Alternating Current Subjective Problems Important Questions for CBSE Class 12 Physics

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CBSE Class 12 Physics Alternating Current Subjective Problems. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Alternating Current – Subjective Problems

Find the value of inductance which should be connected in series with a capacitance of 0.5 μF, a resistance of 10 Ω and an a.c. source of 50 c/s, so that the power factor of the circuit is unity.

Solution:

For unity power factor, the circuit must be in resonance: \(\omega L=\dfrac{1}{\omega C}\), so

\[L=\frac{1}{\omega^2 C}=\frac{1}{(2\pi\times50)^2\times0.5\times10^{-6}}=20.2\ \text{H}\]

Answer: L ≈ 20.2 H.

A coil of resistance 60 Ω and inductance 3 H is connected in series with a condenser of 4 μF and an a.c. supply of 200 V and 50 c/s. Calculate (a) the impedance of the circuit, (b) the phase difference between current and voltage, (c) the p.d. across the inductance coil, and (d) the p.d. across the capacitor.

Solution:

Take \(\omega=2\pi f\approx314\ \text{rad s}^{-1}\). Then

\[X_L=\omega L=314\times3=942\ \Omega,\qquad X_C=\frac{1}{\omega C}=\frac{1}{314\times4\times10^{-6}}=796.2\ \Omega\]

\[Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{60^2+145.8^2}=157.7\ \Omega\]

\[\tan\phi=\frac{X_L-X_C}{R}=\frac{145.8}{60}=2.43\ \Rightarrow\ \phi\approx67.7^\circ\]

\[I=\frac{V}{Z}=\frac{200}{157.8}=1.27\ \text{A}\]

\[V_L=IX_L=1.27\times942\approx1193\ \text{V},\qquad V_C=IX_C=1.27\times796\approx1009\ \text{V}\]

Answer: (a) 157.8 Ω (b) 67.7° (the current lags behind the voltage) (c) about 1193 V (d) about 1009 V.

A 200 km long telegraph wire has a capacity of 0.014 μF/km. If it carries an alternating current of frequency 5 kHz, find the value of the inductance which should be connected in series so that the impedance is minimum.

Solution:

Total capacitance \(C=200\times0.014=2.8\ \mu\text{F}\). The impedance is minimum at resonance, \(\omega L=\dfrac{1}{\omega C}\):

\[L=\frac{1}{\omega^2C}=\frac{1}{(2\pi\times5000)^2\times2.8\times10^{-6}}=3.6\times10^{-4}\ \text{H}\]

Answer: L ≈ 0.36 mH.

A circuit draws a power of 550 W from a source of 220 V, 50 Hz. The power factor of the circuit is 0.8 and the current lags in phase behind the potential difference. If the power factor of the circuit is to be increased to 1.0, find the value of the capacitance which should be connected (in series).

Solution:

The current is \(I=\dfrac{P}{V\cos\phi}=\dfrac{550}{220\times0.8}=3.125\ \text{A}\), so the impedance is \(Z=\dfrac{V}{I}=70.4\ \Omega\).

\[R=Z\cos\phi=56.3\ \Omega,\qquad X=Z\sin\phi=70.4\times0.6=42.2\ \Omega\]

The circuit is inductive with reactance \(X_L=42.2\ \Omega\). For unity power factor, a series capacitor must have \(X_C=X_L\):

\[C=\frac{1}{\omega X_C}=\frac{1}{314\times42.2}=7.5\times10^{-5}\ \text{F}\]

Answer: C ≈ 75 μF (in series).

Prove that in a series LCR circuit, the frequencies \(f_1\) and \(f_2\) at which the current amplitude falls to \(1/\sqrt2\) of the current at resonance are separated by an interval equal to \(R/2\pi L\).

Solution:

The current amplitude is \(I_0=\dfrac{V_0}{\sqrt{R^2+(\omega L-1/\omega C)^2}}\), and at resonance it is \(V_0/R\). It falls to \(1/\sqrt2\) of this value when \(\left|\omega L-\dfrac{1}{\omega C}\right|=R\). So for the two frequencies \(\omega_2>\omega_0>\omega_1\):

\[\omega_2L-\frac{1}{\omega_2C}=R\qquad(1)\qquad\qquad \frac{1}{\omega_1C}-\omega_1L=R\qquad(2)\]

Equating the left sides of (1) and (2): \(L(\omega_1+\omega_2)=\dfrac1C\left(\dfrac{1}{\omega_1}+\dfrac{1}{\omega_2}\right)=\dfrac{\omega_1+\omega_2}{C\,\omega_1\omega_2}\), so \(\omega_1\omega_2=\dfrac{1}{LC}\).

