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Given,
EMF of battery,
Internal resistance of battery, r=
Current flowing in the circuit, I= 0.5 A
Using the forlmula
or
where, R is the external resistance.
is the required resistance.
Terminal voltage,
Given, three resistors 2 are combined in parallel. 
Therefore,
Total resistance of parallel combination,
ii) The combination is connected to a battery of 20 V.
Let the current through resistances 2Ω, 4Ω and 5Ω are I1, I2 and I3 respectively.
Now, using Ohm's law across each resistor we get,
Hence,
Total current is given by
Given,
Resistance of the element,
Resistance of heating element at room temperature,
Temperature coeffient of the material of resistor, = 1.70 x 10–4 °C–1
Now, using the relation
we get,
is the required temperature of the element.
Given,
Now, using the formula is given by,
Resistivity,
Given,
Resitance of silver wire,
Room temperature, t1 = 27.5 °C
Resistance of wire at temperature 100o C, R2= 2.7
Using the relation,
where,
Therefore, temperature coefficient of resistivity of silver is given by,
Here,
Potential applied across the heating element, V= 230 V
Initial current, I1 = 3.2 A
After few seconds, steady current attained is, I2 = 2.8 A
Using ohm's law,
and,
Using the relation,
Therefore,
Given, Emf of cell, E1 = 1.25 V
Balance point is obtained at, l1 = 35 cm
When, the cell is replaced by another cell
New balance point is, l2 = 63 cm
Emf of the second cell can be found out using the relation,
Now, substituting values we get,
Given,
Number density of electrons, n = 8.5 x 1028 m–3 Current carried by the wire, I = 3.0 A
Area of cross-section of the wire, A = 2.0 x 10–6 m2
Length of the wire, l = 3.0 m
Charge on electron, e = 1.6 x 10–19 C
Using the formula for Drift velocity
Therefore,
Time taken by an electron to drift from one end to another end is,
Given,
Surface charge density of earth's surface = 10–9 C/m2
Current flowing across the surface of earth, I = 1800 A
The radius of earth,r= 6370 km = 6.37 x 106 m
Charge on entire surface of the earth, q=
Area of earth's surface, A = 4
This implies,
q=4 (6.37 x 106)2 x 10–9 C
Using the formula , I= we get,
t=
Therefore,
Time required for the flow of entire charge is,
Given,
Emf of secondary cells,
Internal resistance of cell, r= 0.015
Number of secondary cells, n=6
External resistance, R = 8.5
Current drawn from the supply, I =
And, terminal voltage of the supply is, V=IR
Given,
Emf of secondary cell,
Internal resitance of the secondary cell, r = 380
Maximum current drawn from the cell is given by,
Imax = 0.005 A
The amount of current drawn from the cell is enormously low and, this amount of current cannot drive the starting motor of a car. The current required to start the motor is a minimum of 100 A.
Given, two wires of equal length have the same resistance.
Let,
Resistance of Aluminium wire , R1= R and,
Length of wire, l1= l.
Relative density of Al, d1 =2.7
Resistivity of Al,
Resistance of Copper wire, R2 = R
Length of Copper wire, l2 = l
Relative density of Cu, d2 =8.9
Resistivity of Cu,
Using the formula for resistance,
For Aluminium wire,
= and,
Since, mass = volume x density
mass of Al wire, m1 = A1l1 x d1 = A1l1 x 2.7
For Copper wire,
Copper wire is 2.2 times heavier than the aluminium wire. That is why, Aluminium wires are preferred over Copper wires for overhead power cables.
What conclusion can you draw from the following observations on a resistor made of alloy manganin?
|
Current A |
Voltage V |
Current A |
Voltage V |
|
0.2 0.4 0.6 0.8 1.0 2.0 |
3.94 7.87 11.8 15.7 19.7 39.4 |
3.0 4.0 5.0 6.0 7.0 8.0 |
59.2 78.8 98.6 118.5 138.2 158.0 |
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(a) What is the value of ε?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation, if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
(f) Would the circuit work well for determining extremely small emf, say of the order of a few mV (such as the typical emf of a thermocouple)? If not, how will you modify the circuit ?
(a) Given,
Emf of the standard cell, E1 =1.02 V
Distance where balance point is obtained, l1= 67.3 cm
Resistance, R= 600 kΩ
When the standard cell is replaced by another cell, balance point is obtained at, l2= 82.3 cm
Emf of the new cell, E2 = ?
Using the formula for comparison of emf,
(b) The purpose of using high resistance of 600 kΩ is to reduce the amount of current passing through the galvanometer when the movable contact is far from the balance point.
(c) No, the balance point is not affected by the presence of this resistance.
(d) No, the balance point is not affected by the internal resistance of the driver cell.
(e) No, it is necessary that the emf of the driver cell is more than the emf of the cells and hence, the balance point will not be obtained if the emf of the driver cell is 1 V instead of 2 V.
(f) For measurement of small emf, this circuit will not work well.
Modification of circuit:
In order to measure small emf, we can connect a high resistance in series with the cell of 2V. This will decrease the amount of current flowing in the potentiometre wire. Thereby, the potential difference will decrease and small emf can be measured.
OR
The number of potentiometer wires is increased to 11 or 15 to get a potential gradient of 0.1 Vm–1. The purpose discussed above will be served by implementing a single 1 metre long wire, with high series resistance.
Given, standard resistance, S=Y= 10 Ω
Balance point with standard resistor is obtained at, l2= 58.3 cm
Balance point with unknown resistor is at, l1= 68.5 cm
Unknown resistor is, R=X=?
Using the formula,
If we fail to get a balance point with the given emf then, the potential drop across R and X are greater than the potential drop across the potentiometer wire AB. The obvious thing to do is to reduce the current in the outside circuit (i.e., the potential drop across R and X) suitably by putting a series resistor. Alternatively, we can increase the voltage of the driver cell inorder to increase the potential drop across the potentiometre wire.
Effective emf of three cells in parallel, ε = 2 V
Internal resistance, r=?
External resistance, R= 5 ohm
Terminal voltage across the cell, V= 1.5 V
Let, internal resistance of each cell be 'r' ohm.
Effective internal resistance of each three cells in parallel, r= r/3
Therefore,
Total resistance of circuit
Now we have,
Terminal voltage,
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Given a heater and a bulb.
Resistance of heater,
Resistance of bulb,
So, the resistance of bulb is greater than that of heater.
Drift velocity is given by,
So, the drift velocity becomes one-fourth of it's initial value when radius of the conductor is doubled.
C.
high resistance, low melting pointC.
Peltier effectA.
free electron densityD.
the charge crossing in a given time intervalSponsor Area
Voltmeter has high resistance. Therefore, the voltmeter is always connected in parallel so that the voltage of the circuit does not change when it gets connnected.
The drift velocity of electrons is given by
vd=
and, electric field E is equal to V/l.
Therefore, .
This implies,
the drift velocity of electrons would reduce to half of it's initial value when the length of the wire is doubled.
Resistance is given by, R= =
Given,
r' = and
l' =
Therefore,
New resistance, R' =
Implies, there is no change in resistance when radius is halved and length is reduced to one-fourth of it's original length.
Resistivity is defined as the measure of resistance to electrical conduction for a given size of material.
For a particular material at specified temperature, resistivity is the electrical resistance per unit length and per unit of cross sectional area.
Resistivity is the inverse of conductivity.
The resistivity of a material depends upon the material and temperature of the material and is independent of the length and area of cross-section.
i) In the absence of electric field the path of electrons are straight lines between successive collisions.
ii) When an electric field is set up from positive to negative charge, the electrons are accelerated towards the negative charge. Therefore, in the presence of electric field the electrons follow a curved path in between successive collisions.
Consider a conductor of length'l' and of uniform area of cross section 'A'. 
Therefore,
Volume of conductor, V = Al
If, n is the number density of electrons, then
Total number of free electrons in the conductor = Aln
If, e is the total charge on each electron then,
Total charge on electrons in the conductor, q = Alne
Let, a constant potential difference 'V' be applied across the ends of a conductor. Then,
Electric field set up across the conductor, E = V/l
Because of this field, the free electrons present in the conductor begin to move with a drift velocity, vd towards the left hand side.
Therefore,
Time taken by the free electrons to cross the conductor, t = I /Vd
Hence, current, I =
where n is the number of electrons per unit volume.
The slope of the given graph gives us the inverse of resistance. Resistance of a material increases with increasing temperature because the collision between the molecules increases.
In the graph given, T2 has a smaller slope and hence corresponds to higher resistance. Therefore, T2 > T1.
Using the formula,
Therefore,
5 = Ro ( 1+ 50)
6 = Ro ( 1+ 100 )
On solving the above two equations we get,
Now, putting the value of in one of the equations above , we get
5= RO [1+ 50 ] = 4 .
Therefore, the value of resistance at 0o C is 4Ω .
Kirchoff's rule states that :
a) Junction rule : At any junction, the sum of currents entering the junction is equal to the sum of currents leaving the junction.
b) Loop rule : The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero.
Given, the metallic wire is stretched such that, it's length increases to 5%.
Percentage change in length, dl/l = 5%
Resistance of wire is given by R = =
Therefore, percentage change in the resistance of wire is given by,
Resistance is given by R=
The length and resistance of the wires are equal.
Resistivity of Copper wire is less than that of the resistivity of magnanin wire. Therefore, the area of cross- section of copper wire is less than that of magnanin.
Hence, Manganin wire is thicker.
Wheatstone bridge principle states that if four resistances P, Q, R and S are arranged to form a bridge as shown in the fig. below with a cell E and one way key, put between the points A and C and a galvanometer is connected in between the points B and D such that there is no current through G. The balance point is obtained when, galvanometer shows no deflection.
In this case,
It is usually used to find the value of an unknown resistance.
Given,
Length of the potentiometer wire = l
Internal resistance of the wire = r
Emf of the driving cell =E
External resistance of the cell = R
Current drawn by the wire, I=
Now, using Ohm's law V =IR
i.e, V =
Now, potential gradient of the wire is given by
k = V/l
=
The slope of a V-I graph represents resistance. And, the resistance is higher for series combination as compared to that of a parallel combination. In the figure given, we observe that B has higher slope and hence higher resistance. Therefore, A represents the parallel combination.
When switch S is closed,
8 and 12 on the left side of switch S are in parallel. Also, on the right os s are in parallel combination.
Therefore,
Equivalent resistance of each combination =
These two resistance will be in series with each other.
Emf of the circuit = 4 V
Resistors 1, 5 are in series.
equivalent resistance, R1 = 1+5+4 = 10
Now,
10 , 12 are in parallel.
Effective resistance =
5.45 and 2 are also in parallel.
Therefore,
Using ohm's law,
V = IR
This implies,
Current drawn from the circuit, I =
Given,
Length of the potentiometer wire, l= 10 m
Resistance of the potentiometer wire, R = 4Ω
Emf of the accumulator connected in series with the wire, E = 2V
A resistance box in series is connected to the potentiometer wire.
Resistance of the resistance box - ?
i) Required potential gradient, k = 0.1 V/m
Potential drop along the potentiometer wire, V=k.l
=0.1 = 1 V
Therefore, current through the potentiometer wire,
I=
Let, R' be the resistace of the resistance box then,
Using the relation,
Thus, a resiatance of 4Ω has to be introduced in the resistance box.
Given,
Length of the potentiometer wire, l = 100 cm
Resistance of the wire, R = 10 Ω
Emf of the cell, connected in series with it, E = 2V
Balancing length is obtained at 40 cm
Let, I be the current through the potentiometer wire, then
.... (1)
Now, the resistance of the 40 cm length of potentiometer wire is,
R =
Emf of the cell balanced by a length of the wire, E' = 10 mV= 10 10-3 V.
Therefore, using V=IR we have,
Now, putting the value of I obtained in equation 1 we get,
Thus, the value of external resistance is R' = 790Ω .
A potentiometer being a null device, does not draw current from the balance point at the balance point. Therefore, potentiometer measures the actual emf of the cell. Whereas, a voltmeter always draws current from the cell and measures the terminal voltage of the cell instead of the actual emf of the cell .
Let, the drift velocities of the wires be vd and vd' respectively.
Area of cross- section of two wires be A and A'.
Then,
A : A' = 1:2
This implies,
A' = 2A
i) When the two wires are connected in series, the same amount of current flows through both the wires.
Implies,
n A vd e = n A' vd' e
where,
n is the number of electrons per unit volume and,
e is the charge on electron.
Therefore,
ii) When the two wires are connected in parallel, the potential difference across the two wires would be the same.
Then,
where, is the resistivity of the two wires.
Therefore,
(i) Calculate the equivalent resistances of the given electrical network between points A and B.
(ii) Also calculate the current through CD and ACB, if a 10 V d.c. source is connected between A and B, and the value of R is assummed as 2 Ω.
The equivalent circuit of the given circuit is 
The above fig. is a balanced wheatstone bridge.
Resistance across arm CD is 0.
i) Now,
Resistance across arm ACB , R1 = R+R = 2R
Resistance across arm ADB , R2 = R+R = 2R
Effective resistance between AB is ,
Rp = R
ii) The points C and D are at the same potential. Therefore, the current across arm CD is zero.
Resistance across arm ACB = 2R = 2(2) = 4
Current flowing through ACB, I =
Using ohm's law we have,
V = IR
Drift velocity is given by, vd =
But, Electric field, E = V/l
Also, we have I=Anevd
Therefore, I =
i.e.,
On comparing this equation with
We can deduce an expression for resistivity which is given by,
where, is the relaxation time.
From the graph given,
When I = 0A we have emf = 6V.
Given, three identical cells are connected in series.
Therefore,
Emf across each cell = .
Resistivity of a material is given by,
R =
If, length and area is unity then,
R =
Hence, we can say resistivity of a material is equal to the resistance of the conductor having unit length and unit area of cross-section.
Resistivity is dependent on the temperature.
.... (1)
Graph showing the variation of resistivity with temperature.
is the resistivity at temperature T and is the resistivity at a temperature T0 . The mathematical relation (1) says that the the plot of should be a straight line. But, at temperature lower than 00 C , the graph deviates considerably from a straight line.
The formula for terminal potential is given by,
where,
V is the terminal voltage,
E is the emf,
R is the variable resistance and
r is the internal resistance.
when , we have
When two unknown resistances X and y are placed as shown in the fig. then, the null point D is obtained at a distance of 40 cm from A.
i.e., length, l = 40 cm
Therefore,
... (1)
Given that, the null point shifts by 10 cm when a resistance of 10 is connected in series with x.
Length l1 becomes (40+10) cm = 50 cm.
X+10 = Y ... (2)
Now, on solving equation (1) and (2) we get,
Resistance, X = 20
If, a resistance of 10 is connected in series with Y, new null point will be obtained at a distance of l2 from A .
i.e.,
On solving the above equation, we get
We are given E, rx and R in series.
The equivalent resistance of the circuit is given by
Requiv = R + rx
Current flowing through the circuit , I =
Hence,
Voltage drop across R is given by ohm's law,
V=IR
=
which is the required relation.
