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NCERT Exemplar · Class 12 Physics Alternating Current

31 questions

Multiple Choice Questions (Type I) 1–7 (part 1 of 5)

  1. Exercise 1

    If the rms current in a 50\displaystyle 50 Hz ac circuit is 5\displaystyle 5 A, the value of the current 1\displaystyle 1/300\displaystyle 300 seconds after its value becomes zero is
    (a)
    52 A\displaystyle 5 \sqrt{2} \mathrm{~A}
    (b)
    53/2 A\displaystyle 5 \sqrt{3 / 2} \mathrm{~A}
    (c)
    5/6 A\displaystyle 5 / 6 \mathrm{~A}
    (d)
    5/2 A\displaystyle 5 / \sqrt{2} \mathrm{~A}
    NCERT’s answer
    (b)
    (b) \(\displaystyle 5\sqrt{3/2}\ \mathrm{A}\)Current starts from zero, so it is a sine.\[I_0=\sqrt2\,I_{rms}=5\sqrt2\ \mathrm{A},\qquad \omega=2\pi f=100\pi\ \mathrm{rad\,s^{-1}} \]\[I=I_0\sin\omega t=5\sqrt2\,\sin\!\left(100\pi\ \mathrm{rad\,s^{-1}}\times\frac{1}{300}\ \mathrm{s}\right)=5\sqrt2\,\sin\frac{\pi}{3} \]\[I=5\sqrt2\times\frac{\sqrt3}{2}=5\sqrt{\frac32}\ \mathrm{A} \]

  2. Exercise 2

    An alternating current generator has an internal resistance Rg\displaystyle R g and an internal reactance Xg\displaystyle X g. It is used to supply power to a passive load consisting of a resistance Rg\displaystyle R g and a reactance XL\displaystyle X_L. For maximum power to be delivered from the generator to the load, the value of XL\displaystyle X_L is equal to
    (a)
    zero.
    (b)
    Xg\displaystyle X_g.
    (c)
    −Xg\displaystyle -X_g.
    (d)
    Rg\displaystyle R_g.
    NCERT’s answer
    (c)
    (c) \(\displaystyle X_L=-X_g\)For a load resistance \(\displaystyle R\), the power is largest when the net reactance vanishes.\[P=I_{rms}^2R=\frac{V_{rms}^2\,R}{(R_g+R)^2+(X_g+X_L)^2} \]\[(X_g+X_L)^2=0\ \Rightarrow\ X_L=-X_g \]

  3. Exercise 3

    When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220V. This means
    (a)
    input voltage cannot be AC voltage, but a DC voltage.
    (b)
    maximum input voltage is 220V.
    (c)
    the meter reads not v\displaystyle v but <v2>\displaystyle <v^2> and is calibrated to read <v2>\displaystyle \sqrt{<v^2>}.
    (d)
    the pointer of the meter is stuck by some mechanical defect.
    NCERT’s answer
    (c)
    (c) The meter averages \(\displaystyle v^2\) and its scale is calibrated to show the rms value.\[V_{rms}=\sqrt{\langle v^2\rangle}=220\ \mathrm V \]\[V_0=\sqrt2\,V_{rms}=220\sqrt2\ \mathrm V\approx 311\ \mathrm V \]The peak is $\displaystyle 311$ V, not $\displaystyle 220$ V, so (b) fails; a steady reading does not make the input DC.

  4. Exercise 4

    To reduce the reasonant frequency in an LCR series circuit with a generator
    (a)
    the generator frequency should be reduced.
    (b)
    another capacitor should be added in parallel to the first.
    (c)
    the iron core of the inductor should be removed.
    (d)
    dielectric in the capacitor should be removed.
    NCERT’s answer
    (b)
    (b) Another capacitor in parallel adds capacitance, which lowers \(\displaystyle f_0\).\[f_0=\frac{1}{2\pi\sqrt{LC}},\qquad C_{net}=C_1+C_2 \]\[\text{(b) } C\uparrow\ \Rightarrow\ f_0\downarrow \]\[\text{(c) core removed: } L\downarrow\ \Rightarrow\ f_0\uparrow;\qquad \text{(d) dielectric removed: } C\downarrow\ \Rightarrow\ f_0\uparrow \]Generator frequency (a) does not change \(\displaystyle f_0\).