Adding (1) and (2): \((\omega_2-\omega_1)L+\dfrac1C\cdot\dfrac{\omega_2-\omega_1}{\omega_1\omega_2}=2R\). Substituting \(\dfrac{1}{C\omega_1\omega_2}=L\):

\[(\omega_2-\omega_1)\,2L=2R\ \Rightarrow\ \omega_2-\omega_1=\frac{R}{L}\ \Rightarrow\ f_2-f_1=\frac{R}{2\pi L}\]

Answer: Hence proved: \(f_2-f_1=\dfrac{R}{2\pi L}\).

A series circuit consists of a resistance, an inductance and a capacitance. The applied voltage and the current at any time \(t\) are given by \(E=141.4\cos(3000t-10^\circ)\) and \(I=5\cos(3000t-55^\circ)\). If the inductance is 0.01 H, find the values of the resistance and the capacitance.

Solution:

The phase difference is \(\phi=-10^\circ-(-55^\circ)=45^\circ\) (the voltage leads the current). The peak values are \(E_0=141.4\ \text{V}\) and \(I_0=5\ \text{A}\), so

\[Z=\frac{E_0}{I_0}=28.28\ \Omega,\qquad R=Z\cos45^\circ=20\ \Omega,\qquad X=Z\sin45^\circ=20\ \Omega\]

Now \(X=\omega L-\dfrac{1}{\omega C}\) with \(\omega L=3000\times0.01=30\ \Omega\), so \(\dfrac{1}{\omega C}=30-20=10\ \Omega\):

\[C=\frac{1}{3000\times10}=3.33\times10^{-5}\ \text{F}\]

Answer: R = 20 Ω and C ≈ 33.3 μF.

A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply. Find the maximum current in the circuit. Also find the time lag between the maximum current and the maximum voltage.

Solution:

\[X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi\times60\times100\times10^{-6}}=26.5\ \Omega,\qquad Z=\sqrt{40^2+26.5^2}=48\ \Omega\]

\[I_0=\frac{\sqrt2\,V_{rms}}{Z}=\frac{1.414\times110}{48}=3.24\ \text{A}\]

The phase angle is \(\phi=\tan^{-1}\dfrac{X_C}{R}=\tan^{-1}(0.663)=33.5^\circ=0.585\ \text{rad}\) (the current leads the voltage), so the time difference is

\[t=\frac{\phi}{\omega}=\frac{0.585}{2\pi\times60}=1.55\times10^{-3}\ \text{s}\]

Answer: Maximum current 3.24 A; the current leads the voltage by \(1.55\times10^{-3}\ \text{s}\).

A 20 V, 5 W lamp is used on a.c. mains of 220 V, 50 Hz. Find (a) the values of (i) the capacitor and (ii) the inductor to be put in series with the lamp to run it, and (b) the value of the pure resistance which should be connected in series with the lamp to run it.

Solution:

The rated current of the lamp is \(I=\dfrac{P}{V}=\dfrac{5}{20}=0.25\ \text{A}\), and its resistance is \(R_L=\dfrac{20}{0.25}=80\ \Omega\). To run the lamp at this current from the 220 V mains, the total impedance must be \(Z=\dfrac{220}{0.25}=880\ \Omega\).

(a) With a capacitor or an inductor alone in series: \(X=\sqrt{Z^2-R_L^2}=\sqrt{880^2-80^2}=876.4\ \Omega\).

\[C=\frac{1}{\omega X_C}=\frac{1}{314\times876.4}=3.63\ \mu\text{F},\qquad L=\frac{X_L}{\omega}=\frac{876.4}{314}=2.79\ \text{H}\]

(b) With a pure resistance, \(R+80=880\):

\[R=\frac{220-20}{0.25}=800\ \Omega\]

Answer: (a) (i) C ≈ 3.6 μF (ii) L ≈ 2.8 H (b) R = 800 Ω.

A choke coil is required to operate an arc lamp at 160 V r.m.s. and 50 Hz. The arc lamp has an effective resistance of 5 Ω when running at 10 A r.m.s. Find the inductance of the choke coil. If the same arc lamp is operated on 160 V d.c., find the additional resistance required and the power loss if d.c. is used.