Given,
Emf of first cell = E1
Emf of second cell = E2
Internal resistances of two cell are r1 and r2 respectively.
Load resistance = R
Effective emf is given by, Eeff =
Effective internal resistance, reff =
Now, the total resistance of the circuit would be the sum of load resistance and effective internal resistance.
Rtotal = R + reff =
Current through the load will be the same as the current through the circuit.
Thus,
= =
which is the required expression for the current through the load.
Consider, a conductor of length l and area of cross-section A having n electrons per unit length, as shown in the figure. 
Volume of the conductor , V = Al
∴ Total number of electrons in the Conductor = Volume x electron density = Al x n .
If e is the charge of an electron, then total charge contained in the conductor, Q = en.Al
Electric field in the conductor when potential difference V is applied across the conductor, E= V/I
Under the influence of this field E, free electrons begin to drift in a direction opposite to that of the direction of field.
Time taken by electrons to cross-over the conductor is
where, vd is the drift velocity of electrons.
Therefore, current flowing through the conductor is given by
[ n, e A are all constant]
Thus, current density is proportional to drift velocity.
(i) State the principle of working of a meter bridge.
(ii) In a meter bridge balance point is found at a distance l1 with resistors R and S as shown in the figure.
When an unknown resistor X is connected in parallel with the resistor S, the balance point shift to a distance l2. Find the expression for X in terms of l1 l2 and S.
Principle of potentiometer: The potential drop along any length of the wire is directly proportional to that length. When a constant current flows through a wire of uniform cross-section and composition then,
V ∝ l
Comparison of emfs of two primary cells: The circuit diagram is shown in the figure.
When the key K is closed, a constant current flows the potentiometer wire. By closing key K1, the null point can be obtained for cell E1. The jockey is moved along the wire and adjusted till galvanometer shows no deflection.
Suppose AJ1 = l1 is the balancing length for cell E1.
Then E1 = k l1
where, k is the potential gradient.
Now, the null point is obtained for cell E2 by closing key K2.
Let AJ2 = l2 be the balancing length in this case.
Then E2 = kl2
Hence, by increasing the length of a potentiometer's wire the sensitivity of a potentiometer can be increased.
Power of the bulb,
Resistance across the bulb is then,
The current flowing through the lamp
Now,
Power of the lamp when it is connected to a 200 V line is
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Resistivity () of a material in terms of it's relaxation time () , number density (n), mass (m) and charge carriers (e) is given by
(i) Case of metals: As temperature increases, the thermal speed of electrons increases and consequently, free electrons collide more frequently with the positive metal ions thereby, decreasing the relaxation time τ. And, since resistivity has an inverse dependence on relaxation time, resistivity ρ of the metal increases.
(ii) Case of semiconductors: The relaxation time τ does not change with temperature in semiconductor. But, the number density (n) of free electrons increases exponentially with increasing temperature. As a result, the resistivity of semiconductor decreases exponentially with the increase in temperature.
Two cells of emf 1.5 V and 2 V and internal resistance 1 ohm and 2 ohm respectively are connected in parallel to pass a current in the same direction through an external resistance of 5 ohm.
(a) Draw the circuit diagram.
(b) Using Kirchhoff’s laws, calculate the current through each branch of the circuit and potential difference across the 5 ohm resistor.
(ii) Let I1, I2 and I1 + I2 be the currents flowing through the resistors r1, r2 and R respectively.
On applying Kirchhoff’s law to the closed circuit CBAFC, we have
– 2 I2 + 1 I1 = 2.0 – 1.5 = 0
2 I2 – I1 = 0.5 ...(i)
Again, on applying Kirchhoff’s law for closed circuit CFEDC, we have
– 1 I1 – 5 (I1 + I2) + 1.5 = 0
6 I1 + 5 I2 = 1.5 ...(ii)
On solving (i) and (ii), we get
Current through CF,
Current through BA ,
Potential difference across 5 Ω resistor, V= IR
Resistance of potentiometer wire PQ, r = 10 Ω
Length of the potentiometer wire, l = 1m
Current through the wire PQ is
Potential drop across the wire PQ = Vr = Irr
Vr = 2 x 10–3 x 10
= 0.02 V
Potential gradient (k) along the wire PQ is
Potential drop across the wire PS = Potential drop across XY
Thus, emf generated by thermocouple = 0.008 V.
The emf or potential difference between the terminals of the battery is (VP – VQ).
VPQ = Vp – VQ
= 2 + 0.5
= 2.5 V
For electrical network, Kirchhoff’s rules are as follows:
(i) Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction
∴ Σ I = 0
(ii) Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero.
∴ Σ IR + Σ E = 0
According to Kirchhoff’s rule, I1 + I2 = I3
In the given circuit,
Applying loop rule to both the lower and upper loops, we get
40 I3 + 20 I1 = 40 (In loop ABCF)
40 I3 + 20 I2 = 80 + 40 (In loop CDEF)
Adding these two equations, we get
Putting the value of I3 in the above equation to find the value of I1, we get
A potentiometer wire of length 1 m is connected to a driver cell of emf 3 V as shown in the figure. When a cell of 1.5 V emf is used in the secondary circuit, the balance point is found to be 60 cm. On replacing this cell and using a cell of unknown emf, the balance point shifts to 80 cm.
(i) Calculate unknown emf of the cell.
(ii) Explain with reason, whether the circuit works, if the driver cell is replaced with a cell of emf 1V.
(iii) Does the high resistance R, used in the secondary circuit affect the balance point? Justify your answer.
Given,
Length of the potentiometer wire, l = 1 m
Emf of the driver cell , E = 3 V
Emf of the cell in the secondary circuit, E = 1.5 V
Balance point obtained, l1 = 60 cm= 0.60 m
Now,
On replacing the cell by a cell of unknown emf, the balance point obtained is at, l2 = 80 cm = 0.80 m
(i) By using the formula,
Hence, the unknown emf is found to be 2V.
(ii) The circuit will not work.
Reason: Because there will be smaller fall of potential across the potentiometer wire as compared to the emf of the cell in secondary circuit to be determined. Hence, the balance point will not be obtained on the potentiometer wire.
Thus, the condition for obtaining the balance point is that, the emf of the driver cell should be greater than the emf of the cell to be determined.
(iii) High resistance R, used in the secondary circuit does not affect the balance point because, at balance point the galvanometer shows no deflection implying the absence of current in this condition.
Since, capacitor provides infinite resistance to direct current, hence no current flows in capacitor arm.
Therefore, total resistance of current carrying arms
R = 1.2 + 2.8 = 4 Ω
Hence current from battery will be
∴ Potential difference across A and B = IRp = 1.5 x 1.2 = 1.80 V
Current through 2 Ω resistor
Let us consider the extreme right square of the loop.
Resistance across EF = (r1+R+r2) and r3 in parallel
This value should be equal to R, so that by the repeated operation of this type, we will be left with only one square which will be the left extreme one and it will have a value R
i.e., 
Now, Substituting the numerical values
i.e.,
is the required value of R.
Let x be the number of cells in series in each row.
And, consider let there be y such rows in parallel.
Total number of cells = xy = 24
Emf of each cell = 1.5 V
Internal resisatnce of each cell, r= 2Ω
External resistance, R = 12Ω
Now, Resistance of each row in series = 2.x Ω.
Total internal Resistance due to all xy batteries = R
i.e.,
Total internal resistance (because there are y rows in parallel)
We know that the maximum amount of current passes through the circuit when, the internal resistance of the battery of cells equals the external resistance.
Thus,
But, given that
Hence,
i.e., there should be two rows connected parallely, with 12 cells in each row grouped in series.( Fig. below) 
Now, current flowing across the circuit is
I =
Since, there are two rows and the current passing through each arm must be equal therefore,
Current through each arm, I
Therefore, current through each cell = 0.375 A.
And , the potential difference across the external resistance is, R= 12 x 0.75 = 9 V
Applying Kirchhoff’s second law for mesh ABCD,
I1P – I2R = 0
or, I1P = I2R ...(i)
For mesh BCDB,
I1Q – I2S = 0
or, I1Q = I2S ...(ii)
Dividing (i) by (ii), we get
This is the balanced condition of the Wheatstone bridge.
Measurement of specific resistance:
Slide wire or meter bridge is a practical form of Wheatstone bridge.
In the figure above, X is an unknown resistor and R.B is resistance box. After inserting the key k (circuit is closed), jockey is moved along the wire AC till galvanometer shows no deflection (point B).
If k is the resistance per unit length of wire AC.
then, P = resistance of AB = kl
Q = resistance of BC = k(100 - l)
or,
Hence, we can find the value of the unknown resistance.
Resistivity is derived as follows:
If 'r' is the radius of wire and l be its length, then its resistivity will be
Precautions:
(i) The null point should lie in the middle of the wire.
(ii) The current should not be allowed to flow in the wire for a long time or else the wire will get damaged.
Let,
l1, l2, and l3 be lengths of three copper wires.
D1, D2 and D3 be their diameters and,
A1, A2, A3 be their area of cross section.
Given,
l1: l2: l3 = 2: 3: 4
i.e., l1 = 2 l, l2 = 3 l and l3 = 4 l.
Also, given,
D1: D2: D3 = 4: 5: 6
∴ A1: A2:A3 = (4)2: (5)2: (6)2 = 16: 25: 36
That is,
A1 = 16 A, A2 = 25 A and A3 = 36 A
If ρ is the resistivity of copper, then
and,
or
Let I1, I2 and I3 be the currents through the wires of resistances R1, R2 and R3 respectively. Then,
...(i)
and,
and,
Putting these values in equation (i) we get,
I1 + 1.04 I1 + 1.125 I1 = 5
on solving, we get
I1 = 1.58 A
I2 = 1.04 x 1.58 = 1.64 A
and,
I3 = 1.125 x 1.58 = 1.78 A.
An infinite ladder network of resistances is constructed with 1 ohm and 2 ohm resistors as shown in figure below. The 6 volt battery between A and B has negligible internal resistance.(i) Show that effective resistance between A and B is 2 ohms.
(ii) What is the current that passes through the 2 ohm resistance nearest to the battery?
State the working principle of a potentiometer with the help of a circuit diagram. Describe a method to find the internal resistance of a primary cell.
In a potentiometer arrangement, a cell of emf 1.20 volt gives a balance point at 30 cm length of the wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emfs of the two cells is 1.5. Calculate the difference in the balancing length of the potentiometer wire in the two cases.
Difference in the balancing length of potentiometer wire is ,
l1 – l2 = 30 – 20
= 10 cm.
(c) The voltage VAB between A and B is the product of total current between A and B and the equivalent resistance between A and B.
∴ VAB = 2 x 2 = 4V
Similarly,
Voltage drop across B and C, VBC = 2 x 1 = 2V
Voltage drop across C and D, VCD = 2 x 4 =8V
Note that the terminal voltage is 14V. The loss of 2V is due to internal resistance of battery.
Let,
In parallel combination, the combined emf is Eeq and,
Combined internal resistance be req .
and
In series combination,
Let Eeq and req respectively be the equivalent emf and resistance in series combination,
then,
Eeq = E1 + E2 and,
req = r1 + r2
Numerical:
Given,
Emf across cell 1, E1 = 1V
Emf across cell 2, E2 = 2 V
Internal resistance of cell 1, r1 = 2 Ω
Internal resistance of cell 2, r2 = 1Ω
Let the external resistance be R
In series combination,
Now, in parallel combination
Now since, IP = IS we have,
In series combination,
and In parallel combination,
Therefore, heat generated in the series combination will be more than the heat generated in parallel combination.
Eight identical resistors 'r' are connected along the edges of a pyramid and we have to calculate the equivalent resistance.
(i) Without using Kirchhoff’s laws:
Consider a battery connected between A and B. The circuit now has a plane of symmetry. This plane of symmetry passes through the mid-points of AB and CD and the vertex O.
So, the currents are same in (i) AO and OB (ii) DO and OC.
Now, OA and OB can be treated as a series combination which gives resistance 2r.
Also, DO and OC are in series combination giving resistance 2r.
It is in parallel with DC.
This gives a resistance of =
This is in series with resistances AD and CB.
Hence, their combined resistance is =
Now resistor AB (r) and combination of AO and BO (i.e., 2r) are in parallel.
If R is the equivalent resistance, then
(ii) Using Kirchhoff’s laws:
Using Kirchhoff’s second law in loop DOCD, we get
– I3r – I1r + (I2 – I3)r = 0
– 3I3r + I2r = 0
...(i)
Again, using Kirchhoff’s loop law to loop AOBA, we get
– I1r – I2r + (I – I1 – I2) r = 0
3 I1 + I2 = I ...(ii)
Considering loop ADCBA, we get
Using equation (i), we get
Using equation (ii), we get
From equation (ii),
Considering circuit ABEA,
E = 2 l1r ...(iii)
If R is the total resistance, then E = I R
i.e.,
which is the required equivalent resistance using Kirchoff's law.
(a) Find the emf E1 and E2 in the circuit of the following diagram and the potential difference between the points a and b.
(b) If in the above circuit, the polarity of the battery E1, be reversed, what will be the potential difference between a and b?
(a)From the fig. below it is evident that 1 Ampere of current flows in the circuit from B to A.
On applying Kirchhoff’s law to the loop PAQBP,
20 – E2 = (12 x 1) + (1 x 2) + (2 x 2) = 18
Hence, E2 = 2 V .
Thus the potential difference between the points A and B is,
VAB = 18 – 1 – 4 = 13 V.
(b) On reversing the polarity of the battery E1, the current distributions will be changed.
Let the currents be I1 and I2 as shown in the following figure.
Applying Kirchhoff’s law for the loop PABP,
20 + E1 = (6 + 1) I1 – (4 + 1) I2
38 = 7 I1 – 5 I2 ...(i)
Similarly for the loop ABQA,
4I2 + I2 + 18 + 2 (I1 + I2) + (I1 + I2) + 7 = 0
3 I1 + 8 I2 = – 25 ...(ii)
Solving equation (i) and (ii) for I1 and I2, we get
I1 = 2.52 A and I2 = – 4.07 A
Hence,
Vab = – 5 x (4.07) + 18
= – 20.35 + 18
= – 2.35 V.
Let us assume that, m cells are connected correctly and n cells are connected wrongly.
Then, we have
m + n = 12
If, E is the emf of each cell, the total emf of the battery is (m – n) E.
When the battery and the cells add each other, the net emf is = (m – n) E + 2 E
If R is the total resistance of the circuit, the current is given by
I = ...(1)
When the battery and the cells oppose each other, the net emf is, (m – n) E – 2 E.
Then, the current across them is given by,
I= ...(2)
Dividing equation (1) by equation (2) gives us,
m – n = 10
But, m + n = 12
Hence m = 11 and n = 1.
Thus, one cell is wrongly connected.
Given, voltmetres V1 and V2 are connected in series across a D.C line.
Potential across V1 = 80 V
Per volt resistance, R= 200 Ω
The resistance R1 of voltmeter V1 is given by
R1 = 80 x 200 = 16000 Ω = 16 k Ω
Current in the circuit,
Reading of voltmeter, V2= IR2
= (5 x 10–3) x (32 x 103)
= 160 V.
