Physics Part I Chapter 7 Alternating Current
  • Sponsor Area

    NCERT Solution For Class 12 Physics Physics Part I

    Alternating Current Here is the CBSE Physics Chapter 7 for Class 12 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 12 Physics Alternating Current Chapter 7 NCERT Solutions for Class 12 Physics Alternating Current Chapter 7 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 12 Physics.

    Question 9
    CBSEENPH12039716

    Alternating current can not be measured by D.C. ammeter because

    • A.C. cannot pass through D.C.

    • A.C. changes direction

    • average value of current for complete cycle is zero

    • D.C. ammeter will get damaged.

    Solution

    C.

    average value of current for complete cycle is zero

    The full cycle of alternating current consists of two half cycles. For one half, currently is positive and for second half, current is negative. Therefore, for an a.c. cycle, the net value of current average out to zero. While for the half cycle, the value of current is different at different points. Hence, the alternating current cannot be measured by D.C. ammeter.

    Question 10
    CBSEENPH12039729

    In an LCR series a.c. circuit, the voltage across each of the components, L, C and R is 50 V. The voltage across the LC combination will be

    • 50 V

    • 50√2 V

    • 100 V

    • 0 V(zero)

    Solution

    D.

    0 V(zero)

    In an LCR series a.c. circuit, the voltage across inductor L leads the current by 900 and the voltage across capacitor C lags behind the current by 900. Hence, the voltage across LC combination will be zero.

    Question 11
    CBSEENPH12039732

    In a LCR circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to

    • 4L

    • 2L

    • L/2

    • L/4

    Solution

    C.

    L/2

    In the condition of resonance XL = XC
    or
    ωL =1/ ωC  = .................. (i)
    Since, resonant frequency remains unchanged
    So, square root of LC = constant
    or LC = constant
    ∴ L1 C1 = L2 C2 ⇒ L×C = L2× 2C
    L2 = L/2

    Question 15
    CBSEENPH12040128

    An inductor 20 mH, a capacitor 100 µF and a resistor 50 Ω are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is

    • 0.79 W

    • 0.43

    • 1.13

    • 2.74 W

    Solution

    A.

    0.79 W

    Power dissipated in an LCR series connected to an a.c source of emf E

    P = Ermsirms cosϕ = E2rmsRZ2 = Erms2 RR2+ωL-1Cω2=1022 x 50(50)2 +314 x 20 x 10-3 - 1314 x 100 x 10-62solving we get, P = 0.79 W

    Question 16
    CBSEENPH12047709

    An L.C. circuit is in the state of resonance. If C = 0.1F and L = 0.25 Henry. Neglecting ohmic resistance of circuit. What is the frequency of oscillations?

    • 1007 Hz

    • 100 Hz

    • 100 Hz

    • 500 Hz

    Solution

    A.

    1007 Hz

    From the formula the frequency of oscillation is

    f = 12π LC  =12π0.25×0.1×10-6  f=1007 Hz

    Question 17
    CBSEENPH12047711

    When temperature of an ideal gas is increased from 27°C to 227°C, its rms speed is changed from 400 m/s to vs Then, vs is

    • 516 m/s

    • 450 m/s

    • 310 m/s

    • 746 m/s

    Solution

    A.

    516 m/s

    We know that the rms speed is directly proportional to square root of temperature.

    vrms  THence, vrms(1)vrms(2) =T1T2400vrms(2) =27 + 273227 + 273             =300500vrms(2) = 500300×400= 1.29  ×  400            = 516.39 m/s  516 m/s    

    Sponsor Area

    Question 18
    CBSEENPH12047723

    In a L-C-R circuit, the capacitance is made one-fourth, then what should be changed in inductance, so that the circuit remains in resonance?

    • 8 IMES

    • 1/4 times

    • 2 times

    • 4 times

    Solution

    D.

    4 times

    In a RLC circuit , resonance phenomenon occurs when XL =XC or ωo = 1LC. 

    Resonance frequency =12πLCFor the circuit to remain in resonance LC  =constantL1C1 = L2C2 LC = L2 C4L2 = 4L

    Inductance should be changed to four times.

    Question 19
    CBSEENPH12047742

    In a circuit, L, C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45°. The value of C is

    • 12πf 2πf L + R

    • 1πf 2πf L + R

    • 12πf 2πf L - R

    • 1πf 2πf L - R

    Solution

    C.

