential difference is distributed inversely in the ratio of capacitors ie,
V = V1 +V2 +V3 ....
Here, V = 3V
The equivalent capacitance Cs is given by
The two ends f a rod of length L and a uniform cross -sectional area A are kept at two temperatures T1 and T2 (T2 > T1) .The rate of heat transfer, dQ/dt, through the rod in steady state is given by
D.
For a rod of length L and area of cross- section A whose faces are maintained at temperature T1 and T2 respectively.
Then in steady state, the rate of heat flowing from one face to the other face in time t is given by
A galvanometer having a coil resistance of 60 Ω shows full-scale deflection when a current of 1.0 A passes through it. It can be converted into an ammeter to read currents up to 5.0 A by
putting in parallel a resistance of 240 Ω
putting in series a resistance of 15 Ω
putting in series a resistance of 240 Ω
putting in parallel a resistance of 15 Ω
D.
putting in parallel a resistance of 15 Ω
To convert a galvanometer to ammeter a small resistance is connected in parallel to the coil of the galvanometer.
Here G1 = 60 Ω, Ig = 1.0 A, I = 5 A
Ig G1 = (I - Ig) S
Putting 15 Ω resistance in parallel.
The electric potential at a point in free space due to a charge Q coulomb is Q x 1011 V. The electric field at that point is
A.
Search for the relations of electric potential and electric field at particular point. At any point, electric potential due to charge Q is 
where r is the distance of observation point from the charge.
At the same point, electric field is 
The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is
C.
Energy given be the cell
E =CV2
Here, C = capacitance of condenser = 
In an AC circuit, the emf (e) and the current (i) at any instant are given respectively by
e = E0 sin ωt
i = Io sin (ωt -Φ)
The average power in the circuit over one cycle of AC is
EoIo
C.
The power is defined as the rate at which work is being done in the circuit.
Power = rate of work done in one complete cycle.
Or
Pav = W/T
where cos Φ is called the power factor of an AC circuit.
In the circuit shown, the current through the 4 Ω resistor is 1 A when the points P and M are connected to a DC voltage source. The potential difference between the point M and N is 
1.5 V
1.0 V
0.5 V
3.2 V
D.
3.2 V
In parallel resistances, the potential difference across them is same.
Potential difference across PM
V = 4 x 1 = 4 Volt (ie, across 4 Ω)
Equivalent resistance of lower side arm, 
Now the circuit can be shown as
Let current I flow in lower branch, so
1.25 I = 4V
I = 4/1.25 = 3.2 A
Therefore, 3.2 A current flows in 1 Ω resistance, hence potential difference between M and N is
V' = 3.2 x 1 = 3.2 volt
The primary and secondary coil of a transformer have 50 and 1500 turns respectively. If the magnetic flux Φ linked with the primary coil is given Φ = Φo +4t, where Φ is in weber, t is time is second and Φo is a constant the output voltage across the secondary coil is:
90 V
120 V
220 V
30 V
B.
120 V
The magentic flux linked with the primary coil is given by
Φ = Φo +4t
So, voltage across primary
Vp = dΦ/dt = d(Φ+dt) /dt
= 4 volt (Φ = constant)
Also, we have
Np = 50 and Ns = 1500
From relation ,
Vs/Vp = Ns/Np
Vs = 4 x (1500/50)
= 120 V
A charged particle (charge q ) is moving n a circle of radius R with uniform speed v. The associated magnetic moment μ is given by:
qvR/2
qVR2
qVR2/2
qvR
A.
qvR/2
As revolving charge is equivalent to a current so
I = qf = q x ω/ 2π
But ω = v/R
Where R is radius of circle and v is uniform speed of charged particle, therefore,
I= qv/2πR
Now, magnetic moment associated with charged particle is given by
μ = IA = I x πR2
μ = qv/2 πR2
= qvR/2
In mass spectrometer used for measuring the masses of ions, the are initially accelerated by an electric potential V and then made to describe semicircular paths of radius R using a magnetic field B. If V and B are kept constant, the ratio 
will be proportional to:
1/R
1/R2
R2
R
B.
1/R2
Centripetal force os provided by the magentic force qvB.