∴ Line voltage, V = V1 + V2 = 80 + 160 = 240 V.
The distribution of current in the circuit will be as shown in fig. below, following Kirchhoff’s first law.
Here point F is not a true junction, hence is shown separate.
If R’ is the effective resistance of circuit between A and B, then
E = IR’ ...(i)
In a closed circuit EABE
E = (I – I1) R + (I – I1) R
= 2 (I – I1) R ... (ii)
In a closed circuit GDCG,
I2 R/2 + R I2/2 – (I1 – I2) R = 0
I2 = I1/2 ...(iii)
In a closed circuit AGCBA, we have
[from (iii)]
Putting this value in (ii) we get
...(iv)
Comparing (i) and (iv) we get
Given,
Current carried by the wire, I = 1.5 A
Potential difference applied across the wire, V = 2.1 V
Length of the wire, l = 3 m
Area of cross-section of the wire,
Conductance of the wire is G given by,
Conductivity is given by
According to Kirchhoff’s first law, at junction A,
i = i1 + i2 or i2 = i – i1 ...(i)
Let H be the heat produced in the circuit in t seconds, then
H = i12R1t + i22R2t
= i12R1t + (i – i1)2 R2t [from (i)]
In case the heat produced in the circuit is minimum, then
therefore,
2i1R1t + 2(i – i1) (–1) R2t = 0
2i1R1t – 2i2 R2t = 0
i1R1 – i2R2 = 0
which is according to Kirchhoff’s second law in a closed circuit ACDEFA.
Is current density a vector quantity or scalar quantity? Deduce the relation between current density and potential difference across a current carrying conductor of length l, area of cross-section A and number density of free electrons n. How does the current density in a conductor vary with
(a) increase in potential gradient,
(b) increase in temperature,
(c) increase in area of cross-section?
In the given Wheatstone bridge, the current 3R is zero. Find the value of R, if carbon resistor, connected in one arm of the bridge has the colour sequence of red, red and, orange.
The resistance of BC and CD arm are now interchanged and another carbon resistor is connected in place of R so that the current through arm BD is again zero. Write the sequence of colour bands of the carbon resistor. Also find the current through it.
For no current through BD, the Wheatstone bridge is balanced and the resistance of carbon resistor is R’ = R. = 22000 Ω.
When, resistance of BC and CD arms are interchanged and another carbon resistor is connected in place of R, the current through BD is again zero.
Therefore, again for balanced Wheatstone bridge
R’ = 4R
= 4 x 22000 Ω
= 88000 Ω
For 88000 Ω the sequence of colour bands is gray, gray and orange.
Now, net resistance of arm ADC,
= 88000 + 44000
= 132000 Ω
∴ current through carbon resistor,
Applying Kirchhoff's rule for loop ABCDA,
...(i)
For loop DEFAD
...(ii)
Eqs. (i) and (ii) can be written as
and
On solving the above equatuions, we get
Putting the value I2 in equation (i)
Current in R,
Rate of energy dissipation = 2R
(a) Given,
Radius of capacitor plates, r = 12 cm = 0.12 m
distance between the plates, d = 5.0 mm = 5 × 10-3m
Charge carried, I = 0.15 A
Permittivity of medium, 0 = 8.85 × 10-12 C2 N-1 m2
∴ Area of cross-section of plates, A = R2 = 3.14 × (0.12)2 m2
Capacitance of parallel plate capacitor is given by
Now, charge on capacitor plate,
(b) Displacement current is equal to the conduction current i.e., 0.15 A.
(c) Yes, Kirchhoff's first rule is valid at each plate of the capacitor provided. We take the current to be the sum of the conduction and displacement currents.
A parallel plate capacitor (Fig.) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s-1.
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
Given, a parallel plate capacitor made of circular plates.
(a) Here,
Radius of the circular plate, R = 6.0 cm
Therefore,
(b) Yes, the conduction current is equal to the displacement current.
(c) We know that,
This formula goes through even if displacement current, ID (and therefore magnetic field B) oscillates in time. The formula above shows that they oscillate in phase.
Since ID = I, we have
If I = I0, the maximum value of current, then
Amplitude of B = maximum value of B
A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Given,
Frequency band of radio = 7.5 MHz - 12 MHz
Therefore, the corresponding wavelength band is 40 m - 25 m.
Given,
Speed of EM wave,
For electromagnetic waves,
i.e.,
which is the required amplitude of electric field.
Given,
Amplitude of EM waves, Eo = 120 N/C
Frequency of wave, v = 50.0 MHz
(a) Using the relation, we have
which is the required amplitude of magnetic field.
Angular frequency is given by,
Wave vector k is given by,
Wavelength of the EM wave is,
(b) Let the electromagnetic wave travel along +x-axis, and are along y-axis and z-axis respectivelty.
Then,
and,
The above expressions represent electric and magnetic field.
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m-1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field, [c = 3 × 108 m s-1]
Here,
Therefore,
Wavelength of the wave,
Amplitude of oscillating magnetic field,
Energy density in electric field,
Energy density in magnetic field,
Using the relation, we have
But,
Hence, the average energy density of electric field equals the average energy density of magnetic field.
Given,
Electric field, E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]}
The wave is propagating along negative y direction i.e., along -
Given,
Electric field, E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]}
Comparing the given equation with the equation
a
Therefore, we get
.
We know that,
which is the required value of wavelength.
Electric field , E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]}
Frequency of the wave,
electric field , E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]} .
We know that
is the required amplitude of magnetic field.
Electric field E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]
Expression of magnetic field part of the wave
E is along and c is along ,
c is the direction of
Clearly B is in the direction of and
Thus,
is completely represented as,
Answer the following questions:
Long distance radio broadcasts use short-wave bands. Why?
Answer the following questions:
It is necessary to use satellites for long distance TV transmission. Why?
Answer the following questions:
Optical and radiotelescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?
X-rays are absorbed by the atmosphere whereas, visible and radiowaves can penetrate it. Hence, optical and radiowaves can work on the surface of earh but, x-ray astronomical telescopes must be used on satellites orbiting the earth.
Answer the following questions:
The small ozone layer on top of the stratosphere is crucial for human survival. Why?
Answer the following questions:
If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
Answer the following questions:
Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter' with a devastating effect of life on earth. What might be the basis of this prediction?
(i) Microwave
(ii) Gamma rays
(a) Range of microwaves vary from 0.3 to 10-3 m
(b) Range of X-rays vary from 3 × 10-8 to 1 × 10-13 m
(i) Microwave
(ii) Gamma rays
Oscillating electric field is given by,
On comparing the above equation with we get,
which is the required value of the wavelength.
Oscillating electric field,
Here, EO = 30
Hence, Bo = Eo/c =
Oscillating magnetic field is given by,
Given, a plane electromagnetic wave travelling in y-direction.
(i) Ratio of the magnitude of electric to magnetic field is,
where, c is the speed of light.
(ii) For an electromagnetic wave travelling along y-direction its electric and magnetic field vectors are along x-axis and z-axis respectively.
TV signals have frequencies of 100-200 MHz, which penetrate ionosphere (frequencies > 30 MHz cannot be used), hence sky wave propagation cannot be used for TV transmission.
Two methods by which range of TV transmission can be increased:
(a) use of tall antenna
(b) use of repeaters between transmitters and receivers (line-of-sight transmission).
What is the wavelength of a photon whose energy is 1 eV? In which part of the electromagnetic spectrum is it?
Energy of photon, E = 1 eV
Using the formula,
Energy
This is Infrared rays.
Application of X-rays are:
(i) for the detection of explosives, opium and gold in the body of the smugglers.
(ii) in detecting fractures, diseased organs in human body.
The following table gives the wavelength range of some constituents of this electromagnetic spectrum:
|
S. No. |
Wavelength range |
|
1 |
1 mm to 700 nm |
|
2 |
400 nm to 1 nm |
|
3 |
1 nm to 10-3 nm |
|
4 |
< 10-3 nm |
Select the wavelength range and name the electromagnetic waves that are:
(i) widely used in the remote switches of household electronic devices.
(ii) produced in nuclear reactions.
(i) Infrared waves are used in the remote switches of household electronic devices and their wavelength vary from 1 mm to 700 nm.
(ii) Gamma rays are produced in nuclear reaction and it's wavelength range from 10-10 to 10-14 m.
Only accelerated charge can produce electromagnetic waves. Therefore, charge moving in a circular orbit will produce EM waves as circular motion is accelerated motion.
(i) γ-rays
(ii) Radio waves
The ratio of their velocities in glass will be the same and is given by,
Write any four characteristics of electromagnetic waves. Give two uses of (i) radio-waves (ii) Microwaves.
Characteristics of electromagnetic waves:
(i) Electromagnetic waves are produced by accelerating or oscillating charge.
(ii) E.M. waves do not require any material medium for their propagation.
(iii) E.M. waves travel in free space with a velocity which is equal to the velocity of light (c = 3 × 108 m/s).
(iv) E.M. waves are transverse in nature.
Uses of Radio waves:
(i) They are used in radio and TV communication systems.
(ii) Cellular phones use radio waves to transmit voice communication in the ultrahigh frequency (UHF) band.
Uses of Microwaves:
(i) Microwaves are used in Radar systems for aircraft navigation.
(ii) Microwave ovens are used for cooking purposes.
Wavelength of mercury light = 5.5 × 10-5 cm
Frequency,
Time period,
Transverse nature of electromagnetic waves means the electric and magnetic fields in an electromagnetic wave are perpendicular to each other and to the direction of propagation.
Below, is the diagram showing the propagation of an electromagnetic wave along x-direction. 
Given,
Speed of electromagnetic waves, v = 2 108 m/s
Relative permeability of the medium = 1
Speed of light,
Using the formula of speed of electromagnetic wave in a medium,
Therefore,
Relative permittivity,
A parallel-plate capacitor with rectangular plates is being discharged. Consider a rectangular loop centred on the plates and between them. The loop measures a by 2a, the plate measures 2a by 4a. What fraction of the displacement current is encircled by the loop?
Given, a plane electromagnetic wave propagating in the x-direction.
Wavelength of EM wave = 10 nm
Direction of electric field = y direction.
Magnitude of electric field = 60 V/m
Thus, equations for electric and magnetic field is given by,
Using this, a relation between their magnitude is derived as,
Now, on substituting values , we get
which is the required maximum value of magnetic field.
The intensity of the sunlight reaching the earth is 1380 Wm-2. Calculate the amplitudes of electric and magnetic field in the light wave. Assume the light to be a plane monochromatic wave.
Two students A and B prepare the following Table about the electromagnetic waves. Rewrite the table in its corrected form.
|
Direction of |
Peak Value of |
||||
|
Student |
Electric field |
Magnetic field |
Propagation |
Electric field |
Magnetic field |
|
A |
Along X-axis |
Along Y-axis |
Along Y-axis |
E |
B = CE |
|
B |
Along Y-axis |
Along Z-axis |
Along X-axis |
E = CB |
B |
The velocity of propagation (in vacuum) and the frequency of (i) X-rays and (ii) radio waves are denoted by (v1, n1) and (v2, n2) respectively. How do the value of (a) v1 and v2(b) n1 and n2 compare with each other?
Velocity and frequency of X-rays = (v1 , n1)
Velocity and frequency of radio waves = (v2 , n2 )
(a)
(b)
The above expression compares the values of v1, v2, n1 and n2
Name the electromagnetic waves used for the following and arrange them in increasing order of their penetrating power.
(a) Water purification
(b) Rewrite sensing
(c) Treatment of cancer.
(a) Ultra violet waves are required for water purification.
(b) Micro waves are used for rewrite sensing.
(c) Gamma rays are used for the treatment of cancer.
In increasing order of penetration powers:
microwaves < ultra violet waves < gamma rays
Given, a plane electromagnetic wave of angular frequency is propagating with velocity c along the Z- axis.
Thus,
Equations for oscillating and electric magnetic field is given by,
and
where, is the wave vector.
The figure given below shows these fields diagramatically. 
D.
None of theseA.
yellow light is scattered less by the fog particlesThe scientists who are connected with the history of electeromagnetic waves are Maxwell, Hertz, Bose and Marconi.
The velocity of all light waves in vacuum is same (c= 3 108 m/s ). Therefore, the ratio of speed of infrared and ultraviolet rays is 1:1 .
For frequency value higher than 1500 kHz, the absorption of signals by the ground increases and, transmission is disrupted.
Since, microwaves have smaller wavelength as compared to radiowaves they can be transmitted as a beam signal in a particular direction much better than radiowaves. Microwaves do not bend around the corners of any obstacle in between their path.
Microwaves have wavelength in the range of 10-2 m.
Microwaves are used in :
i) RADAR
ii) microwave oven
The electromagnetic waves produced are x-rays.
Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance?
Internal resistance of the circuit is negligible.
So, total resistance is R.
Therefore, current across the circuit is given by, 
Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic radiations. Name the radiations and write the range of their frequency.
The radiations are ultraviolet radiations.
The frequency range of UV rays is 1015 –1017Hz.
Explain the term ‘drift velocity’ of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of ‘drift velocity’.
The velocity gained by the accelerating electrons in uniform electric field inside the conductor is drift velocity. The average velocity, acquired by free electrons along the length of a metallic conductor, due to existing electric field is called drift velocity.
Let ‘n’ be the number density of free electrons in a conductor of length ‘l’ and area of cross-section ‘A’.
Total charge in the conductor, Q = Ne
= (nAl)e
Time taken, t is given by, 
Therefore, the current flowing across the conductor is given by, 
That is, 
, which is the amount of current flowing through a conductor in terms of drift velocity.
Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell.
Potentiometer can be used to measure the internal resistance of the cell.
A cell of emf E is connected across the resistance box through key K1.
When key K1 is opened galvanometer shows deflection at the balancing length l1.
So, E = k
If both keys are closed, then balancing point is obtained at length l2 (l2 < l1).
So, V = k
Now, using the relation, 
Therefore, we have
Define the current sensitivity of a galvanometer. Write its S.I. unit. Figure shows two circuits each having a galvanometer and a battery of 3 V. When the galvanometers in each arrangement do not show any deflection, obtain the ratio R1 / R2. 
Current sensitivity is defined as the ratio of deflection produced in the galvanometer to the current flowing through it.
Mathematically it is given as, 
SI unit is radian per ampere.
No current flows through the galvanometer for balanced Wheatstone bridge.
Graph showing the variation of current versus voltage for a material GaAs is shown in the figure. Identify the region of
i) negative resistance
ii) where Ohm’s law is obeyed.
i) The region of negative resistance is DE because, the slope is negative for this part of curve.
ii) BC is the part of the curve where Ohm’s law is obeyed because here, current is varying linearly with with the voltage. This gives us direct proportionality between current and voltage.
A Wheatstone bridge arrangement is shown as below: 
Using Kirchoff’s second law to the loop ABDA, we get 
Applying Kirchoff’s law to loop BCDB, we get
When the bridge is balanced, 
Then, the equations can be written as,
... (1) 
... (2)
On dividing equation (1) by (2), we get
, which is the balanced condition of a Wheatstone bridge.