    12πf 2πf L - R

    tan ϕ = ωL - 1ωCR 

    ϕ being the angle by which the current leads the voltage.Given:- ϕ = 45o tan 45o = ωL - 1ωCR      1   =  ωL - 1ωCR                  ..... tan 45o = 1      R  =    ωL - 1ωC ωC    =   1ωL - R     C  =  1ω ωL - R  C    =    12πf 2πf L -R

    In series resonance, L-C-R circuit 1ωCRis greater than unity.

    Question 20
    CBSEENPH12047804

    In a series L-C-R circuit, resistance R = 10 Ω and the impedance Z = 10 Ω. The phase difference between the current and the voltage is

    • 0o

    • 30o

    • 45o

    • 60o

    Solution

    A.

    0o

    Impedance (Z ) is an expression of the opposition that an electronic component, circuit, or system offers to alternating and/or direct electric current. Impedance is a vector ( two- dimensional ) quantity consisting of two independent scalar ( one dimensional ) phenomena: resistance and reactance.

    Impedance,

            z = R2 + XC - XL2

    ∴        10 = 102+ XC - XL2

    ⇒        100 = 100 + XC - XL2

    ⇒        XL - XC = 0                            ..... (i)

    Let ϕ is the phase difference between current and voltage

    ∴        tan ϕ = XC - XLR

               tan ϕ = XC - XLR

    ⇒              ϕ = 0                    [ from equation (i) ]

    Question 21
    CBSEENPH12047805

    The impedance of a circuit, when a resistance R and an inductor of inductance L are connected in series in an AC circuit of frequency f, is

    • R + 2 π2 f2 L2

    • R + 4π2 f2 L2

    • R2 + 4 π2 f2 L2

    • R2 + 2π2 f2 L2

    Solution

    C.

    R2 + 4 π2 f2 L2

    In L-R circuit,

         Z = R2 + X L2

    Here, XL = ω L = 2 π f L 

    ⇒    Z = R2 + 4 π2 f2 L2

    Question 22
    CBSEENPH12047817

    A choke is preferred to a resistance for limiting current in AC circuit because

    • choke is cheap

    • there is no wastage of power

    • choke is compact in size

    • choke is a good absorber of heat

    Solution

    B.

    there is no wastage of power

    In an AC circuit, the coil of high inductance and negligible resistance used to control current, is called the choke coil. The power factor of such a coil is given by

                     cosϕ = RR2 + ω2 L2

                             ≈ RωL                         ...( as R<< ωL )

    As R << ωL, cosϕ is very small.

    Thus, the power absorbed by the coil is very small. The only loss of energy is due to hysteresis in the iron core, which is much less than the loss of energy in the resistance that can also reduce the current if placed instead of the choke coil.

    Question 23
    CBSEENPH12047825

    Induced emf in the coil depends upon

    • conductivity of coil

    • amount of flux

    • rate of change of linked flux

    • resistance of coil

    Solution

    C.

    rate of change of linked flux

    According to Faraday's law, "the induced emf in a closed loop equals the time rate of change of magnetic flux through the loop".

    i.e                e = -Bdt

    Hence, induced emf in a coil depends on the rate of change of flux.

    Question 24
    CBSEENPH12047837

    In an L-C-R circuit inductance is changed from L to L/2.To keep the same resonance frequency, C should be changed to

    • 2 C

    • C2

    • 4 C

    • C4

    Solution

    A.

    2 C

    An LCR circuit is an electrical circuit consisting of Resistor (R), an inductor (L) and a capacitor (C), connected in series or in parallel.

           

           ω = 1L C

    When L is changed, C must be changed to 2C to keep to ω same i.e to balanced equation.

    Question 25
    CBSEENPH12047850

    220 V, 50 Hz, AC source is connected to an inductance of 0.2 Hand a resistance of 20 Ω in series. Whatis the current in the circuit?

    • 3.33 A

    • 33.3 A

    • 5 A

    • 10 A

    Solution

    A.

    3.33 A

    From the formula

                 I =  VR2 + ωL2

                I = 200202 + 2π × 50 × 0.22

               I = 20066

               I = 3.33 A

    Question 26
    CBSEENPH12047869

    If we change the value of R, then

           

    • voltage does not change on L

    • voltage does not change on LC combination

    • voltage does not change on C

    • voltage changes on LC combination

    Solution

    D.

    voltage changes on LC combination

    If we change the value of R, then voltage on LC combination changes.