The radius of the orbit in which ions moving is determined by the relation as given below.
where m is the mass, v is velocity, q is charge of ion and B is the flux density of the magnetic field, so that qvB is the magnetic force acting on the ion, and mv2/R is the centripetal force on the ion moving in a curved path of radius R. The angular frequency of rotation of the ions about the vertical field B is given by
If ions are accelerated by electric potential V, then energy attained by ions
E =qV ..... (ii)
From Eqs. (i) and (ii), we get
If V and B are kept constant, then
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain of the amplifier is:
500
1000
1250
100
C.
1250
AC power gain is ratio of change in output power to the change in input power.
AC power gain
The core of a transformer is laminated because
energy losses due to eddy currents may be minimised
the weight of the transformer may be reduced
rusting of the core may be prevented
ratio of voltage in primary and secondary may be increased
A.
energy losses due to eddy currents may be minimised
When magnetic flux linked with a coil changes, induced emf is produced in it and the induced current flows through the wire forming the coil. In 1895, Focault experimentally found that these induced currents are set up in the conductor in the form of closed loops. These current look like eddies or whirl pools and likewise are known as Focault's current. These current oppose the cause of their origin, therefore, due to eddy currents, a great amount of energy is wasted in form of heat energy. If core of transformer is laminated, then their effect can be minimised.
A coil of inductive reactance 31 Ω has a resistance of 8Ω. It is placed in series with a condenser of capacitative reactance 25Ω. The combination is connected to an a.c. soruce of 110 V. The power factor of the circuit is
0.56
0.64
0.80
0.33
C.
0.80
Power factor 
is the ratio of resistance and impedance of a.c. circuit.
Power factor of a.c. circuit is given by
...(i)
where R is resistance employed and z the impedance of the circuit.
...(ii)
Eqs. (i) and (ii) meet to give,
...(iii)
Given, 
An electric dipole of moment 
is lying along a uniform electric field 
The work done in rotating the dipole by 
2pE
pE
D.
pE
When an electric dipole in an electric field 
a torque 
acts on it. This torque tries to rotate the dipole through an angle.
If the dipole is rotated from an angle 
, then work done by external force is given by
...(i)
putting 
A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting thecharging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates
decreases
does not change
becomes zero
increases
D.
increases
Charge remains constant after charging.
If the battery is removed after charging then the charge stored in the capacitor remains constant.
q = constant
Change in capacitance
As 
Hence, 
Hence, potential difference between the plates
or 
As capacitance decreases, so potential difference increases.
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system
Increases by a factor of 4
Decreases by a factor of 2
Increases by a factor of 2
Remains the same
B.
Decreases by a factor of 2
Charge on capacitor
q = CV
when it is connected with another uncharged capacitor.
The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is
Independent of the distance between the plates
Linearly proportional to the distance between the plates
Inversely proportional to the distance between the plates
Proportional to the square root of the distance between the plates
A.
Independent of the distance between the plates
The electrostatic force between the metal plates
For isolated capacitor Q = constant
Hence, F is Independent of the distance between plates
Two identical conducting balls A and B have positive charges q1 and q2 respectively. But q1 ≠ q2. The balls are brought together so that they touch each other and then kept in their original positions. The force between them is
less than that before the balls touched
greater than that before the balls touched
same as that before the balls touched
zero
B.
greater than that before the balls touched
According to Coulomb's law, the force of repulsion between them is given by
when the charged spheres A and B are brought in contact, each sphere will attain equal charge q'.
Now, the force of repulsion between them at the same distance r is
A positively charged ball hangs from a silk thread. we put a positive test charge qo at a point and measure F/qo, then it can be predicted that the electric field strength E
>F/qo
=F/q
<F/qo
Cannot be estimated
A.
>F/qo
Due to the presence of test charge qo in front of the positively charged ball, there would be a redistribution of charge on the ball. In the redistribution of charge, there will be less charge on the front half surface and more charge on the back half surface. As a result, the net force F between ball and charge will decrease, i.e., the electric field is decreased. Thus, the actual electric field will be greater than F/qo
Capacitor C1 of capacitance 1μF and capacitor C2 of capacitance 2 μF are separately charged fully by a common battery. The two capacitors are then separately allowed to discharged through equal resistors at time t =0
the current in each of the two discharging circuits is zero at t = 0
the currents in the two discharging circuits at t= 0 are equal but non-zero
the currents in the two discharging circuits at t=0
Capacitor C1 loses 40% of its initial charge sooner than C2 loses 40% of the initial charge.