Name the parts of the electromagnetic spectrum which is
(a) suitable for radar systems used in aircraft navigation.
(b) used to treat muscular strain.
(c) used as a diagnostic tool in medicine.
Write in brief, how these waves can be produced.A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus (i) R and (ii) the current I.
It is found that when R = 4
, the current is 1 A and when R is increased to 9 
, the current reduces to 0.5 A. Find the values of the emf E and internal resistance r.
Plot for V vs. R 
Plot for V vs. I
Current in the circuit is given by,
Define the term 'Mobility' of charge carriers in a conductor. Write its S.I. unit
Drift velocity per unit applied electric field is known as the mobility of charge carriers in a conductor.
Mathematically,
Mobility, 
S.I. unit of mobility: 
or 
Show variation of resistivity of copper as a function of temperature in a graph.
The graph below shows variation of resistivity of copper with temperature. The graph is parabolic in nature.
'
State Kirchhoff's rules. Explain briefly how these rules are justified.
Kirchhoff’s First Law or Junction Rule states that “The sum of the currents flowing towards a junction is equal to the sum of currents leaving the junction.”
This is in accordance with the conservation of charge which is the basis of Kirchhoff’s current rule. 
Here, I1, I2 I3, and I4 are the currents flowing through the respective wires.
Convention: The current flowing towards the junction is taken as positive and the current flowing away from the junction is taken as negative.
I3 + (− I1) + (− I2) + (− I4) = 0
Kirchhoff’s Second Law or Loop Rule states that In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistances and the currents flowing through them.
OR
“The algebraic sum of all the potential drops and emfs along any closed path in a network is zero.”
For the closed loop BACB:
E1 − E2 = I1R1 + I2R2 − I3R3
For the closed loop CADC:
E2 = I3R3 + I4R4 + I5R5
This law is based on the law of conservation of energy.
A potentiometer wire of length 1 m has a resistance of 10 Ω. It is connected to a 6 V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives a balance point at 40 cm.
Given,
Total length of the potentiometer wire, L = 1m
Resistance of the wire, R = 10 Ω
Voltage of the battery = 6 V
Resistance of the battery = 5 Ω
Therefore, total resistance of the circuit, R = (RAB + 5) Ω = 15 Ω
Using the figure given above, we have
Current in the circuit, I = V R = 6 x 15 A
Therefore,
Voltage across AB, VAB = i.RAB = 4 V
Emf of the cell, e = 
... (1)
Here,
Balance point is obtained at, l = 40 cm
Total length, AB = L = 1 m = 100 cm
Putting the values in equation (1), we have
Answer the following:
(a) Name the EM waves which are suitable for radar systems used in aircraft navigation. Write the range of frequency of these waves.
(b) If the earth did not have atmosphere, would its average surface temperature be higher or lower than what it is now? Explain.
(c) An EM wave exerts pressure on the surface on which it is incident. Justify.(a) Microwaves are suitable for radar systems used in aircraft navigation. The range of frequency for these waves is 109 Hz to 1012 Hz.
(b) There would be no greenhouse effect on the surface of the Earth in the absence of atmosphere. As a result, the temperature of the Earth would decrease rapidly, making it difficult for human survival.
(c) When the wave is incident on the metal surface, it is completely absorbed. Energy U and hence momentum (p = 
is delivered to the surface of the earth. The momentum delivered becomes twice when, the wave is totally reflected because momentum is changed from p to –p. Thus, force and thereby pressure is exerted on the surface of the earth by EM waves.
Show on a graph, the variation of resistivity with temperature for a typical semiconductor.
For a semiconductor, resistivity decreases rapidly with increase in temperature.
Name the physical quantity which remains same for microwaves of wavelength 1 mm and UV radiations of 1600 Å in vacuum.
Both microwaves and UV radiations are EM waves. Therefore, their speed ( c = 
) will remain same in vacuum.
A cell of emf E and internal resistance r is connected to two external resistances R1 and R2 and a perfect ammeter. The current in the circuit is measured in four different situations:
(i) without any external resistance in the circuit.
(ii) with resistance R1 only
(iii) with R1 and R2 in series combination
(iv) with R1 and R2 in parallel combination.
The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in that order. Identify the currents corresponding to the four cases mentioned above.The current in the circuit to corresponding situations is given by:
i) When there is no external resistance in the circuit,
The current in this case will be maximum because effective resistance is minimum. So, I1 = 4.2 A
ii) In the presence of resistance R1 only, we have
Here, effective resistance is more than (i) and (iv) but less than (iii).
So, I2 = 1.05 A
iii) When R1 and R2 are in series combination, we have
In this case, effective resistance is maximum, so current is minimum.
Thus, I3 = 0.42 A
iv) When R1 and R2 are in parallel combination, we have
Here, effective resistance is more than (i) but less than (ii) and (iii).
So, I4 = 1.4 A
Using Kirchhoff’s rules determine the value of unknown resistance R in the circuit so that no current flows through 4 resistance. Also find the potential difference between A and D.
Applying Kirchhoff's loop rule for loop ABEFA,
–9 + 6 + 4 × 0 + 2I = 0
I = 1.5 A … (i)
For loop BCDEB,
3 + IR + 4 × 0 – 6 = 0
Therefore, IR = 3
Putting the value of I from equation (i), we have
Potential difference between A and D through path ABCD,
A resistance R is connected across a cell of emf 
and internal resistance r. A potentiometer now measures the potential difference between the terminals of the cell as V. Write the expression for 'r' in terms of 
, V and R.
The expression for internal resistance is given by,
How are infrared waves produced? Why are these referred to as 'heat waves’? Write their one important use.
Infrared waves are produced by hot bodies and molecules.
They are referred as heat waves because they are readily absorbed by water molecules in most materials, which increase their thermal motion, so they heat up the material.
Infrared waves are used for therapeutic purpose and long distance photography.
Define the terms (i) drift velocity, (ii) relaxation time.
A conductor of length L is connected to a dc source of emf ε. If this conductor is replaced by another conductor of same material and same area of cross-section but of length 3L, how will the drift velocity change?i) Drift velocity: The electrons drift towards the direction of positive potential, whenever a potential difference is applied across the conductor. The small average velocity of free electrons along the direction of positive potential is called the drift velocity.
Drift velocity is denoted by vd.
ii) Relaxation time: The time for which an electron moves freely between two successive collisions of electron with lattice ions/atoms is called the relaxation time.
Drift velocity is given by, 
where L is the length of the conductor,
is the emf of the DC source,
L is the length of the conductor.
When, the conductor is replaced by another conductor of length 3L, drift- velocity will become one-third because both of them are inversely proportional to each other.
In the circuit shown, R1 = 4 Ω, R2 = R3 = 15 Ω, R4 = 30 Ω and E = 10 V. Calculate the equivalent resistance of the circuit and the current in each resistor.
Given, R1 = 4 Ω, R2 = R3 = 15 Ω, R4 = 30 Ω and E = 10 V
Here, R2 , R3 and R4 are connected parallel to each other.
Therefore, equivalent resistance is given by,
Now, R1 is in series with R. So, equivalent resistance is given by, 
Current I1 is given by, 
... (1)
This current is divided at A into three parts I2, I3, I4 .
... (2)
Also, 
Now, putting the values of 
and 
in (2), we get
Thus, I1 = 1A, I2 = I3 = 0.4 A and I4 = 0.2 A.
A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38 
as shown in the figure. Find the value of the current in circuit. 
Since the cells are in opposition,
E net = E1 – E2 = (200-10) = 190 V
So, current in the circuit is given by, I = 
The emf of a cell is always greater than its terminal voltage. Why? Give reason.
V= E – I r;
The emf of a cell is greater than terminal voltage because there is some potential drop across the cell due to its small internal resistance.
(a) State the working principle of a potentiometer. With the help of the circuit diagram; explain how a potentiometer is used to compare the emf’s of two primary cells. Obtain the required expression used for comparing the emf’s.
(b) Write two possible causes for one sided deflection in a potentiometer experiment.
OR
(a) State Kirchhoff’s rules for an electric network. Using Kirchhoff’s rules, obtain the balance condition in terms of the resistances of four arms of Wheatstone bridge.
(b) In the meter bridge experimental set up, shown in the figure, the null point ’D’ is obtained at a distance of 40 cm from end A of the meter bridge wire. If a resistance of 10W is connected in series with R1, null point is obtained at AD=60 cm. Calculated the values of R1 and R2.
(a) Working Principle of Potentiometer
Principle:
Consider a long resistance wire AB of uniform cross-section. It’s one end A is connected to the positive terminal of battery B1 whose negative terminal is connected to the other end B of the wire through key K and a rheostat (Rh).
The battery B1 connected in circuit is called the driver battery and this circuit is called the primary circuit. By the help of this circuit a definite potential difference is applied across the wire AB; the potential falls continuously along the wire from A to B. The fall of potential per unit length of wire is called the potential gradient. It is denoted by ‘k’.
A cell e is connected such that it’s positive terminal is connected to end A and the negative terminal to a jockey J through the galvanometer G. This circuit is called the secondary circuit. In primary circuit the rheostat (Rh) is so adjusted that the deflection in galvanometer is on one side when jockey is touched on wire at point A and on the other side when jockey is touched on wire at point B.
The jockey is moved and touched to the potentiometer wire and the position is found where galvanometer gives no deflection. Such a point P is called null deflection point. VAB is the potential difference between points A and B and L meter be the length of wire, then the potential gradient is given by,
If the length of wire AP in the null deflection position be l, then the potential difference
between points A and P,
VAP = kl
Therefore, Emf of the cell = VAP= kl
In this way the emf of a cell may be determined by a potentiometer.
Comparison of two emfs’ of a cell:
First of all the ends of potentiometer are connected to a battery B1, key K and rheostat Rh such that the positive terminal of battery B1 is connected to end A of the wire. This completes the primary circuit.
Now the positive terminals of the cells C1 and C2 whose emfs’ are to be compared are connected to A and the negative terminals to the jockey J through a two-way key and a galvanometer (fig). This is the secondary circuit.
Distance if P1 from A is l1.
So, AP1 = l1
Emf of cell C1 is given by, 
1 = kl1 … (1)
Now, plug is taken from terminals 1 and 3 and inserted between the terminals 2 and 3 to bring cell C2 in the circuit. Null deflection point is obtained at P2.
Distance from P2 to A is l2.
Emf of cell C2, 
2 = kl2 … (2)
Now, dividing equation (1) by (2), we get
Out of these cells if one is standard cell.
OR
Kirchhoff’s rule states that,
(i) At any junction, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction.
(ii) The algebraic sum of the charges in potential around any closed loop involving resistors and cells in the loop is zero.
Conditions of balance of a Wheatstone bridge:
P, Q, R and S are four resistance forming a closed bridge, called Wheatstone bridge.
A battery is connected across A and C, while a galvanometer is connected B and D. Current is absent in the galvanometer’s balance point.
Derivation of Formula: Let the current given by battery in the balanced position be I. This current on reaching point A is divided into two parts I1 and I2. At the balanced point, current is zero.
Applying Kirchhoff’s I law at point A,
I - I1 - I 2 = 0 or
I = I1 + I 2 ...(i)
Applying Kirchhoff’s II law to closed mesh ABDA,
- I1P + I 2R = 0 or
I1P = I 2 R ...(ii)
Applying Kirchhoff’s II law to mesh BCDB,
- I1Q + I 2S = 0 or
I1Q = I 2S ...(iii)
Dviding equation (ii) by (iii), we get,
; which is the required condition of balance for Wheatstone bridge.
b)
Distinguish between emf and terminal voltage of a cell.
|
Emf |
Terminal Voltage |
|
The maximum potential difference that can be delivered by a cell when no current flows through the circuit. |
Potential difference across the terminals of the load when the circuit is switched on and current flows through it. |
|
It is represented by E and remains constant for a cell. |
It is represented by V and depends on the internal resistance. |
In a meter bridge shown in the figure, the balance point is found to be 40 cm from end A. If a resistance of 10 is connected in series with R, balance point is obtained 60 cm from A. Calculate the values of R and S. 
For null point, balance length l1 = 40 cm
So, 
... (1)
If resistance 10 
is connected in series of R, balance point shifts towards AD = 60 cm.
From equations (1) and (2), we have
From equation (1), we have
State the underlying principle of a potentiometer. Write two factors by which current sensitivity of a potentiometer can be increased. Why is a potentiometer preferred over a voltmeter for measuring the emf of a cell?
Underlying principle of potentiometer: Potential difference across a uniform wire is directly proportional to the length of the part across which the potential is measured when, a steady current flow through the wire.
Sensitivity of potentiometer can be increased by:
(i) Increasing the length of the wire
(ii) Decreasing the current in the wire using a rheostat
Potentiometer is preferred over voltmeter because potentiometer used the null method. During the balanced condition of potentiometer, no current is drawn by the galvanometer. Voltmeter measures the voltage across the terminals of a cell when the cell is in closed circuit, that is, when current is flowing through the cell.
During a thunderstorm the 'live' wire of the transmission line fell down on the ground from the poles in the street. A group of boys, who passed through, noticed it and some of them wanted to place the wire by the side. As they were approaching the wire and trying to lift the cable, Anuj noticed it and immediately pushed them away, thus preventing them from touching the live wire. During pushing some of them got hurt. Anuj took them to a doctor to get them medical aid.
Based on the above paragraph, answer the following questions:
(a) Write the two values which Anuj displayed during the incident.
(b) Why is it that a bird can sit on a suspended 'live' wire without any harm whereas touching it on the ground can give a fatal shock?
(c) The electric power from a power plant is set up to a very high voltage before transmitting it to distant consumers. Explain, why.a) Anuj displayed concern for others lives, presence of mind and a selfless attitude.
b) When the bird perches on a live wire, body becomes charged for a moment and has a same voltage as the live wire. However, no current flows in its body. Body is a poor conductor of electricity as compared to the copper wire. So the electrons do not travel through the bird’s body.
On the other hand, if the bird touches the ground while being in contact with the high voltage live wire, then the electric circuit gets complete and high current flows through the body of the bird to the ground, giving it a fatal shock.
c) Inorder to reduce the loss of power transmission, the power plant is set up at high voltages. Power loss during the transmission is I2R. Therefore, by reducing the value of current, power loss can be minimized. So, the value of the voltage should be kept high.
a) State Kirchhoff's rules and explain on what basis they are justified.
(b) Two cells of emfs E1 and E2 and internal resistances r1 and r2 are connected in parallel.
Derive the expression for the
(i) Emf and
(ii) internal resistance of a single equivalent cell which can replace this combination.
OR
(a) "The outward electric flux due to charge +Q is independent of the shape and size of the surface which encloses is." Give two reasons to justify this statement.
(b) Two identical circular loops '1' and '2' of radius R each have linear charge densities -
and +
C/m respectively. The loops are placed coaxially with their centre distance apart. Find the:
a)
i) Junction Rule: The algebraic sum of currents meeting at a point in an electrical circuit is always zero.