  5. Exercise 5

    Which of the following combinations should be selected for better tuning of an LCR circuit used for communication?
    (a)
    R=20Ω,L=1.5H,C=35μ F\displaystyle R=20 \Omega, L=1.5 \mathrm{H}, C=35 \mu \mathrm{~F}.
    (b)
    R=25Ω,L=2.5H,C=45μ F\displaystyle R=25 \Omega, L=2.5 \mathrm{H}, C=45 \mu \mathrm{~F}.
    (c)
    R=15Ω,L=3.5H,C=30μ F\displaystyle R=15 \Omega, L=3.5 \mathrm{H}, C=30 \mu \mathrm{~F}.
    (d)
    R=25Ω,L=1.5H,C=45μ F\displaystyle R=25 \Omega, L=1.5 \mathrm{H}, C=45 \mu \mathrm{~F}.
    NCERT’s answer
    (c)
    (c) Sharpest tuning needs the largest quality factor.\[Q=\frac{\omega_0L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}} \]\[\text{(a) } Q=\frac{1}{20\ \Omega}\sqrt{\frac{1.5\ \mathrm H}{35\times10^{-6}\ \mathrm F}}\approx 10.4 \]\[\text{(b) } Q=\frac{1}{25\ \Omega}\sqrt{\frac{2.5\ \mathrm H}{45\times10^{-6}\ \mathrm F}}\approx 9.4 \]\[\text{(c) } Q=\frac{1}{15\ \Omega}\sqrt{\frac{3.5\ \mathrm H}{30\times10^{-6}\ \mathrm F}}\approx 22.8 \]\[\text{(d) } Q=\frac{1}{25\ \Omega}\sqrt{\frac{1.5\ \mathrm H}{45\times10^{-6}\ \mathrm F}}\approx 7.3 \]

  6. Exercise 6

    An inductor of reactance 1Ω\displaystyle 1 \Omega and a resistor of 2Ω\displaystyle 2 \Omega are connected in series to the terminals of a 6\displaystyle 6 V (rms) a.c. source. The power dissipated in the circuit is
    (a)
    8\displaystyle 8 W.
    (b)
    12\displaystyle 12 W.
    (c)
    14.4\displaystyle 4 W.
    (d)
    18\displaystyle 18 W.
    NCERT’s answer
    (c)
    (c) $\displaystyle 14.4$ W\[Z=\sqrt{R^2+X_L^2}=\sqrt{2^2+1^2}=\sqrt5\ \Omega \]\[I_{rms}=\frac{V_{rms}}{Z}=\frac{6\ \mathrm V}{\sqrt5\ \Omega}=\frac{6}{\sqrt5}\ \mathrm A \]\[P=I_{rms}^2R=\frac{36}{5}\ \mathrm A^2\times2\ \Omega=14.4\ \mathrm W \]Only the resistor dissipates power; do not use \(\displaystyle V^2/R\) with the full $\displaystyle 6$ V.

  7. Exercise 7

    The output of a step-down transformer is measured to be 24\displaystyle 24 V when connected to a 12\displaystyle 12 watt light bulb. The value of the peak current is
    (a)
    1/2 A\displaystyle 1 / \sqrt{2} \mathrm{~A}.
    (b)
    2 A\displaystyle \sqrt{2} \mathrm{~A}.
    (c)
    2\displaystyle 2 A.
    (d)
    22 A\displaystyle 2 \sqrt{2} \mathrm{~A}.
    NCERT’s answer
    (a)
    (a) \(\displaystyle 1/\sqrt2\ \mathrm A\)The $\displaystyle 24$ V is an rms value and the bulb is resistive.\[P=V_{rms}I_{rms}\ \Rightarrow\ I_{rms}=\frac{12\ \mathrm W}{24\ \mathrm V}=0.5\ \mathrm A \]\[I_0=\sqrt2\,I_{rms}=\frac{\sqrt2}{2}\ \mathrm A=\frac{1}{\sqrt2}\ \mathrm A \]