B.
the currents in the two discharging circuits at t= 0 are equal but non-zero
Here Vo and R in the two CR circuits are same, hence, the currents in the two discharging circuits at t = 0 (V0/R) will be the same.
We know that time constant of a resistance and capacitance circuit is CR. Here time constant of the first circuit (Capacity C1 = 1μF) is half of the time constant of second circuit (Capacity C2 = 2μF). Hence, C1 loses 50% its initial charge sooner.
Two condensers, one of capacity C and the other of capacity C/2, are connected to a V- volt battery, as shown.
The work done in charging fully both the condensers is
CV2
C.
The two condensers in the circuit are in parallel order, hence
The work done in charging the equivalent capacitor is stored in the form of potential energy.
Hence,
A charge Q is distributed uniformly in a sphere (solid). Then, the electric field at any point r, where, r < R (r is the radius of sphere) varies as:
r -1
r
r -2
C.
r
Electric field intensity at a point inside the sphere
E ∝ r
As per this diagram a point charge +q is placed at the origin O. Work done in taking another point charge -Q from the point A [coordinates (O, a)] to another point B [coordinates (a, O)] along the straight path AB is
zero
A.
zero
The work done in carrying a test charge consists in product of difference of potentials at points A and B and value of test charge.
Potential at A
Potential at B
Thus, work done in carrying a test charge-Q from A to B
W = (VA - VB) ( -Q ) = 0
A charged particle (charge q) is moving in a circle of radius R with uniform speed v. The associated magnetic moments given by
qνR2
qνR
A.
As revolving charge is equivalent to a current, so
I = qf
where R is radius of circle and v is uniform speed of charged particle.
μ = IA
= I × πR2
A steady current of 1.5 A flows through a copper voltameter for 10 min. If the electrochemical equivalent of copper is 30x 10-5 g C-1 , the mass of copper deposited on the electrode will be
0.40 g
0.50 g
0.67 g
0.27 g
D.
0.27 g
If m is the mass of a substance deposited or liberated at an electrode during electrolysis when a charge q passes through electrolyte, then according to Faraday's first law of electrolysis,
m ∝ q
⇒ m = zq
where z is a constant of proportionality and is called electrochemical equivalent (ECE) of the substance.
If an electric current I flows through the electrolyte, then
q = (I.t)
m = zIt
Here , I = 1.5 A,
t = 10 min = 10 × 60 s
z = 30 × 10-5 gC-1
Hence, mass of copper deposited on the electrode
m = 30 × 10-5 × 1.5 × 10 × 60
= 27 × 10-2
m = 0.27 g
When a charged particle moving with velocity is subjected to a magnetic field of induction , the force on it is non- zero. This implies that
angle between is necessarily 90o
angle between can have any value other than 90o
angle between can have any value other than 180o
angle between is either zero or 180o
C.
angle between can have any value other than 180o
When a charged particle q is moving in a uniform magnetic field with velocity such that angle between be θ, then due to an interaction between the magnetic field produced due to moving charge and magnetic force applied, the charge q experiences a force which is given by
F = qv sinθ
If θ = 0o or 180o
then sinθ = 0
∴ F = qvB sinθ
F = 0
Since, force on charged particle is non-zero, so angle between can have any value other than zero and 180°.
Note:- Force experienced by the charged particle is Lorentz force.
What is not true for equipotential surface for uniform electric field?
Equipotential surface is flat
Equipotential surface is spherical
Electric lines are perpendicular to equipotential surface
Work done is zero
B.
Equipotential surface is spherical
An equipotential surface is a surface with a constant value of potential at all points on the surface. For a uniform electric field, say, along the x-axis, the equipotential surfaces are planes normal to the x-axis, ie, plane parallel to the y-z plane. Equipotential surfaces for a dipole and its electric field lines are shown in figure.
As said above that on equipotential surface, potential at all points is constant, this means that on equipotential surface work done in moving a test charge from one point to other point is zero.
A capacitor having capacitance 1 uF with air, is filled with two dielectrics as shown. How many times capacitance will increase
12
6
8/3
3
B.
6
After filling with dielectrics the two capacitors will be in parallel order. As shown, the two capacitors are connected in parallel. Initially the capacitance of capacitor
C =
where A is area of each plate and d is the separation between the plates.
After filling with dielectrics, we have two capacitors of capacitance.
Hence, their equivalent capacitance
Ceq = C1 + C2
= 4C +2C
C = 6C
ie, new capacitance will be six times of the original.