This law is in accordance with law of conservation of charge.
ii) Loop Rule: In a closed loop, the algebraic sum of emfs is equal to the algebraic sum of the products of the resistances and the current flowing through them.
This law is based on the conservation of energy.
b) Consider the circuit, 
Here, E1 and E2 are the emf of two cells,
r1 and r2 are the internal resistance of cell,
I1 and I2 current due to two cells.
Terminal potential difference across the first cell is given by, 
For the second cell, terminal potential difference will be equal to that across the forst cell.
So, 
Let E be the effective emf and r the resultant internal resistance.
Consider, I as the current flowing through the cell.
Therefore, 
Now, comparing the equation with V = E –Ir, we have
OR
a) The outward electric flux due to the charge enclosed inside a surface is the number of electric field lines coming out of the surface. Outward flux is independent of the shape and size of the surface because:
i) Number of electric field lines coming out from a closed surface is dependent on charge which does not change with the shape and size of the conductor.
ii) Number of electric lines is independent of the position of the charge inside the closed surface.
b) Magnitude of electric field at any point on the axis of a uniformly charged loop is given by, 
Electric field at the centre of loop 1 due to charge present on loop 1 = 0
Electric field at the centre of loop1 due to charge present on loop 2 is given by,
, is the required electric field.
To which part of the electromagnetic spectrum does a wave of frequency 5 x 1019 Hz belong?
The part of the electromagnetic spectrum is X-rays.
A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable resistor ‘R’. Plot a graph showing variation of terminal voltage ‘V’ of the cell versus the current ‘I’. Using the plot, show how the emf of the cell and its internal resistance can be determined.
The relation between V and I is given by,
V = E – Ir
Thus, the graph between V and I is as shown below:
Emf is given by the intercept on the vertical axis i.e., the V axis.
Internal resistance is given by the slope of the line i.e.. slope of V vs. I graph.
Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 x 10–7 m2 carrying a current of 1.5 A. Assume the density of conduction electrons to be 9 x 1028 m–3.
Area of cross-section, A = 1.0 x 10–7 m2
Current = 1.5 A
Density of conduction electrons = 9 x 1028 m–3
Drift velocity of electrons is given by, 
(a) Why are the connections between the resistors in a meter bridge made of thick copper strips ?
(b) Why is it generally preferred to obtain the balance point in the middle of the meter bridge wire?
(c) Which material is used for the meter bridge wire and why?
OR
A resistance of R draws current from a potentiometer as shown in the figure. The potentiometer has a total resistance Ro
. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer. 
a) The resistivity of copper wire is very low. The connections between the resistors are made of thick wires so as to increase the rate of cross-section. Therefore, the resistance of wires is almost negligible.
b) Balance point is obtained in the middle of the meter bridge wire so as to increase the sensitivity of the meter bridge.
c) Constantan is used for meter bridge wire because its temperature coefficient of resistance is almost negligible due to which the resistance of the wire does not change with increase in temperature of the wire due to flow of current.
OR
Total resistance is given by, Rtot = 
= 
Total current through the device is given by, Itotal = V / Rtotal
Current through resistance R is given by, I2 = Itotal x 
Two wires of equal length, one of copper and the other of manganin have the same resistance. Which wire is thicker?
Resistance of a material is given by, 
Resistance for copper wire and manganin wire is given by, 
That is, manganin wire is thicker.
A capacitor of capacitance of ‘C’ is being charged by connecting it across a dc source along with an ammeter. Will the ammeter show a momentary deflection during the process of charging? If so, how would you explain this momentary deflection and the resulting continuity of current in the circuit? Write the expression for the current inside the capacitor.
When capacitor is charged by a dc source, ammeter will show a momentary deflection because of the presence of displacement current.
Conduction current is equal to the displacement current.
i.e., 
Resulting continuity of current becomes zero because, is maximum when, capacitor is fully charged.
Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0.2 A. What would be the potential difference between points B and E? 
We have,
Effective resistance between B and E is,
On applying Kirchoff's law, we have
Hence,
Define relaxation time of the free electrons drifting in a conductor. How is it related to the drift velocity of free electrons? Use this relation to deduce the expression for the electrical resistivity of the material.
The average time elapsed between two successive collisions is known as the relaxation time of free electrons drifting in a conductor.
Relation between 
and vd is given by,
Consider a conductor of length ‘l’, area of cross-section A and current density n.
Current flowing through the conductor is given by, 
Electric field applied across the ends is given by, E = V/l
So current flowing through the conductor becomes,
Then, 
Using ohm's law, we get
Therefore,
In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine the potential at point B.
Current in 2 resistor is 1 A.
Now, applying Kirchhoff’s law ACDB,
VA + 1 + 2 x 1 - 2 = VB
As, VA = 0
Therefore,
VB = 1 + 2 - 2 = 1 V
A parallel plate capacitor is being charged by a time varying current. Explain briefly how Ampere’s circuital law is generalized to incorporate the effect due to the displacement current.
When AC current is passed through capacitor, current moves from one plate to another. According to Ampere’s circuital law, a magnetic field has to exist between the plates of the capacitor for the current transfer to happen. So, Maxwell modified Ampere’s circuital law with the help of continuity equation.
The modified Ampere’s law by Maxwell is given by,
where, iD = 
, is called the Displacement current.
In the meter bridge experiment, balance point was observed at J with AJ = l.
(i) The values of R and X were doubled and then interchanged. What would be the new position of balance point?
(ii) If the galvanometer and battery are interchanged at the balance position, how will the balance point get affected?
i) Balance point is obtained by, 
When both R and X are doubled and then interchanged, new balance length becomes l’ which is given by,
Now, from the above two equations, we have
ii) The balance point will not get affected if galvanometer and balance point are interchanged at this position.
Given,
EMF, E1 = 1.5 V ; E2 = 2 V
Internal resistance, r = 0.2 
Effective emf of two cells connected in parallel is,
Eeff = 
This implies, 
The effective resistance can be calculated as:
Reff = 
That is,
Reff = 
i)
Drift velocity is the average velocity of the free electrons in the conductor with which they get drifted towards the positive end of the conductor under the influence of an external electric field.
ii)
Free electrons are in continuous random motion. They undergo changethey undergo change in direction at each collision and the thermal velocities are randomly distributef in all directions. 
Electric field, E = -eE
Acceleration of each electron, 
... (2)
Here,
m = mass of an electron
e = charge on an electron
Drift velocity is given by, 
Electrons are accelerated because of the external electric field.
They move from one place to another and current is produced.
For small interval dt, we have
I dt = -q ; where q is the total charge flowing
Let, n be the free electrons per unit area. Then, total charge crossing area A in time dt is given by,
Idt = neAvd dt
Substituting the value of vd, we get
I.dt = neA 
Current density, J = 
From ohm's law, we have
J = 
Here, 
is the conductivity of the material through whic the current is flowing,
Thus, 
iii)
Alloys like constantan and manganin are used for making standard resistors because:
a) they have high value of resistivity
b) temperature coefficient of resistance is less.
i)
Principale of potentiometer: The potential difference across any two points of current carrying wire, having uniform cross-sectional area and material, of the potentiometer is directly proportional to the length between the two points.
That is, V 
Proof:
V = IR = I 
i.e., V = 
For unifrom current and cross- sectional area, we have
ii)
Given,
E = 2 V; R = 15 
; RAB = 10 
Potential difference across the wire, = 
Therefore, potential gradient = 0.8/1 = 0.8 V/ m
Potential difference across AO = 
Therefore,
Length, AO = 
= 
; which is the required balance length of the wire.
The plot of the variation of potential difference across a combination of three identical cells in series, versus current is shown below. What is the emf and internal resistance of each cell?
Potential difference, E = 6/3 = 2 V
Internal resistance, r = 
Given, three cells are connected in series,
r' = r/3 = 6/3 = 2 ohm
A battery of emf 12V and internal resistance 2 ohm is connected to a 4 ohm resistor as shown in the figure.
a) Show that a voltmeter when placed across the cell and across the resistor, in turn, gives the same reading.
b) To record the voltage and the current in the circuit, why is voltmeter placed in parallel and ammeter in series in the circuit?
a) Emf, E = 12 V
Internal resistance, r = 2V
Now, using the formula,
E = V + Ir
When the voltmeter is connected across the cell,
I = 
V1 = 12 - 2(2) = 8 V
When the voltmeter is connected across the resistor,
V2 = IR
= 2 x 4 = 8 V
That is, V1 = V2
Hence proved.
b) Voltmeter has very high resistance to ensure that it is connection does not alter the flow of current in the circuit. Current chooses the low resistance path. Therefore, voltmeter is connected in parallel to the load across which potential difference is to be measured.
Ammeter measures the value of current flowing through the circuit. Ammeter has a very low value of resistance to ensure that all the current flows through it. Hence, it should be connected in series.
(i) Derive an expression for drift velocity of free electrons.
(ii) How does drift velocity of electrons in a metallic conductor vary with increase in temperature ? Explain.
i) Expression for the drift velocity of electrons:
When a potential difference is applied across a conductor, an electric field is produced and free electrons are acted upon by an electric force (=-eE).
As a result, electrons accelerate and keep colliding with each other and acquire a constant average velocity (vd).
Therefore,
Fe = -Ee
ii) As the temperature is increased, drift velocity of electrons in a metallic conductor increases.
From the above relation,
Therefore, as the temperature of the metallic conductor increases, the collision between the electrons and ions increases, resulting in decrease in the relaxation time.
Thus, the drift velocity decreases.
Does the charge given to a metallic sphere depend on whether it is hollow or solid? Give a reason for your answer.
No, Because the charge resides only on the surface of the conductor.
How is the speed of em-waves in vacuum determined by the electric and magnetic field?
The speed of the em-waves in a vacuum is determined by the ratio of the peak value of electric and magnetic fields.
c =E0/B0
A resistance of R draws current from a potentiometer. The potentiometer wire, AB, has a total resistance of Ro. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of potentiometer wire.
When the slide is in the middle of the potentiometer, only half of its total resistance i.e. R0/2 (since resistance is directly proportional to length) will be between A and point of contact (C), say R1, will be given by the following expression.
The total resistance between A and B will be sum of the resistance between A & C and C&B i.e R1+R0/2
Current flowing through the potentiometer will be
The voltage V1 taken from the potentiometer will be the product of current I and the resistance R1
Derive an expression for drift velocity of electrons in a conductor. Hence deduce Ohm's law.
Let an electric field E be applied the conductor. Acceleration of each electron is
Velocity gained by the electron
Let the conductor contain n electrons per unit volume. The average value of time't', between their successive collisions, is the relaxation time, 't'.
Hence average drift velocity 
The amount of charge, crossing area A, in time Δt is
=neAvdΔt = IΔt
Substituting the value of vd, we get
But I = JA, where J is the current density
A wire whose cross-sectional area is increasing linearly from its one end to the other, is connected across a battery of V volts.
Which of the following quantities remain constant in the wire?
(a) drift speed
(b) current density
(c) electric current
(d) electric field
The electric current will remain constant in a wire whose cross-sectional area is increased linearly from its one end to the other, is connected across a battery of V volts.
Because current is the only quantity that does not depend on the area of cross- sections of the wire.
I=dq/dt, that is the rate of flow of charge, where as drift speed, current density and electric field depending on the increasing area of the cross-section of the following relations:
Drift speed: νd=I/Ane
Current density = I/A
Electric field = J/σ
State the two Kirchhoff's laws. Explain briefly how these rules are justified.
Kirchhoff’s First Law or Junction Rule states that “The sum of the currents flowing towards a junction is equal to the sum of currents leaving the junction.”
This is in accordance with the conservation of charge which is the basis of Kirchhoff’s current rule. 
Here, I1, I2 I3, and I4 are the currents flowing through the respective wires.
Convention: The current flowing towards the junction is taken as positive and the current flowing away from the junction is taken as negative.
I3 + (− I1) + (− I2) + (− I4) = 0
Kirchhoff’s Second Law or Loop Rule states that In a closed loop, the algebraic sum of the emf is equal to the algebraic sum of the products of the resistances and the currents flowing through them.
OR
“The algebraic sum of all the potential drops and EMFs along any closed path in a network is zero.”
For the closed loop BACB:
E1 − E2 = I1R1 + I2R2 − I3R3
For the closed loop CADC:
E2 = I3R3 + I4R4 + I5R5
This law is based on the law of conservation of energy.
The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r as shown in the figure. Obtain the expression for (i) the current draw from the cell and (ii) the power consumed in the network.
From the above figure, we can use the horizontal symmetry to deduce the current.
As both resistance in the circuit, as well as the internal resistance in the circuit, have the same resistance 'r'
The resistance of the loop I will be
Because the circuit I and II are same we can say
RII = 2r/3
Combined resistance will be
R = RI + RII
Now this circuit is series with internal resistance 'r'
Resultant resistance will be
The current drawn from the cell will be
I=3E/4r
And power consumed by the network
Two electric bulbs P and Q have their resistances in the ratio of 1: 2. They are connected in series across a battery.Find the ratio of the power dissipation in these bulbs.
Let ratio be x
Rp = x (Resistance of bulb P)
Rq = 2x (Resistance of bulb Q)
We know, P = VI = V.V/R
A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 Ω as shown in the figure. Find the value of current in the circuit.
The positive terminals are connected to the offsite ends of the resistor, they send the current in opposite directions.
Hence, the net emf = 200-10 = 190 V
∴ current in the circuit I = E/R
= 190V / 38 Ω
= 5 A
In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 ohm is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.
Potentiometer at open circuit ℓ1 = 350
R = 9
ℓ2 = 300
Define the term 'conductivity' of a metallic wire. Write its SI unit.
The conductivity of a metallic wire is defined as the degree to which a specified material conduct electricity calculated as the ratio of current density in the material to the electric field which causes the flow of current,
The SI unit of conductivity is ohm-1 metre-1 or mho metre-1 or siemen metre-1
When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10–4 ms–1. If the electron density in the wire is 8 × 1028 m–3, the resistivity of the material is close to:
1.6 x 10-8Ωm
1.6 x 10-7Ωm
1.6 x 10-6Ωm
1.6 x 10-5Ωm
D.
1.6 x 10-5Ωm
In the circuit shown, the current in the 1Ω resistor is:
1.3 A from P to Q
0 A
0.13 A, from Q to P
0.13 A, from P to Q
C.
0.13 A, from Q to P
Connect Point Q to ground and by applying Kirchhoff's laws
consider the grounded circuit as shown below,
Applying Kirchhoff's law at point Q,
Incoming current at Q = outgoing current from Q
Thus, current in the 1Ω resistance is 0.13 A, from Q to P
Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle ‘θ’ with the vertical. If wires have mass λ per unit length then the value of I is: (g = gravitational acceleration)
B.
consider free body diagram of the wire
As the wires are in equilibrium. they must carry current in opposite direction.
Here, 
, where l is the length of each wire are d is a separation between wires.