Solution:

\[Z=\frac{160}{10}=16\ \Omega,\qquad X_L=\sqrt{Z^2-R^2}=\sqrt{16^2-5^2}=15.2\ \Omega\]

\[L=\frac{X_L}{2\pi f}=\frac{15.2}{314}=0.048\ \text{H}\approx0.05\ \text{H}\]

On d.c., the total resistance needed is \(160/10=16\ \Omega\), so the additional resistance is \(16-5=11\ \Omega\), and the power lost in it is

\[P=I^2R=10^2\times11=1100\ \text{W}\]

Answer: L ≈ 0.05 H; additional resistance 11 Ω; power loss 1100 W.

A series LCR circuit with L = 0.12 H, C = 480 nF and R = 23 Ω is connected to a 230 V variable frequency supply. (a) What is the source frequency for which the current amplitude is maximum? Find this maximum value of current. (b) What is the source frequency for which the average power absorbed by the circuit is maximum? Obtain the value of the maximum power. (c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? (d) What is the Q factor of the circuit?

Solution:

(a) The current is maximum at resonance:

\[\omega_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.12\times480\times10^{-9}}}=4167\ \text{rad s}^{-1}\ (\nu_0\approx663\ \text{Hz})\]

\[I_0=\frac{V_0}{R}=\frac{230\sqrt2}{23}=14.14\ \text{A}\]

(b) The power is also maximum at resonance, i.e. at \(\omega_0=4167\ \text{rad s}^{-1}\):

\[P_{max}=I_{rms}^2R=\frac{V_{rms}^2}{R}=\frac{230^2}{23}=2300\ \text{W}\]

(c) The power is half the maximum when the current amplitude is \(I_0/\sqrt2\), i.e. at \(\omega=\omega_0\pm\dfrac{R}{2L}\):

\[\frac{R}{2L}=\frac{23}{2\times0.12}=95.8\ \text{rad s}^{-1}\ \Rightarrow\ \omega=4167\pm95.8=4263\ \text{and}\ 4071\ \text{rad s}^{-1}\]

(d) \(Q=\dfrac{\omega_0L}{R}=\dfrac{4167\times0.12}{23}=21.74\)

Answer: (a) 4167 rad/s, 14.14 A (b) 4167 rad/s, 2300 W (c) 4263 rad/s and 4071 rad/s (d) Q = 21.74.

A box P and a coil Q are connected in series with an a.c. source of variable frequency. The e.m.f. of the source is constant at 10 V. Box P contains a capacitance of 1 μF in series with a resistance of 32 Ω. Coil Q has a self-inductance of 4.9 mH and a resistance of 68 Ω. The frequency is adjusted so that the maximum current flows in P and Q. Find the impedance of P and Q at this frequency. Also find the voltage across P and Q respectively.

Solution:

The current is maximum at resonance:

\[\omega=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{4.9\times10^{-3}\times10^{-6}}}=14286\ \text{rad s}^{-1}\]

\[X_C=\frac{1}{\omega C}=70\ \Omega,\qquad X_L=\omega L=70\ \Omega\]

\[Z_P=\sqrt{32^2+70^2}=77\ \Omega,\qquad Z_Q=\sqrt{68^2+70^2}=97.6\ \Omega\approx98\ \Omega\]

At resonance the total impedance is \(32+68=100\ \Omega\), so \(I=\dfrac{10}{100}=0.1\ \text{A}\):

\[V_P=0.1\times77=7.7\ \text{V},\qquad V_Q=0.1\times97.6\approx9.8\ \text{V}\]

Answer: Z_P ≈ 77 Ω, Z_Q ≈ 98 Ω; V_P ≈ 7.7 V, V_Q ≈ 9.8 V.

An inductor of 20 × 10⁻³ H, a capacitor of 100 μF and a resistor of 50 Ω are connected in series across a source of e.m.f. \(V=10\sin314t\). Find the energy dissipated in the circuit in 20 minutes. If the resistance is removed from the circuit and the value of the inductance is doubled, then find the variation of current with time in the new circuit.

Solution:

\[X_L=314\times0.02=6.28\ \Omega,\qquad X_C=\frac{1}{314\times100\times10^{-6}}=31.85\ \Omega\]

\[Z=\sqrt{50^2+(31.85-6.28)^2}=56.2\ \Omega,\qquad I_0=\frac{10}{56.2}=0.178\ \text{A}\]

\[E=\frac12I_0^2R\,t=\frac12(0.178)^2\times50\times1200\approx951\ \text{J}\]

In the new circuit, \(R=0\) and \(L=40\ \text{mH}\): \(X_L=314\times0.04=12.56\ \Omega\), so

\[Z=|X_L-X_C|=|12.56-31.85|=19.3\ \Omega,\qquad I_0=\frac{10}{19.3}=0.518\ \text{A}\]

The circuit is purely capacitive in effect (\(X_C>X_L\)), so the current leads the voltage by \(\pi/2\):

\[I=0.518\sin\left(314t+\frac{\pi}{2}\right)=0.518\cos314t\ \text{A}\]

Answer: Energy dissipated ≈ 951 J; the new current is \(I=0.518\cos314t\) A.