Figure shows three spherical and equipotential surfaces A, B and C round a point charge q. The potential difference If t1 and t2 be the distances between them, then 
t1 = t2
t1 > t2
t1 < t2
t1 ≤ t2
C.
t1 < t2
Potential difference between two equipotential surfaces A and B,
VA - VB =
=
⇒
⇒
Similarly,
Since, therefore
∴
Three capacitors of capacitances 1 µF, 2µF and 4μF are connected first in a series combination and then in a parallel combination. The ratio their equivalent capacitances will be
2 : 49
49 : 2
4 : 49
49 : 4
C.
4 : 49
Capacitors are connected together in series when they are daisy chained together in single
In series combination
=
⇒
Capacitors are connected together in parallel when both of its terminals are connected to each terminal of another capacitor.
Parallel capacitors equation
CT = C1 + C2 + C3 + ...
In parallel combination,
C1 = 1 + 2 + 4
= 7 μF
∴
A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be
A.
The condenser with air as the dielectric has capacitance
C1 =
Similarly, the condenser with K as the dielectric constant has capacitance
Since C1 and C2 are in parallel
Cnet = C1 + C2
=
Cnet =
Two equal negative charges - q are fixed at the point (0, a) and (0, - a) on the y-axis. A positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge wil
execute SHM about the origin
move to the origin and remain at rest
move to infinity
execute oscillatory but not SHM
D.
execute oscillatory but not SHM
Component of force on charge of + Q at P, along x-axis,
Fx =
=
=
Which is not directly proportional to x. So, motion is oscillatory but not SHM.
A charged particle is accelerated by a potential of 200 V. If its velocity is 8.4 × 108 m/s, then value of e/m for that particle is
17.6 × 1016
14.5 × 1012
1.76 × 1015
1.45 × 1015
C.
1.76 × 1015
If electron is accelerated by voltage V, then its kinetic energy
= qV
Where V is the velocity of accelerated particles
q is the charge on particle
m is the mass
⇒
=
= 1.76 × 1015
Two capacitors of capacities C1 and C2 are charged upto the potential V1 and V2 then condition for not flowing the charge between on connected them in parallel is
C1 = C2
C1V1 = C2V2
V1 =V2
B.
C1V1 = C2V2
The Capacitor Charge calculator calculates the charge of a capacitor with a capacitance of C, and voltage of V
If charge on both the capacitors is the same, then transfer of energy does not take place
Q = CV
Where C is Capacitance
V is Voltage
C1V1 = C2V2
It is the required condition.
If V =ary, then electric field at a point will be proportional to
r
r -1
r -2
r2
D.
r2
Electric field E ∝ r2
Electric field at point 20 cm away from the centre of dielectric sphere is 100 V/m, radius of sphere is 10 cm, then the value of electric field at a distance 3 cm from the centre is
100 V/m
125 V/m
120 V/m
0
D.
0
Inside a dielectric sphere, electric field is zero.
Dielectric, insulating material or a very poor conductor of electric current. When dielectrics are placed in an electric field, practically no current flows in them because, unlike metals, they have no loosely bound, or free, electrons that may drift through the material.
Magnetic field of the earth is H =0.3 g. A magnet is vibrating 5 oscillations per min then the dippreciation required in the magnetic field of the earth of increase time period upto 10 oscillations per minute is
2.25 g
0.6 g
0.9 g
0.12 g
A.
2.25 g
The dipole in a uniform magnetic field
T =
Where I is the moment of inertia
H is the magnetic field
We know that
T
∴
⇒
∴ H2 = 0.75
Dippreciation in field
= 3 - 0.75
= 2.25 g
Two small balls, each carrying a charge q are suspended by equal insulator strings of length l m from the hook of a stand. This arrangement is carried in a satellite in space. The tension in each string will be
B.
Suppose, T be the tension in the spring T. Also in the satellite, two balls will be in a straight line.
So, force acting on the balls
F = ......(i)
This force will be balanced by tension in the string
T = F ......(ii)
Now from equatin (i) and (ii) we get
T =
The concentric, conducting spherical shells X, Y and Z with radii r, 2r and 3r, respectively. X and Z are connected by a conducting wire and Y is uniformly charged to charge Q as shown in figure. Charges on shells X and Z will be
B.