From the diagram, d = 2L sin θ
T = cos θ = mg = λlg
(in vertical direction) ..... (i)
(In horizontal direction) ..... (ii)
From eqs. (i) and (ii)
An inductor (L = 0.03 H) and a resistor (R = 0.15 kΩ) are connected in series to a battery of 15V EMF in a circuit shown. The key K1 has been kept closed for a long time. Then at t = 0, K1 is opened and key K2 is closed simultaneously. At t = 1 ms, the current in the circuit will be:
(
)
100 mA
67 mA
6.7 mA
0.67 mA
D.
0.67 mA
After long time inductor behaves as short-circuit.
At t = 0, the inductor behaves as short circuit.The current,
As K2 is closed, current through the indicator starts decay which is given at any time t as
The current voltage relation of the diode is given by I = e(1000V/T-1) mA, where the applied voltage V is in volt and the temperature T is in kelvin. If a student makes an error measuring ± 0.01V while measuring the current of 5mA at 300K, what will be the error in the value of current in mA?
0.2 mA
0.02 mA
05 mA
0.05 mA
A.
0.2 mA
Given, I =(e1000V/T-1)mA
dv =±0.01 V
T = 300 K
I = 5 mA
I = e1000V/T-1
I +1 = e1000V/T
Taking log on both sides, we get
log(l+T) = 1000V/T
On differentiating, 
So, error in the value of current is 0.2 mA
The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 x 103Am-1.The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is
30 mA
60 mA
3 A
6 A
C.
3 A
For solenoid, the magnetic field needed to be magnetised the magnet,
B =μonl
Where, n = 100
l = 10 cm = 10/100 m = 0.1m
During the propagation of electromagnetic waves in a medium:
Electric energy density is equal to the magnetic energy density.
Both electric and magnetic energy densities are zero.
Electric energy density is double of the magnetic energy density.
Electric energy density is half of the magnetic energy density.
C.
Electric energy density is double of the magnetic energy density.
The supply voltage to a room is 120 V. The resistance of the lead wires is 6 Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?
zero
2.9V
13.3 V
10.04 V
D.
10.04 V
Resistance of bulb =(120x 120)/60 = 240Ω
Resistance of Heater =(120x 120)/240 = 60Ω
Voltage across bulb before heater is switched on,
Voltage across bulb after heater is switched on,
A decrease in the voltage is V1 − V2 = 10.04 (approximately)
This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement- I: Higher the range, greater is the resistance of ammeter.
Statement- II: To increase the range of ammeter, additional shunt needs to be used across it.
Statement – I is true, Statement – II is true, Statement – II is the correct explanation of statement- I.
Statement – I is true, Statement – II is true, Statement – II is not the correct explanation of Statement–I.
Statement – I is true, statement – II is false.
Statement – I is false, Statement – II is true
D.
Statement – I is false, Statement – II is true
For Ammeter, S = 
So for I to increase, S should decrease, so additional S can be connected across it.
The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is :
3 V/m
6 V/m
9 V/m
12 V/m
B.
6 V/m
Ε = cB
Where E = electric field,
B = magnetic field
c = speed of EM waves
= 3 × 108 × 20 × 10–9 = 6 V/m
Two electric bulbs marked 25W – 220V and 100W – 220V are connected in series to a 440Vsupply. Which of the bulbs will fuse?
Both
100 W
25 W
Neither
C.
25 W
D.
Neither
Resistances of both the bulbs are
Hence, R1 >R2
When connected in series, the voltages divide in them in the ratio of their resistances. The voltage of 440 V divide in such a way that voltage across 25 w bulb will be more than 220 V. Hence 25 w bulb will fuse.
If a wire is stretched to make it 0.1% longer, its resistance will
increase by 0.2%
decreases by 0.2%
decrease by 0.05 %
increase by 0.05%
A.
increase by 0.2%
The resistance,
A rectangular loop has a sliding connector PQ of length
C.
A moving conductor is equivalent to a battery of emf
= vBI
Equivalent circuit I = I1 + I2
Applying Kirchhoff's law
I1R + IR -vBl = 0 .... (i)
I2R + IR -vBl = 0 .... (ii)
Adding Eqs. (i) and (ii), we get
2IR + IR = 2vBI
I = 2VBI/3R
I1 = I2 = VBI/3R
In the circuit shown below, the key K is closed at t = 0. The current through the battery is
B.
At t = 0, inductor behaves like an infinite resistance so , at t = 0,
i = V/R2
and at t = ∞, inductor behaves like a conducting wire i.e., resistance less wire
then, 
Which of the following statements is false?
A rheostat can be used as a potential divider
Kirchhoff's second law represents energy conservation
Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude.
In a balanced Wheatstone bridge, if the cell and the galvanometer are exchanged, the null point is disturbed.
D.
In a balanced Wheatstone bridge, if the cell and the galvanometer are exchanged, the null point is disturbed.
When a current of 5 mA is passed through a galvanometer having a coil of resistance 15Ω, it shows full-scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into to voltmeter of range 0 – 10 V is
2.535 × 103 Ω
4.005 × 103 Ω
1.985 × 103 Ω
2.045 × 103 Ω
C.
1.985 × 103 Ω
10 = (5 × 10–3) (15 + R)
r = 1985 Ω
In the above circuit the current in each resistance is
0.5 A
0 A
0.25
1 A
B.
0 A
Taking voltage of point A as = 0
Then voltage at other points can be written as shown in figure. Hence voltage across all resistance is zero.
Hence current = 0
The Kirchhoff’s first law (∑ i= 0) and second law (∑iR = ∑E), where the symbols have their usual meanings, are respectively based on
conservation of charge, conservation of energy
conservation of charge, conservation of momentum
conservation of energy, conservation of charge
conservation of momentum, conservation of charge
A.
conservation of charge, conservation of energy
Kirchhoff's Ist law or KCl states that the algebraic sum of current meeting at any junction is equal to zero. Thus, no charge has been accumulated at any junction i.e., the charge is conserved, and hence KCl is based on conservation of charge.
The resistance of a bulb filament is 100 Ω at a temperature of 100°C. If its temperature coefficient of resistance be 0.005 per °C, its resistance will become 200 Ω at a temperature of
200°C
300°C
400°C
500°C
B.
300°C
200 = 100[1 +(0.005 × ∆t)]
T− 100 = 200
T = 300° C
Two thin, long parallel wires separated by a distance ‘d’ carry a current of ‘i’ A in the same direction. They will
attract each other with a force of µ0i2/(2πd)
repel each other with a force of µ0i2/(2πd)
attract each other with a force of µ0i2(2πd2 )
repel each other with a force of µ0i2/(2πd2)
A.
attract each other with a force of µ0i2/(2πd)
An energy source will supply a constant current into the load of its internal resistance is
equal to the resistance of the load.
very large as compared to the load resistance.
zero
non-zero but less than the resistance of the load.
B.
very large as compared to the load resistance.
The total current supplied to the circuit by the battery is
1A
2 A
4 A
6 A
C.
4 A
The given circle is written as
I = V/R
I = 6 V / 1.5 Ω
= 4 A
The resistance of the series combination of two resistances is S. When they are joined in parallel through total resistance is P. If S = nP, then the minimum possible value of n is
4
3
2
1
A.
4
Let the resistance R1 and R2
So, S = R1+ R2;
In a metre bridge experiment null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y?
50 cm
80 cm
40 cm
70 cm
A.
50 cm
We have from meter bridge experiment,
The electrochemical equivalent of a metal is 3.3 × 10−7 kg per coulomb. The mass of the metal liberated at the cathode when a 3 A current is passed for 2 seconds will be
19.8 × 10−7 kg
9.9 × 10−7 kg
6.6 × 10−7 kg
1.1 × 10−7 kg
A.
19.8 × 10−7 kg
Mass of substance liberated at cathode m = zit
where, z = electro-chemical equivalent = 3.3× 10-7 kg/C
i = current flowing = 3A.
t = 2 sec
∴ m = 3.3× 10-7× 3× 2
= 19.8× 10-7 kg
The reading of the ammeter for a silicon diode in the given circuit is
13.5 mA
0 mA
15 mA
11.5 mA
D.
11.5 mA
Clearly, from fig. given in the question, Silicon diode is in forward bias.
therefore,
The potential barrier across the diode
ΔV = 0.7 volts
Current,
Two batteries with e.m.f 12 V and 13 V are connected in parallel across a load resistor of 10Ω. The internal resistance of the two batteries are 1Ω and 2Ω respectively. The voltage across the load lies between :
11.7 V and 11.8 V
11.6 V and 11.7 V
11.5 V and 11.6 V
11.4 V and 11.5 V
C.
11.5 V and 11.6 V
Using Kirchhoff's law at P we get
On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 kΩ. How much was the resistance on the left slot before interchanging the resistances?
910Ω
990Ω
505Ω
550 Ω
D.
550 Ω
R1 + R2 = 1000
⇒ R2 = 1000-R1
On blancing condition
R1(100 - ℓ) = (1000-R1)ℓ ... (1)
on Inter-changing resistance
On balancing condition
(1000-R1)(110-ℓ) = R1(ℓ-10)
Or
R1(ℓ-10)=(1000-R1)(110-ℓ)....(2)
divide equation (1) / (2)
⇒ (100 -ℓ)(110 -ℓ) = ℓ(ℓ-10)
⇒ 11000 - 100ℓ -110 ℓ + ℓ2 = ℓ2-10ℓ
⇒ 11000 = 200ℓ
ℓ = 55
put in eq (1)
R1(100-55) = (1000-R1)55
R1(45) = (1000-R1)55
R1(9) = (1000-R1)11
20R1 = 11000
R1 = 550
In a potentiometer experiment, it is found that no current passes through the galvanometer when the
terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a
resistance of 5Ω, a balance is found when the cell is connected across 40 cm of the wire. Find the internal
resistance of the cell.
2.5 Ω
1 Ω
1.5 Ω
2 Ω
C.
1.5 Ω
When the cell is shunted by a resistance of 5Ω
A non -conducting ring of radius 0.5m carries a total charge of 1.11 x 10-10 C distributed non-uniformly on its circumference producing on its circumference on the electric field. E everywhere is space.
The value of the line integral (l = 0 being centre of ring) in volts is
+2
-1
-2
zero
A.
+2
From the definition of potential difference = potential difference at infinity and at centre of ring = Vcentre -Vinfinity
But by convention Vinfinity = 0 and Vcentre =
=
a potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf is,
5:4
3:4
3:2
5:1
C.
3:2
According to question, emf of the cell is directly proportional to the balancing length.
i.e., E
... (i)
Now, in the first case, cells are connected in series
That is,
Net EMF = E1 + E2
From equation (i),
E1 + E2 = 50 cm (given) ... (ii)
Now, the cells are connected in series in the opposite direction,
Net emf = E1 -E2
From equation (i)
A square loop of ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be
D.
In the above fig., the direction of currents in a long straight conductor XY and arm AB of a square loop ABCD are in the same direction.
So, there exists a force of attraction between the two conductors, which will be experienced by FAB as,
In the case of XY and arm CD, the direction of currents are in the opposite direction.
There exists a force of repulsion which will be experience by CD as,
Therefore, net force on the loop ABCD is,
Floop = FBA - FCD = 
Floop = 
A, B and C are voltmeters of resistance R, 1.5 R and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and Vc respectively then,
VA =VB =Vc
A.
VA =VB =Vc
The power dissipated in the circuit shiwn in the figure is 30 Watt. The value of R is
20 Ω
15Ω
10Ω
5Ω
C.
10Ω
Here,
R1 = R2= ?
R2 = 5Ω, V = 10 V
and P= 30 W
Hence,
A cell having an emf and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by
C.
Here, E = I (R+r)
E = IR+I
and
E= V+Ir
This equation represents option (c).
In an electrical circuit R, L, C and AC voltage source are all connected in series when L is removed from the circuit, the phase difference between the voltage and the current in the circuit is π/3. If instead, Cis removed from the circuit, the phase difference is again π/3. The power factor of the circuit is
1/2
1
C.
1
Here, phase difference
It is a condition of resonance, therefore, phase difference between voltage and current is zero and power factor is cos Φ = 1.
A millivoltmeter of 25 mV range is to be converted into an ammeter of 25 A range. The value (in ohm) of necessary shunt will be
0.001
0.01
1
0.05
A.
0.001
The full-scale deflection current
Where G is the resistance of the meter. The value of shunt required for converting it into ammeter of range 25 A is 
The resistances in the two arms of the meter bridge are 5 ohm and R ohm, respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6 l1. The resistance R is,
10 ohm
15 ohm
20 ohm
25 ohm
B.
15 ohm
Case 1:
... (i)
Now, by shunting resistance R by an equal resistance R, new resistance in that arm become R/2
So, 
... (ii)
From equations (i) and (ii),
160 - 1.6 l1 = 200 - 3.2 l1
1.6 l1 = 40
From equation (i),
A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of i) infinity ii) 9.5 ohm the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is,
0.25 ohm
0.95 ohm
0.5 ohm
0.75 ohm
C.
0.5 ohm
Given,
e.m.f., e = 2 Volt
length, l = 4 m
Potential drop per unit length, 
Case 1:
... (i)
where, e' is the emf of the cell
Case 2:
V = 
... (@)
From equations (i) and (ii), we have
e'/V = l1 / l2
e' =l (r+R) and
In an ammeter 0.2 % of the main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be,
1/499 G
499/500 G
1/500 G
500/499 G
A.
1/499 G
For ammeter,
0.0002 I x G = 0.998 I x rg
rg = 
That is,
Resistance of ammeter is approximately equal to resistance of shunt rB.
In the circuit shown the cells A and B have negligible resistances. For VA = 12V, R1 = 500Ω and R = 100Ω the galvanometer (G) shows no deflection. The value of VB is
4V
2V
12 V
6 V
B.
2V
Applying Kirchoff's law
500I +100I = 12
So, I = 12 x 10-2/6 = 2 x 10-2
Hence, VB = 100 (2 x 10-2) = 2V
The condition under which a microwave oven heats up a food item containing water molecules most efficiently is
the frequency of the microwave must match the resonant frequency of the water molecules
the frequency of the microwave has no relation with natural frequency of water molecules
microwave are heat waves, so always produce heating
infrared waves produce heating in a microwave oven
A.
the frequency of the microwave must match the resonant frequency of the water molecules
It is an electromagnetic wave, therefore the correct choice is (a).
A wire of resistance 4Ω is stretched to twice its original length. The resistance of stretched wire would be
2Ω
4Ω
8Ω
16Ω
D.
16Ω
As R' = n2R = 22 x4 = 4 x 4 = 16 Ω
The resistance of the four arms P, Q, R and S a Wheatstone's bridge are 10Ω, 30Ω and 90Ω respectively. The emf and internal resistance of the cell are 7V and 5Ω respectively. If the galvanometer resistance is 50Ω, the current drawn from the cell will be
1.0 A
0.2 A
0.1 A
2.0 A
B.
0.2 A
Effective resistance,
If power dissipated in the 9Ω resistor in the circuit shown is 36 W, the potential difference across the 2Ω resistor is
8 V
10 V
2 V
4 V
B.