    Question 27
    CBSEENPH12047872

    Real power consumption in a circuit is least when it contains.

    • high R, low L

    • high R, high L

    • low R, high L

    • high R, low C

    Solution

    C.

    low R, high L

    Power is the voltage times the current

    Power consumption = lrms Vrms cos ϕ

    Where, cosϕ is power factor.

                  cosϕ = RZ

    For cosϕ to be minimum, R should be minimum and Z should be higher. So, higher Z (impedance), L (inductance) should be high.

    Question 28
    CBSEENPH12047886

    The rms speed (in m/s) of oxygen molecules of the gas at temperature 300 K, is

    • 483

    • 504

    • 377

    • 346

    Solution

    A.

    483

    The rms speed is given by

              Vrms3 RTM0

    where R = gas constant

              T = Temperature of the gas

       and  M0 = Molecular weight of the gas

                   Vrms3 × 8.3 × 30032 × 10-3

                   Vrms = 483 m/s

    Question 29
    CBSEENPH12047889

    An inductor (L = 20 H), a resistor (R = 100 Ω) and a battery (E = 10 V) are connected in series. After a long time, the circuit is short-circuited and then the battery is disconnected. Find the current in the circuit at 1 ms after short circuiting.

    • 4.5 x 105 A

    • 3.2x 10-5 A

    • 9.8 × 10-5 A

    • 6.7 × 10-4 A

    Solution

    D.

    6.7 × 10-4 A

    The initial current i = i0ER

                                       = 10100

                                  i = 0.10 A

    the time constant

                       τ = LR

                          = 20 mH100 Ω

                     τ = 0.20 ms

    The current  at  t = 1 ms is 

             i = i0-t /Τ

    ⇒       i = ( 0.10 ) e- 1 ms0.20 ms

    ⇒       i = ( 0.10 ) e-5

    ⇒       i = (0.10) × 0.0067

    ⇒       i = 6.7 × 10-4 A

    Question 30
    CBSEENPH12047904

    The magnetic flux through each turn of a coil having 200 turns is given as (t2 - 1) x 10-3 Wb, where t is in second. The emf induced in the coil at t = 3s is

    • 0.7 V

    • 1.2 V

    • 0.8 V

    • 0.9 V

    Solution

    C.

    0.8 V

    Given magnetic flux

             ϕ = t2 - 2t  × 10-3

    On differentiating with respect to time t

           dt =  2t - 2  × 10-3

     At time t = 3s

         dt t = 3s = 2 × 3 - 2  ×10-3 

                      = 4 × 10-3 Wb s-1

    Now 

           e = N dt

                = 200 × 4 × 10-3

                 = 800 × 10-3

    ⇒     e = 0.8 V

    Question 31
    CBSEENPH12047908

    A beam of light travelling along X-axis is described by the electric field Ev = 600 Vm. sinω t - xc, the maximum magnetic force on a charge q = 2e, moving along Y-axis with the speed of 3 × 108 m/s is 

    • 19.2 × 10-17 N

    • 1.92 × 10-17

    • 0.192

    • None of these

    Solution

    A.

    19.2 × 10-17 N

    Given, electric field

        Ey600 vm. sinω t - xc                       ......(i)

    We know that

        E = E0 sinω t - xc                              .........(ii)

    On comparing Eqs. (i) and (ii), we get

        E0 = 600 V/m

    and C = 3 × 108 m/s

    Maximum magnetic field

        B0E0C

             = 6003 × 108

       B0 = 2 × 10-6 T

    ∴  Fm = q V B0

              = 2 ev B0

              = 2 × 1.6 × 10-19 × 3 × 108 × 2 × 10-6

         Fm = 19.2 × 10-17

    Question 32
    CBSEENPH12047912

    An L-C-R series circuit with a resistance of 100 Ω is connected to 200 V (AC source) and angular frequency 300 rad/s. When only the capacitor is removed, then the current lags behind the voltage by 60°. When only the inductor is removed the current leads the voltage by 60°. The average power dissipated in original L-C-R circuit is

    • 50 W

    • 100 W

    • 200 W

    • 400 W

    Solution

    D.