Given:- spherical shells X , Y and Z having radii r, 2r and 3r.
Let charges on X and Z be qx and qz
Law of conservation says that net charge in an isolated system will always remain constant.
By conservation of charge
Vx =
Vz =
Since X and Z are connected.
Vx = Vz
∴ qx =
qx + qz = 0
⇒ qz = -qx
⇒ qz =
In the given figure, the capacitors C1, C3, C4, C5 have a capacitance 4 µF each. If the capacitor C2 has a capacitance 10 μF, then effective capacitance between A and B will be 
2 μF
6 μF
4 μF
8 μF
C.
4 μF
When a battery is applied across A and B, then the points b and c will be at the same potential.
( C1 = C4 = C3 = C5 = 4 μF )
Therefore, no charge flows through C2.
As C1 and C5 are in series.
∴ Their equivalent capacitance,
C' =
=
C' = 2 μF
Similarly, C4 and C3 are in series. Therefore, their equivalent capacitance.
C'' =
=
C'' = 2 μF
Now, C' and C" are in parallel. Therefore, effective capacitance between A and B
= C' + C"
= 2 + 2
= 4 μF
A capacitor is charged and then made to discharge through a resistance. The time constant is t. In what time will the potential difference across the capacitor decrease by 10 %?
τ ln 0.1
τ ln 0.9
τ ln
τ ln
C.
τ ln
As, we know that
q = q0 e-t/λ
According to the question
VC =
= e-t/λ
e-t/λ
et/λ =
⇒ t = τ ln
As shown in figure, two vertical conducting rails separated by distance 1.0 m are placed parallel to z-axis. At z = 0, a capacitor of 0.15 F is connected between the rails and a metal rod of mass 100 gm placed across the rails slides down along the rails. If a constant magnetic fields of 2.0 T exists perpendicular to the plane of the rails, what is the acceleration of the rod? ( take g = 9.8 m/s2)
2.5 m/s2
1.4 m/s2
9.8 m/s2
0
B.
1.4 m/s2
Due to motion of rod, emf induced across capacitor,
e = B l v
∴ Charge stored in capacitor,
Q = C ( Blv )
Current
I =
= C ( Blv)
I = CBl α
Force opposing the downward motion,
Fm = BI l
∴ Fm = B (CBla) l
Fm = B2 l2 Ca
Net force on rod
Fnet = W Fm
= mg B2 l2Ca
∴ m = mg B2 l2C
⇒
so
= 1.4 m/s2
A large solid sphere with uniformly distributed positive charge has a smooth narrow tunnel through its centre. A small particle with negative charge, initially at rest far from the sphere, approaches it along the line of the tunnel, reaches its surface with a speed v, and passes through the tunnel. Its speed at the centre of the sphere will be
0
v
v
v
D.
v
Potential at infinity = V∞ = 0
Potential at the surface of the sphere
Vs = k
Potential at the centre of the sphere,
Vc =
Let m and -q be the mass and the charge of the particle respectively.
Let v0 = speed of the particle at the centre of the sphere.
= qk ....(i)
......(ii)
Dividing eqn. (ii) by eqn. (i),
= 1.5
v0 = v
In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is however not constant, but increases uniformly along the positive z-direction at the rate of 105 N C-1m-1 The force experienced by the system having a total dipole moment equal to 10-7 C m in the negative z-direction is
10-2 N
10-2 N
10-4 N
10-4 N
A.
10-2 N
Consider an electric dipole with -q charge at A and +q charge at B, placed along z-axis, such that its dipole moment is in negative z-direction.
i.e pz = 10-7 C m
The elecric field is along positive direction of z-axis, such that
= 105 N C-1 m-1
From F = qdE
= ( q × dz ) ×
F = p
Force experienced by the system in the negative z-direction,
F =
= 10-7 × ( 10-5 )
F = 10-2 N
An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge upto a radius R. The atom as a whole is neutral. The electric field at a distance r from the nucleus is (r < R)
A.
Charge on nucleus = + Ze
Total negative charge = - Ze
( atom is electrically neutral)
Negative charge density
ρ =
=
⇒ ρ = ...(i)
Consider a Gaussian surface with radius r.