10 V
Electric power, P = i2R
The decreasing order of wavelength of infrared, microwave, ultraviolet and gamma rays is
gamma rays, ultraviolet, infrared, microwaves
microwaves, gamma rays, infrared, ultraviolet
infrared, microwave, ultraviolet, gamma rays.
microwave infrared, ultraviolet, gamma rays
D.
microwave infrared, ultraviolet, gamma rays
Decreasing order of the wavelength of various rays.
Microwave > infrared > Ultraviolet > Gamma
A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along direction of fields, then the electron
speed will decrease
speed will increase
will turn towards the left of the direction of motion
will turn towards the right of direction a motion
A.
speed will decrease
Field B is not applied only force. Field E will apply a force opposite to velocity of the electron hence, speed will decrease.
The energy of the EM waves is of the order of 15 keV. To which part of the spectrum does it belong?
X - rays
infrared rays
Ultraviolet rays
gamma -rays
A.
X - rays
Given, energy of EM waves is of the order of 15 keV
A potentiometer wire of length L and a resistance r are connected in series with battery of e.m.f. Eo and a resistance r1. An unknown e.m.f is balanced at a length l of the potentiometer wire. The e.m.f E will be given by
B.
Consider a potentiometer wire of length L and resistance r are connected in series to a battery of emf Eo and a resistance r1as shown in a figure.
Current in wire AB = Eo/r1 +r
A square loop, carrying a steady current I, is placed in a horizontal plane near a long straight conductor carrying a steady current I1 at a distance d from the conductor as shown in the figure. The loop will experience
a net repulsive force away from the conductor
a net torque acting upward perpendicular to the horizontal plane
a net torque acting downward normal to the horizontal plane
a net attractive force towards the conductor
D.
a net attractive force towards the conductor
In the circuit shown in the figure, if potential at point A is taken to be zero,the potential at point B is
-1V
+2 V
-2 V
+1 V
D.
+1 V
By KVL along path ACDB
VA + 1 (1)(2) - 2 = VB
0+1 = VB
VB = 1V
In the given circuit the reading of voltmeter V1 and V2 are 300 V each. The reading to the voltmeter V3 and ammeter A are respectively
150 V, 2.2 A
220 V , 2.2 A
220 V, 2.0 A
100 V, 2.0 A
B.
220 V , 2.2 A
For series LCR circuit
Voltage, 
Consider the following two statements
A) Kirchhoff junction law follows from the conservation of charge.
B) Kirchhoff's loop law follows from the conservation of energy
Which of the following is correct?
Both (A) and (B) are wrong
(A) Is correct and (B) is wrong
(A) is wrong and (B) is correct
Both (A) and (B) are correct
C.
(A) is wrong and (B) is correct
Kirchhoff's first law follows from the conservation of charge.
Kirchhoff's second law follow from the conservation of energy.
The voltage gain of an amplifier with 9% negative feedback is 10. The voltage gain without feedback will be
90
10
1.25
100
D.
100
The process of injecting a fraction of output energy of some device back to the input is known as feedback. when the feedback energy (voltage or current) is out of phase with the input signal and thus opposes it, it is called negative feedback.
Voltage gain with feedback is
A current of 3 A flows through the 2 Ω resistor shown in the circuit. The power dissiated in the 5 Ω resistor is
diagram
4 W
2 W
1 W
5 W
D.
5 W
Voltage across 2 Ω is same as voltage across arm containing 1 Ω and 5 Ω resistance.
Voltage across 2 Ω resistance,
V = 2 x 3 = 6 V
So, voltage across lowest arm,
V1 = 6 V
Current across 5 Ω, I = 6/ 1+6 = 1 A
Thus, power across 5 Ω,
P = I2R = (1)2 x 5 = 5 W
A wire of a certain material is stretched slowly by ten percent. Its new resistance and specific resistance become respectively
1.2 times, 1.1 times
1.21 times, same
both remain the same
1.1 times, 1.1 times
B.
1.21 times, same
In stretching, specific resistance remains unchanged.
After stretching, specific resistance (rho) will remain same.
Original resistance of the wire,
The velocity of electromagnetic radiation in a medium of permittivity 
and permeability 
is given by
C.
The velocity of electromagnetic radiation is the velocity of light (c), ie 
where, μo is the permeability and is 
the permittivity of free space.
A steady current of 1.5 A flows through a copper voltameter for 10 min. If the electrochemical equivalent of copper is 30 x 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
Apply first law of electrolysis during deposition fo charge on an electrode through electrolysis.
If m is the mass of 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 directly proportional q
m =zq
where z is a constant of proportionality and is called electrochemical equivalent (ECE) of the the substance.
If an electric current I flows through the electrolytem, then
q = It
m= zit
Here, i = 1.5 A, t= 10 min= 10 x 60 s
z = 30 x 10-5 g C-1
Hence, the mass of copper deposited on the electronde
m= 30 x 10-5 x 1.5 x 10 x 60
= 27 x 10-2 = 0.27 g
In the circuit shown, if a conducting wire is connected between points A and B, the current in this wire will
flow from A to B
flow in the direction which will be decided by the value of V
be zero
flow from B to A
D.
flow from B to A
Current will flow from higher to lower potential.
Resistances 
are connected in series, so their effective resistance is 
Similarly, 
and 
are in series
So, 
Now R' and R" will be in parallel, hence effective resistance
Current through the circuit, from Ohm's law
Let current 
flow in the branches as shown:
Since, drop of potential is greater in 
resistance so. It will be at lower potential than B, hence, on connecting wire between points A and B, the current will flow from B to A.
Two cells, having the same emf, are connected in series through an external resistance R. Cells have internal resistances r1 and r2 (r1 > r2) respectively. When the circuit is closed, the potential difference across the first cell is zero, The value of R is
A.
Current in the circuit is given by Ohm's law.
Net resistance of the circuit = 
Net emf in series = E + E = 2E
Therefore, from Ohm's law, current in the circuit.
It is given that, as circuit is closed, potential difference across the first cell is zero. That is,
Equating Eqs. (i) and (ii), we get
Kirchhoff's first and second laws for electrical circuits are consequences of
conservation of energy
conservation of electric charge and energy respectively
conservation of electric charge
conservation of energy and electric charge respectively
B.
conservation of electric charge and energy respectively
Kirchhoff's first law is junction rule, according to which the algebraic sum of the currents into any junction is zero. The junction rule is based on conservation of electric charge. No charge can accumulate at a junction, so the total charge entering the junction per unit time must equal to charge leaving per unit time.
Kirchhoff's second law is loop rule according to which the algebraic sum of the potential difference in any loop including those associated emf's and those of resistive elements, must equal zero.
This law is basically the law of conservation of energy.
A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F, because the method involves :
Cells
Potential gradients
A condition of no current flow through the galvanometer
A combination of cells, galvanometer and resistances
C.
A condition of no current flow through the galvanometer
Reading of potentiometer is accurate because during taking reading it does not draw any current from the circuit.
A carbon resistor of (47 ± 4.7) kΩ is to be marked with rings of different colours for its identification. The colour code sequence will be
Violet – Yellow – Orange – Silver
Yellow – Violet – Orange – Silver
Green – Orange – Violet – Gold
Yellow – Green – Violet – Gold
B.
Yellow – Violet – Orange – Silver
Colour code for carbon resistor,
| 0 | Black |
| 1 | Brown |
| 2 | Red |
| 3 | Orange |
| 4 | Yellow |
| 5 | Green |
| 6 | Blue |
| 7 | Violet |
| 8 | Grey |
| 9 | White |
Tolerance:
±5% Gold
± 10% Silver
± 20% No colour
(47±4.7) kΩ = 47 x 103 ± 10%
Therefore, Yellow - Violet - Orange -Silver
A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of 'n' is
10
11
9
20
A.
10
In series grouping equivalent resistance Rseries = nR
In parallel grouping equivalent resistance Rparallel = R/n
A battery consists of a variable number 'n' of identical cells (having internal resistance 'r' each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n?
A.
Short circuited current,
So, I is independent of n and I is constant
You are given resistance wire of length 50 cm and a battery of negligible resistance. In which of the following cases is the largest amount of heat generated?
When the wire is connected to the battery directly
When the wire is divided into two parts and both the parts are connected to the battery in parallel.
When the wire is divided into four parts and all the four parts are connected to the battery in parallel.
When only half of the wire is connected to the battery
C.
When the wire is divided into four parts and all the four parts are connected to the battery in parallel.
Let R be the resistance of the wire
b) The resistance of each part will be R/2. When they are connected in parallel, the resistance will be R/4.
Hence, d)
d)
Hence, the largest amount of heat will be generated in the case of four parts connected in parallel.
The potential difference between point A and B is
zero
D.
zero
From the given figure, current through lower branch of resistances which are joined in series
Again current through upper branch of resistances which are also connected in series
When a certain current is passed in the circuit as shown in figure, 10 kcal of heat is produced in 5 Ω resistance. How much heat is produced in 4 Ω resistance?
4 kcal
2 kcal
5 kcal
3 kcal
B.
2 kcal
Heat produced in 5Ω = 10 kcal
Now from the relation
....(1)
Now heat produced in 4Ω resistance is
In the given circuit, what will be the equivalent resistance between the points A and B ?
10 / 3 Ω
20 / 3 Ω
10 / 5 Ω
5 Ω
A.
10 / 3 Ω
∴ Given circuit is a balanced Wheatstone's Bridge. So, 10Ω resistance of upper arms will be ineffective. Equivalent resistance of upper arms
R1 = 2+3 =5Ω
∴ The equivalent resistance of lower arms
R2 = 4+6 = 10Ω
A uniform wire of length I and having resistance R is cut into n equal parts and all parts are connected in parallel, then the equivalent resistance will be :
R
R/n
R/n2
n2R
C.
R/n2
Two or more resistors are said to be parallel if one end of all the resistors are joined together and simillarly the other ends are joined together.
Resistance of each part R' = R/n
When these parts are connected in parallel.
Then,
For the network shown in the figure, the value of the current i is
B.
The circuit given resembles the balanced Wheatstone Bridge as
Thus, a middle arm containing 4 Ω resistance will be ineffective and no current flows through it. The equivalent circuit is shown as below:
Net resistance of AB and BC
R' = 4 + 2 = 6 Ω
Net resistance of AD and DC
R'' = 6 + 3 = 9 Ω
Thus parallel combination of R' and R'' gives
When a wire of uniform cross-section a, length land resistance R is bent into a complete circle, resistance between two of dimetrically opposite points will be
4R
A.
When wire is bent to form a complete circle then
Thus, net resistance in parallel combination of two semicircular resistance
The resistance of an ammeter is 13 Ω and its scale is graduated for a current upto 100 A. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto 750 A by this meter. The value of
shunt resistance is
20 Ω
2 Ω
0.2 Ω
2 KΩ
B.
2 Ω
Let ia is the current flowing through ammeter and 'i' is the total current. So, a current ( i - ia ) will flow through shunt resistance.
Potential difference across ammeter and shunt resistance is same
i.e ia × R = ( i - ia ) × S
Given:- ia = 100 A, i = 750 A, R = 13 Ω
Hence
S = 2Ω
Three resistances P, Q, R each of 2Ω and an unknown resistance S form the four arms of a Wheatstone's bridge circuit. When a resistance of 6Ω is connected in parallel to S the bridge gets balanced. What is the value of S?
2 Ω
3 Ω
6 Ω
1 Ω
B.
3 Ω
The situation can be depicted as shown in figure
As resistances S and 6Ω are in parallel their effective resistance is:
As the bridge is balanced, hence it is balanced Wheatstone's bridge.
For balancing condition
⇒S = 3 Ω
For a metallic wire, the ratio ( V = applied potential difference and i = current flowing) is
independent of temperature
increases as the temperature rises
decreases as the temperature rises
increases or decreases as temperature rises depending upon the metal
B.
increases as the temperature rises
The resistance of a metallic wire at temperature t°C is given by
Rt = Ro ( 1 + )
Where is coefficient of expansion.
Hence, resistance of wire increases on increasing the temperature. Also, from Ohm's law, ratio is equal to R ie
Hence, on increasing the temperature the ratio increases.
A galvanometer having a resistance of 8 Ω is shunted by a wire of resistance 2 Ω.1f the total current is 1 A, the part of it passing through the shunt will be
0.25 A
0.8 A
0.2 A
0.5 A
B.
0.8 A
Potential difference across galvanometer should be equal to potential difference across shunt.
The shunt and galvanometer are connected as shown figure.
Let total current through the parallel combination is i, the current through the galvanometer is ig and the current through the shunt is ( i - ig )
The potential difference Vab ( = Va - Vb ) is the same for both paths, so
ig G = ( i - ig ) S
⇒ ig ( G + S ) = i S
The fraction of current passing through shunt:
The fraction of current passing through shunt
= 0.8 A
At room temperature, copper has free electron density of 8.4 x 1028 m-3, The electron drift velocity in a copper conductor of cross-section area of 10-6 m2 and carrying a current of 5.4 A will be
4 ms-1
0.4 ms-1
4 ms-1
0.4 mm s-1
D.
0.4 mm s-1
Drift velocity in copper conductor
= 0.4 × 10-3 ms-1
A uniform wire of resistance R and length L is cut into four equal parts, each of length L/4, which are then connected in parallel combination. The effective resistance of the combination will be
R
4 R
D.
According to ohm's law
R =
ρ is a constant of proportionality and depends on the material of a conductor but not on its dimensions. ρ is called resistivity
R ∝ l
i.e resistance is proportional to length
∴
= 4
....( R1 = R )
In parallel combination of such four resistances
⇒
⇒
⇒
⇒
A capacitor of capacitance 5µF is connected as shown in the figure. The internal resistance of the cell is 0.5 Ω . The amount of charge on the capacitor plates is
80 μC
40 μC
20 μC
10 μC
D.
10 μC
In steady state, there will be no current in the capacitor branch
Net resistance of the circuit
R = 1 + 1 + 0.5
R = 2.5 Ω
Current drawn from the cell
i =
=
i = 1 A
Potential drop across two parallel branches
V = E - i r
= 2.5 - 1 × 0.5
V = 2.0 V
So charge on the capacitor plates
q = CV
= 5 × 2
q = 10 μC
The equivalent resistance across A and B is 
2 Ω
3 Ω
4 Ω
5 Ω
C.
4 Ω
Wheatstone Bridge is an application of Kirchoff's rule.
The equivalent circuit can be redrawn as
We have,
i.e
So, the given circuit is a balanced Wheatstone's bridge. Hence, the equivalent resistance
RAB =
RAB =
=
RAB = 4 Ω
In the circuit shown the value of I in ampere is
1
0.60
0.4
1.5
C.
0.4
We can simplify the network as shown
So net resistance,
R = 2.4 + 1.6 = 4.0 Ω
Therefore, current from the battery,
=
i = 1A
Now from the circuit (b),
4 I' = 6I
⇒ I' = I
But, i = I + I'
= I + I
1 = I
⇒ I =
I = 0.4 A
A current of 5A is passing through a metalic wire of cross-sectional area 4 × 10-6 m2 . If the density of charge carriers of the wire is 5 × 1026 m-3, then the drift velocity of the electrons will be
1 × 102 m/s
1.56 × 10-2 m /s
1.56 × 10-3 m/s
1 × 10-2 m/s
B.