Two alternating-current generators in series are connected to a coil and a resistor as shown in the figure. The voltages across generators 1 and 2 are 6 V and 8 V respectively. If the voltage between the points a and b is 10 V and the current in the circuit is 2 A, calculate the phase difference between the voltages supplied by the two generators. (a) What is the impedance of the circuit? (b) If the power supplied to the coil is 12 W, calculate the resistance R and the inductive reactance \(X_L\) of the coil.

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Solution:

The voltage between a and b is the phasor sum of the two generator voltages:

\[V^2=V_1^2+V_2^2+2V_1V_2\cos\phi\ \Rightarrow\ 100=36+64+96\cos\phi\ \Rightarrow\ \cos\phi=0\ \Rightarrow\ \phi=90^\circ\]

(a) \(Z=\dfrac{V}{I}=\dfrac{10}{2}=5\ \Omega\).

(b) The power is dissipated only in the resistance: \(P=I^2R\), so

\[R=\frac{12}{2^2}=3\ \Omega,\qquad X_L=\sqrt{Z^2-R^2}=\sqrt{25-9}=4\ \Omega\]

Answer: Phase difference 90°; (a) Z = 5 Ω; (b) R = 3 Ω and X_L = 4 Ω.

A coil of radius 100 cm has 490 turns and it lies in a vertical plane. The ends of the coil are connected to an external resistor of 72 Ω. If the coil rotates about its vertical diameter at a speed of 3600 rpm, calculate the average power delivered to the resistor. The horizontal component of the earth's magnetic field is 0.2 oersted. Neglect the resistance and the inductance of the coil.

Solution:

Here \(N=490\), \(A=\pi r^2=\pi(1)^2=3.14\ \text{m}^2\), \(B_H=0.2\ \text{oersted}=0.2\ \text{gauss}=2\times10^{-5}\ \text{T}\) and \(\omega=2\pi\times\dfrac{3600}{60}=377\ \text{rad s}^{-1}\).

\[E_0=NBA\omega=490\times2\times10^{-5}\times3.14\times377=11.6\ \text{V}\]

\[P_{av}=\frac{E_0^2}{2R}=\frac{(11.6)^2}{2\times72}\approx0.94\ \text{W}\]

Answer: The average power is about 0.94 W.

An alternating e.m.f. of angular frequency 400 rad per second is applied to a coil of inductance 20 mH and resistance 6 Ω. (a) What should be the capacitance of a condenser placed in the circuit so as to make the power factor unity? (b) Calculate the change in current due to the insertion of this condenser.

Solution:

(a) \(X_L=\omega L=400\times0.02=8\ \Omega\). For unity power factor, \(X_C=X_L\):

\[C=\frac{1}{\omega X_C}=\frac{1}{400\times8}=3.125\times10^{-4}\ \text{F}=312.5\ \mu\text{F}\]

(b) Before inserting the condenser, \(Z_1=\sqrt{6^2+8^2}=10\ \Omega\); afterwards, \(Z_2=R=6\ \Omega\). If the applied voltage is \(V\):

\[\Delta I=\frac{V}{6}-\frac{V}{10}=\frac{V}{15}\ \Rightarrow\ \frac{I_2}{I_1}=\frac{10}{6}=\frac53\]

Answer: (a) C = 312.5 μF. (b) The current increases by \(V/15\) (an increase of about 67%, i.e. it becomes 5/3 times the earlier value).

An alternating voltage of 220 V r.m.s. at a frequency of 40 cycles/s is supplied to a circuit containing a pure inductance of 0.01 H and a pure resistance of 6 Ω in series. Calculate (a) the current, (b) the potential difference across the resistance, (c) the power factor.

Solution:

\[X_L=2\pi\times40\times0.01=2.51\ \Omega,\qquad Z=\sqrt{6^2+2.51^2}=6.5\ \Omega\]

\[I=\frac{220}{6.5}=33.8\ \text{A},\qquad V_R=IR=33.8\times6=203\ \text{V},\qquad \cos\phi=\frac{R}{Z}=\frac{6}{6.5}=0.92\]

Answer: (a) 33.8 A (b) about 203 V (c) 0.92.