By Gauss's theorem
E (r) × .....(ii)
Charge enclosed by Gaussian surface,
q = Ze + ρ
= Ze [ using (i) ]
From (ii),
E (r ) =
=
⇒ E (r) =
A deflection magnetometer is adjusted in the usual way. When a magnet is introduced, the deflection observed is θ, and the period of oscillation of the needle in the magnetometer is T. When the magnet is removed, the period of oscillation is T0. The relation between T and T0 is
T2 = cosθ
T = T0 cosθ
T =
A.
T2 = cosθ
In the usual setting of deflection magnetometer, field due to magnet (F) and
horizontal component (H) of earth's field are perpendicular to each other. Therefore, the net field on the magnetic needle is
∴ T = 2 ....(i)
When the magnet is removed
T0 = 2 .....(ii)
Also, = tanθ
Dividing (i) by (ii), we get
=
=
=
= cosθ
∴ T0 = cosθ
A parallel plate capacitor with air as a dielectric has capacitance C. A slab of dielectric constant K, having same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be
A.
The two condensers with Kand with air are in parallel.
With air
C1 =
C1 =
With medium
C2 =
C2 =
∴ C' = C1 + C2
=
C' =
⇒ C' =
Assertion: If a dielectric is placed in external field then field inside dielectric will be less than applied field.
Reason: Electric field will induce dipole moment opposite to field direction.
If both assertion and reason are true and reason is the correct explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false.
C.
If assertion is true but reason is false.
Dipole moment will be in the same direction as the external field. The collective effect of dipole moments produces a field that opposes the external field and hence, the net electric field inside the dielectric is less than the external electric field.
Two charged spheres separated by a distance 'd' exert some force on each other. If they are immersed in a liquid of dielectric constant 2, then what is the force exerted, if all other conditions are same?
F/2
F
2F
4F
A.
F/2
Given:-
Distance between two charges = d
and dielectric constant (k) = 2
Force between charged spheres
(F) =
where ε0 is permitivity of free space.
Force for a medium with dielectric constant k
F' =
=
F' =
A hexagon of side 8 cm has a charge 4 µC at each of its vertices. The potential at the centre of the hexagon is
2.7 × 106 V
7.2 × 1011 V
2.5 × 1012 V
3.4 × 104 V
A.
2.7 × 106 V
As shown in the figure, 0 is the centre of hexagon ABCDEF of each side 8 cm. As it is a regular hexagon OAB, OBC, etc are equilateral triangles.
OA = OB = OC = OD = OE = OF = 8 cm
= 8 x 10-2 m
∴ VO = VA + VB + VC + VD + VE + VF
VO =
VO =
VO = 9 × 109 ×
VO = 9 × 109×
where = 9 × 109
VO = 2.7 × 106 V
A 5 µF capacitor is charged by being connected to a 3 V battery. The battery is then disconnected. If the resistance of the dielectric material between the capacitor plates is 109 Ω, what is the charge remaining on the capacitor 1 h after it has been disconnected ?
1.5 × 10-5 C
7.4 × 10-6 C
5.4 × 10-6 C
8.4 × 10-5 C
B.
7.4 × 10-6 C
T = R C
= ( 109 Ω ) ( 5 × 10-6 F )
= 5 × 103 s
=
T =
T = 1.4 h
The initial charge on the capacitor is
Q0 = C V
= ( 5 × 10-6 F ) ( 3V )
Q0 = 1.5 × 10-5 C
After t = 1h
= 0.71
and Q = Q0 e-t/T
= (1.5 × 10-5 C ) ( e-0.71 )
= ( 1.5 × 10-5 C ) ( 0.49 )
Q = 7.4 × 10-6 C
A metallic solid sphere is placed in a uniform electric field. Which path, the lines of force follow as shown in figure?
1
2
3
4
D.
4
When metallic solid sphere is placed in a uniform electric field, the electrons of the sphere will move against the direction of the electric field. Consequently, the left face of the sphere acquires negative charge while the right face attains positive charge. The field lines will terminate at the left face of sphere and restart from its right face. The electric field inside the sphere is zero. On the surface of the sphere, the field lines are normal at every point i.e. directed towards the centre. Therefore, the correct field line is represented by line 4.
Sponsor Area
Sponsor Area
at the center O is
, which makes angle θ with respect to the x-axis. When subjected to an electric field 
, it experiences a torque 
. When subjected to another electric field 
it experiences torque 
. The angle θ is
and (2, 0) respectively. The potential difference between the points A and B will be
100t volt is connected to a capacitor of capacity 1μF. The rms value of the current in the circuit is 