1.56 × 10-2 m /s
Drift velocity
Vd =
Where Vd is the drift velocity
I is the current
A is the cross-sectional area of the conductor
n is charge density (m-3 ). This is the number of charge carriers that can move per m3.
⇒ =
Vd = 1.56 × 10-2 ms-1
conservation of energy
conservation of charge
conservation of energy as well as charge
conservation of momentum
B.
conservation of charge
Kirchhoff's first law states that in an electric circuit, the algebraic sum of the currents meeting at any junction in the circuit is zero.
i.e Σ i = 0
Hence, according to Kirchhoff's law
i1 + i2 = i3 + i4 + i5
Thus, the sum of currents flowing towards the junction is equal to the sum of the currents flowing away from the junction. In other words, when a steady current flows in a circuit, then there is neither any accumulation of charge at any point in the circuit, nor any charge is removed from there.
Thus, Kirchhoff's first law expresses the conservation of charge.
A battery of emf 10 V and internal resistance of 0.5 ohm is connected across a variable resistance R. The maximum value of R is given by
0.5 Ω
1.00 Ω
2.0 Ω
0.25 Ω
A.
0.5 Ω
V = E - I r
IR = E - I r
Where E is the open circuit e.m.f
I =
=
=
P = I2 R
=
P = 400 = 0
( 2R + 1 )2 = 4R
⇒ 2R = 1
⇒ R = 0.5
The internal resistance of a primary cell is 42. It generates a current of 0.2 A in an external resistance of 21 n. The rate at which chemical energy to consumed in providing current is
1 J/s
5 J/s
0.42 J/s
0.8 J/s
A.
1 J/s
Rate energy
(But total Resistance R = 21 + 4 = 25 Ω)
So,
The equivalent resistance across A and B is
2 Ω
3 Ω
4 Ω
5 Ω
C.
4 Ω
The equivalent circuit can be redrawn as
We have
So, the given circuit is a balanced Wheatstone's bridge.
Hence, the equivalent resistance
=
RAB = 4 Ω
At room temperature, copper has free electron density of 8.4 x 1028 m-3, The electron drift velocity in a copper conductor of cross-sectional area of 10-6 m2 and carrying a current of 5.4 A, will be
4 ms-1
0.4 ms-1
4 cm s-1
0.4 mm s-1
D.
0.4 mm s-1
Drift velocity in a copper conductor
vd =
where, i ⇒ the current
vd ⇒ the drift velocity
e ⇒ the charge
n ⇒ the free electron density
⇒ vd =
⇒ vd = 0.4 × 10-3 ms-1
⇒ vd = 0.4 mms-1
A uniform wire of resistance R and length L is cut into four equal parts, each of length L/4 which are then connected in parallel combination. The effective resistance of the combination will be
R
4R
D.
Resistivity of materials is the resistance to the flow of an electric current with some materials resisting the current flow more than others
where R is resistance in ( Ω )
ρ is resistivity
L is length in meters (m)
A is the area in square meters (m2)
⇒
=
⇒ R2 = .....( since R1 = R )
In parallel combination of such four resistance
⇒
⇒
⇒ R' =
A 5 amp fuse wire can withstand a maximum power of 1 W in circuit. The resistance of the fuse wire is
0.2 Ω
5 Ω
0.4 Ω
0.04 Ω
D.
0.04 Ω
Electric power = volltage difference × current
P = i V
= i × iR
⇒ Power P = i2 R
⇒ R =
Given:- P = 1W and i = 5A
∴ R =
R = 0.04 Ω
Find equivalent resistance between X and Y
R
R/L
2R
5R
A.
R
The given circuit can be shown in the following way. No, current will be flown in the middle resistance.
The equivalent resistance of R and R' = 2R
Total resistance
R' = R
A engine pumps up 100 kg of water through a height of 10 m in 5s. Given that, the efficiency of engine is 60%. If g = 10 ms-2, the power of this engine is
3.3 kW
0.33 kW
0.033 kW
33 kW
A.
3.3 kW
Efficiency of engine η = 60%
Power is equal to the work divided by time.
Efficiency tells how efficient something is and gives a number between 0 and 1.
Power =
=
Given:- m2 = 100 kg
h = 10 m
t = 5s
and g = 10 ms-2
∴ Power =
= 3.3 × 103 W
Power P = 3.3 kW
A wire of resistance 10Ω is bent to form a complete circle. Find its resistance between two diametrically opposite point.
5 Ω
2.5 Ω
1.25 Ω
Ω
B.
2.5 Ω
The wires ADC and ABC will have resistance 5 Ω each. These two are joined in parallel between A and C. The equivalent resistance R between A and C is, therefore give n by each. These two are joined in parallel between A and C. The equivalent resistance R between A and C is, therefore given by
R =
R = 2.5 Ω
Find the resistance of a hollow cylindrical conductor of length 1.0 mm and 2.0 mm respectively. The resistivity of the material is 2.0 × 10-8 Ωm
2.1 × 10-3 Ω
1.3 × 10-4 Ω
3.2 × 10-4 Ω
4.6 × 10-2 Ω
A.
2.1 × 10-3 Ω
The area of cross-section of the conductor through which charges will be flow is
A =
A = 3
The resistance of the wire is therefore
R =
=
R = 2.1 × 10-3 Ω
The ammeter shown in figure consists of a 480 Ω coil connected in parallel to a 20 Ω shunt. The reading of ammeter is
0.125 A
1.67 A
0.13 A
0.67 A
A.
0.125 A
Given:- The ammeter coil of resistance 480Ω and shunt of to a 20Ω are connected in parallel
The equivalent resistance of the ammeter is
R' =
R' = 19.2 Ω
Now, resistance of 140.8 Ω and 19.2 Ω are in series, hence total resistance in the circuit,
The equivalent resistance of the circuit is
R = 140.8 Ω + 19.2 Ω = 160 Ω
The current I =
=
I = 0.125 A
Hence, ammeter reading is 0.125 A.
A copper rod of length 20 cm and cross-sectional area 2 mm is joined with a similar aluminium rod as shown below
The resistance of pair of rods is (ρAl =2.6 x 10-8 Ω-m and ρCu =1.7 x 10-8 Ω-m)
1.0 mΩ
2.0 mΩ
3.0 mΩ
None of these
A.
1.0 mΩ
Given
Cross-sectional area a = 2 mm2
a = 2 × 10-6 m2
Length of rod l = 20 cm
= 20 × 10-2 m
The resistance of the Copper rod
R =
R1 =
R1 = 1.7 × 10-3 Ω
The net resistance of the Aluminium rod
R2 =
R2 = 2.6 × 10-3 Ω
The equivalent resistance
⇒ R = × 10-3
⇒ R = 1.0 × 10-3 Ω
⇒ R = 1.0 mΩ
The electron of H - atom is revolving around the nucleus in circular orbit having radius with . The current produced due to the motion of electron is
zero
D.
Given:-
radius r =
velocity v =
As we know that time period
T =
∴ T =
=
∴ Current i =
=
curent i =
i =
The galvanometer resistance is 30 Ω and it is connected to 2 V battery along with a resistance 2000 Ω in series. A full scale deflection of 25 divisions is obtained. In order to reduce this deflection to 20 divisions, the resistance in series should be
2470 Ω
2370 Ω
2180 Ω
2210 Ω
A.
2470 Ω
Total initial resistance = RG + R
= 30 + 2000
= 2030Ω
And connected with battery E = 2V
Current i = A
as the full-scale deflection scale for
1 mA = 25 divisions
Let x be the effective resistance of the circuit, then
2V = 2030 × × 103
= x ×
⇒ x =
⇒ x = 2500 Ω
∴ Resistance to be added= 2500 - 30 = 2470 Ω
In the given figure, potential difference between A and B is
0 V
5 V
10 V
15 V
C.
10 V
Equivalent resistance, R = 10 +
= 15 Ω
∴ Current I =
=
= 2 × 10-3 A
Hence potential difference between A and B,
V = × 10 × 103
V = 10 V
Two batteries of emf 3V and 6 V with internal resistances 2 Ω and 4 Ω are connected in a circuit with resistance of 10 Ω as shown in figure. The current and potential difference between the points P and Q are
D.
Kirchoff's voltage law states that for a closed loop series path, the algebraic sum of all voltages around any closed loop in a circuit is equal to zero.
Applying Kirchhoff's voltage law in the given loop
-- 4I + 6 - 3 - 2I - 10I = 0
-16 I = -3
I = A
Potential difference across PQ
vPQ = V
vPQ
A rod made up of metal is 1.2 m long and 0.8 cm in diameter. Its resistance is 3.5 x 10-3 Ω. Another disc made of the same metal is 2.0 cm in diameter and 1.25 mm thick. What is the resistance between the round faces of the disc?
1.35x 10-8 Ω
2.70 x 10-7 Ω
5.82 × 10-7 Ω
8.10 × 10-5 Ω
C.
5.82 × 10-7 Ω
Given:-
l = 12 m,
d = 0.8 cm,
R = 3.5 x 10-3 Ω ,
thickness= 1.25 mm = 1.25 × 10-3 m
Resistivity of the material of the rod
=
= 46.6 × 10-9 Ω m
Resistance of disc
R =
=
= 5.82 × 10-7Ω
A body is moving along a rough horizontal surface with an initial velocity of 10 ms-1. If the body comes to rest after travelling a distance of 12m, then the coefficient of sliding friction will be
0.5
0.2
0.4
0.6
C.
0.4
Given:-
u = 10 ms-1
s = 12 m
v = 0
By third equation of motion,
v2 = u2 - 2 a s
Where v - final velocity
u - initial velocity
a - acceleration
s - dispalcement
0 = (10)2 - 2 × a × 12
24 a = 100
a = 4.17 ms-2
Coefficient of sliding friction is given by
μ =
=
μ = 0.41
A galvanometer of resistance 25 Ω shows a deflection of 5 divisions when a current of 2 mA is passed through it. If a shunt of 4 Ω is connected and there are 20 divisions on the scale, then the range of the galvanometer is
1 A
58 A
58 mA
30 mA
C.
58 mA
Given: - Galvanometer resistance Rg = 25 Ω
There are 20 divisions on scale.
Initially, current for 20 divisions
i =
i = 8 mA
∴ ig = 8 mA
Let I be the maximum current that galvanometer can read.
⇒ I=
⇒ =
⇒ = 29 × 2
⇒ I = 58 mA
The length of a given cylindrical wire is increased by 150%. Due to the consequent decrease in diameter the change in the resistance of the wire will be
200 %
525 %
300 %
400 %
B.
525 %
If suppose initial length, l1 = 100,
then the length of the cylindrical wire increased by 150%
l2 = l1 + 150
l2 = 100 + 150
l2 = 250
∴
=
⇒
⇒ R2 = R1
⇒
=
⇒ × 100 = 525 %
The measurement of voltmeter in the following circuit is
2.25 V
4.25 V
2.75 V
6.25 V
A.
2.25 V
Resistance of voltmeter and 60 Ω are in parallel.
∴ Equivalent resistance =
=
= 24 Ω
For the whole circuit
Current I =
=
I = A
The voltmeter reads potential difference across 60 Ω
∴ Voltmeter reading = I × Req
=
= V
= 2.25 V
A meter bridge is set-up as shown, to determine an unknown resistance X using a standard 10 ohm resistor. The galvanometer shows null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determined value of
10.2 ohm
10.6 ohm
10.8 ohm
11.1 ohm
B.
10.6 ohm
Wheatstone bridge is an electrical circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component.
Applying the condition of balanced Wheatstone bridge,
we get
=
X = 10 Ω ×
X = 10.6 Ω
A and B are two points on a uniform ring of resistance R. The ∠ ACB = 0, where C is the centre of the ring. The equivalent resistance between A and B is
A.
Resistance per unit length ρ =
Lengths of sections APB and AQB are rθ and r
Resistances of sections APB and AQB are
R1 = ρr θ
=
R1 =
and R2 = r( 2θ )
R2 =
As R1 and R2 are in parallel between A and B , their equivalent resistance is
Req =
=
Req =
⇒ Req =
In the circuit shown in figure, if A the diode forward voltage is 0.3 V, the voltage difference between A and B is
1.3 V
2.3 V
0
0.5 V
B.
2.3 V
Let V be the potential difference between A and B, then
V 0.3 = ( 5 + 5 ) × 103 × ( 0.2 × 10-3 )
V = 10 × 0.2 + 0.3
V = 2.3 V
Assertion: An electrical bulb starts glowing instantly as it is switched on.
Reason: Drift speed of electrons in a metallic wire is very large.
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.
As the conductor is full of electrons, with slight drift anywhere the queue starts moving. As a result, bulb starts glowing instantly. The drift velocity of electrons in a good conductor is very low.
A voltmeter of resistance 20000 Ω reads 5 volt. To make it read 20 volt, the extra resistance required is
40000 Ω in parallel
60000 Ω in parallel
60000 Ω in series
40000 Ω in series
C.
60000 Ω in series
⇒ R = 3 G
= 3 × 20000
⇒ R= 6000
Extra resistance = 60000 Ω in series
Ten identical cells each of potential E and internal resistance r are connected in series to form a closed circuit. An ideal voltmeter connected across three cells, will read
10 E
3 E
13 E
7 E
B.
3 E
Given:-
Initial number of cells (n1 ) =10
Potential of each cell = E
Internal resistance of each cell = r
and Final number of cells ( n2 ) = 3
We know from the Ohm's law
total voltage of ten cells = 10 × E = 10 E
total resistance in ten cells = 10 × r = 10 r
Therefore,
current in the circuit
I =
I =
Potential difference across three cells
V = I × 3r
= × 3r
V = 3E
(Since the voltmeter is ideal, therefore it will read 3E).
A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor?
Electric field
Drift velocity
Current
Current density
C.
Current
The current flowing through a conductor of non-uniform cross-section remain same in the whole of the conductor.
On increasing the temperature of a conductor, its resistance increases because the
relaxation time increases
electron density decreases
relaxation time decreases
relaxation time remains constant
C.
relaxation time decreases
Resistance of a conductor
R =
where the symbols have their usual meaning. As the temperature increases, the relaxation time τ decreases.
What is the equivalent resistance between A and B in given figure?
50 Ω
25 Ω
75 Ω
100 Ω
A.
50 Ω
The Wheatstone bridge is the interconnection of four resistances forming a bridge. The four resistance in circuit ars referred as arms of the bridge. The bridge is used for finding the value of an unknown resistance connected with two known resistors, one variable and a galvanometer.
Given circuit is a balanced Wheatstone bridge. So circuit becomes
Equivalent resistance
R =
R =
R = 50 Ω
Sponsor Area
Sponsor Area
and 
which are always perpendicular to each other. the direction of polarisation is given by 
and that of wave propagation by 
then