A circuit consists of three branches in parallel across the source of an alternating e.m.f. Through the first branch, which has a pure resistor, flows a current of 10 A. The second branch has an inductance and a resistance and carries a current of 8 A, which lags behind the impressed e.m.f. by 60°. The third branch has an inductance and a resistance and carries a current of 15 A lagging 45° behind the e.m.f. Calculate the total current drawn from the source and its phase with respect to the e.m.f.

Solution:

Resolve the currents along the e.m.f. (in-phase) and perpendicular to it:

\[I_x=10+8\cos60^\circ+15\cos45^\circ=10+4+10.61=24.61\ \text{A}\]

\[I_y=8\sin60^\circ+15\sin45^\circ=6.93+10.61=17.54\ \text{A (lagging)}\]

\[I=\sqrt{I_x^2+I_y^2}=\sqrt{24.61^2+17.54^2}=30.2\ \text{A},\qquad \tan\phi=\frac{17.54}{24.61}=0.713\ \Rightarrow\ \phi=35.5^\circ\]

Answer: The total current is about 30.2 A, lagging behind the e.m.f. by about 35.5°.

A circuit with R = 70 Ω in series with a parallel combination of L = 1.5 H and C = 30 μF is driven by a 230 V supply of angular frequency 300 rad/s. Find (a) the impedance of the circuit, (b) the r.m.s. value of the total current, (c) the current amplitudes in the L and C arms of the circuit, (d) the total power input to the circuit and the power factor, (e) how the circuit will behave if \(\omega=1/\sqrt{LC}\).

Solution:

\[X_L=\omega L=300\times1.5=450\ \Omega,\qquad X_C=\frac{1}{\omega C}=\frac{1}{300\times30\times10^{-6}}=111.1\ \Omega\]

The parallel LC combination has the admittance \(Y=\dfrac{1}{X_C}-\dfrac{1}{X_L}=0.009-0.00222=0.00678\ \text{S}\) (capacitive), so its effective reactance is \(X_p=\dfrac{1}{0.00678}=147.5\ \Omega\) (capacitive).

(a) \(Z=\sqrt{70^2+147.5^2}=163.3\ \Omega\).

(b) \(I_{rms}=\dfrac{230}{163.3}=1.41\ \text{A}\).

\[I_{L0}=\frac{293.8}{450}=0.65\ \text{A},\qquad I_{C0}=\frac{293.8}{111.1}=2.64\ \text{A}\]

(d) \(P=I_{rms}^2R=(1.41)^2\times70=139\ \text{W}\) and the power factor is \(\cos\phi=\dfrac{R}{Z}=\dfrac{70}{163.3}=0.43\).

(e) If \(\omega=1/\sqrt{LC}\) (about 149 rad/s), the parallel combination is in resonance and its impedance is infinite (anti-resonance). The current falls to zero, and the circuit behaves like an open circuit.

Answer: (a) 163 Ω (b) 1.41 A (c) 0.65 A and 2.64 A (d) 139 W, power factor 0.43 (e) the current becomes zero.

A 60 μF capacitor, a 0.3 H inductor and a 50 Ω resistor are connected in series with a 120 V, 60 Hz a.c. source. Calculate (a) the impedance of the circuit, (b) the current in the circuit, (c) the power dissipated in the circuit.

Solution:

\[X_L=2\pi\times60\times0.3=113.1\ \Omega,\qquad X_C=\frac{1}{2\pi\times60\times60\times10^{-6}}=44.2\ \Omega\]

\[Z=\sqrt{50^2+(113.1-44.2)^2}=\sqrt{2500+4747}=85.1\ \Omega\]

\[I=\frac{120}{85.1}=1.41\ \text{A},\qquad P=I^2R=(1.41)^2\times50\approx99.4\ \text{W}\]

Answer: (a) 85.1 Ω (b) 1.41 A (c) about 99 W.

A 12 Ω resistance and an inductance of \(0.05/\pi\) H with negligible resistance are connected in series. Across the ends of this circuit is connected a 130 V a.c. voltage of frequency 50 per second. Calculate the a.c. current in the circuit and the potential difference across the resistance and that across the inductance.

Solution:

\[X_L=2\pi\times50\times\frac{0.05}{\pi}=5\ \Omega,\qquad Z=\sqrt{12^2+5^2}=13\ \Omega\]

\[I=\frac{130}{13}=10\ \text{A},\qquad V_R=IR=120\ \text{V},\qquad V_L=IX_L=50\ \text{V}\]

Answer: I = 10 A, V_R = 120 V, V_L = 50 V.

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Useful Resources and Notes for Class 12 Physics Chapter 07 Alternating Current

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