    400 W

    Given:-

        R = 100 Ω,

        V = 200 V,

        θ = 60°,

        ω = 300 rad/s

    ∴ Phase angle 

           tanθ = XLR = XCR

    ⇒     tan 60oXLXR = XCR

    ∴     XL = XC3 R

          Z = R2 + 3 R - 3 R 2

          Z = R

    So average power

        P = V2R

          = 200 × 200100

       P = 400 W

    Question 33
    CBSEENPH12047939

    In an LCR circuit as shown in figure, both switches are open initially. Now switch S1 is closed, S2 kept open. (q is charge on the capacitor and τ = RC is capacitive time constant). Which of the following statement is correct?       

                                    

    • At t = τ2, q = CV ( 1 - e-1 )

    • Work done by the battery is half of the energy dissipated in the resistor

    • At t = τ, q = CV/2

    • At t = τ, q = CV ( 1 - e-2 )

    Solution

    D.

    At t = τ, q = CV ( 1 - e-2 )

    As switch S1 is closed and switch S2 is kept open. Now, capacitor is charging through a resistor R. Charge on a capacitor at any time t is

            q = q0 ( 1 - e-t/τ

    And  q0 = CV

     At  t = τ2,

          q = CV ( 1 - e-t/2τ )

              = CV ( 1 - e-1/2 )

    At t = τ

          q = CV ( 1 - e-τ/τ )

              = CV ( 1 - e-1 )

    At t = 2τ

          q = CV ( 1 - e-2t /τ )

              = CV ( 1 - e-2 )

    Question 34
    CBSEENPH12047954

    A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. The
    current induced in the coil is

    (Horizontal component of the earth's magnetic field at that place is 3.0 x 10-5 T)

    • 1.9 × 10-3 A

    • 2.9 × 10-3 A

    • 3.9 × 10-3 A

    • 4.9 × 10-3 A

    Solution

    A.

    1.9 × 10-3 A

    Initial magnetic flux through the coil

       ϕi = BH Acosθ

           = 3.0 × 10-5 × ( π × 10-2 ) × cos0o

        ϕi = 3 π × 10-7 Wb

    Final magnetic flux after the rotation

        ϕf = 3.0 × 10-5 × ( π × 10-2 )×cos180o

           = 3π × 10-7 Wb

    lnduced  emf, 

           ε = - Ndt

              = - N ϕf - ϕit

        ε = 500 - 3 π × 10-7 - 3π ×10-7 0.25

            = 500 × 6 π × 10-70.25

         = 3.8 × 10-3 V

     Current is given by

      I = εR

        = 3.8 × 10-32 Ω

      I = 1.9 × 10-3 A

    Question 35
    CBSEENPH12047979

    A series resonant LCR circuit has a quality factor ( Q-factor) 0.4. If  R = 2 kΩ,  C = 0.1 μF, then the value of inductance is

    • 0.1 H

    • 0.064 H

    • 2 H

    • 5 H

    Solution

    B.

    0.064 H

    The amplitude of the current is maximum at resonant frequency ω. The quantity factor is given by

          Q = ωoLR = 1ωoCR

    where ωo = resonant frequency

    Given:- R = 2kΩ = 2 × 103 Ω

                C = 0.1 μF = 0.1 × 10-6 F

    The quantity factor Q defined by

                Q = 1R LC

                LC = (QR)2

    ∴          L = ( 0.4 × 2 × 103 )2 × 0.1 × 10-6

                L = 0.064 H

    Question 36
    CBSEENPH12048000

    Assertion:  At resonance, LCR series circuit have a minimum current. 

    Reason:  At resonance, in LCR series circuit, the current  and  e.m.f are not in phase with each other.

    • 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.

    Solution

    D.

    If both assertion and reason are false.

    At resonance

              X= XC or   ωL = 1ω C

    Because of this impedance of LCR series circuit be equal to resistance of circuit

               Z = R2 + XL - XC2 

    Therefore from I = EZ = ER, at resonance, current in LCR circuit is maximum. Correspondingly phase angle is also equal to zero. Therefore emf and current are in phase in LCR series circuit.

    Mock Test Series

    Sponsor Area

    Sponsor Area

    NCERT Book Store

    NCERT Sample Papers

    Entrance Exams Preparation