Exercises — Alternating Current

1. 13 marksNCERT Exercises, Chapter 7

   A $100\Omega$ resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle?

   Hint. For a purely resistive circuit Ohm's law applies to rms values directly, and voltage and current stay in phase.

   For a purely resistive circuit the current and voltage remain in phase, so Ohm's law applies to the rms values without any phase factor.

   (a) $I_{\text{rms}}=\frac{V_{\text{rms}}}{R}=\frac{220}{100}=2.2$ A

   (b) Since the phase difference is zero, the power factor $\cos \varphi =1$ and the average power is simply:

   $P=V_{\text{rms}}{I}_{\text{rms}}\cos \varphi =220\times 2.2\times 1=484\text{W}$

   Unlike an inductor or capacitor, a resistor dissipates energy throughout the cycle, so the net power over a full cycle is non-zero.

   ✦ (a) I_rms = 2.2 A (b) P = 484 W

2. 22 marksNCERT Exercises, Chapter 7

   (a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current?

   Hint. The rms and peak values differ by a factor of the square root of two.

   For a sinusoidal ac quantity the rms and peak values are related by a factor of $\sqrt{2}$:

   $V_{\text{rms}}=\frac{V_0}{\sqrt{2}},I_0=\sqrt{2}{I}_{\text{rms}}$

   (a) $V_{\text{rms}}=\frac{300}{\sqrt{2}}\approx 212$ V

   (b) $I_0=10\times \sqrt{2}\approx 14.1$ A

   The factor arises because the rms value averages the square of a sinusoid over a cycle, which gives half the squared amplitude. Note the direction of the conversion differs between the two parts, so it is worth writing down which quantity is being sought.

   ✦ (a) V_rms = 212 V (b) I_peak = 14.1 A

3. 33 marksNCERT Exercises, Chapter 7

   A 44 mH inductor is connected to a 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

   Hint. Find the inductive reactance first, remembering it grows with frequency.

   An inductor opposes ac through its inductive reactance:

   $X_L=2\pi fL=2\pi \times 50\times 44\times 10^{-3}$

   $X_L\approx 13.8\Omega$

   The rms current then follows from the ac form of Ohm's law:

   $I_{\text{rms}}=\frac{V_{\text{rms}}}{X_L}=\frac{220}{13.8}\approx 15.9\text{A}$

   Since $X_L$ is proportional to frequency, an inductor passes low frequencies readily and increasingly blocks high ones — the opposite of a capacitor.

   ✦ X_L = 13.8 ohm; I_rms = 15.9 A

4. 43 marksNCERT Exercises, Chapter 7

   A $60\mu$F capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

   Hint. Capacitive reactance falls as the frequency rises, the reverse of an inductor.

   A capacitor opposes ac through its capacitive reactance:

   $X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi \times 60\times 60\times 10^{-6}}$

   $X_C\approx 44.2\Omega$

   The rms current is therefore:

   $I_{\text{rms}}=\frac{V_{\text{rms}}}{X_C}=\frac{110}{44.2}\approx 2.49\text{A}$

   Because $X_C$ is inversely proportional to frequency, a capacitor blocks dc entirely and passes high frequencies easily — exactly opposite to the inductor of the previous question.

   ✦ X_C = 44.2 ohm; I_rms = 2.49 A

5. 53 marksNCERT Exercises, Chapter 7

   In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle? Explain your answer.

   Hint. Consider the phase difference between current and voltage in a pure inductor and a pure capacitor.

   In both circuits the net power absorbed over a complete cycle is zero.

   The reason is the phase relationship. In a pure inductor the current lags the voltage by $90^{\circ }$; in a pure capacitor it leads by $90^{\circ }$. In either case the phase difference is $\varphi =90^{\circ }$, so the power factor is:

   $\cos \varphi =\cos 90^{\circ }=0$

   and the average power $P=V_{\text{rms}}{I}_{\text{rms}}\cos \varphi$ vanishes.

   Physically, energy is not consumed but exchanged. During one quarter cycle the inductor stores energy in its magnetic field, or the capacitor in its electric field, and during the next quarter it returns that energy to the source. Over a full cycle the two contributions cancel exactly.

   This is why ideal inductors and capacitors are called non-dissipative elements, in contrast with the resistor of Exercise 7.1 which consumed 484 W.

   ✦ Zero in both cases, because the phase difference is 90 degrees so the power factor is zero; energy is stored and returned each cycle rather than dissipated

6. 62 marksNCERT Exercises, Chapter 7

   A charged $30\mu$F capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

   Hint. An LC circuit oscillates at its natural frequency, determined only by L and C.

   An LC circuit undergoes free oscillations as energy shuttles between the capacitor's electric field and the inductor's magnetic field.

   The angular frequency of these oscillations is:

   $\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{27\times 10^{-3}\times 30\times 10^{-6}}}$

   $\omega =\frac{1}{\sqrt{8.1\times 10^{-7}}}=\frac{1}{9.0\times 10^{-4}}$

   $\omega \approx 1.1\times 10^3{\text{rad s}}^{-1}$

   The frequency depends only on $L$ and $C$, not on how much charge the capacitor initially carried, since that affects the amplitude rather than the rate of oscillation.

   ✦ omega = 1.1 x 10^3 rad/s

7. 73 marksNCERT Exercises, Chapter 7

   A series LCR circuit with $R=20\Omega$, $L=1.5$ H and $C=35\mu$F is connected to a variable-frequency 200 V ac supply. When the frequency equals the natural frequency of the circuit, what is the average power transferred in one complete cycle?

   Hint. At resonance the reactances cancel, so the impedance reduces to the resistance alone.

   When the supply frequency equals the circuit's natural frequency, the circuit is at resonance.

   At resonance $X_L=X_C$, so the two reactances cancel exactly and the impedance falls to its minimum value:

   $Z=\sqrt{R^2+(X_L-X_C{)}^2}=R=20\Omega$

   The circuit therefore behaves as if purely resistive, with the current in phase with the voltage and a power factor of 1.

   $P=\frac{V_{\text{rms}}^2}{R}=\frac{(200{)}^2}{20}=2000\text{W}$

   This is the maximum power the circuit can absorb at any frequency, since the impedance is smallest at resonance. Note the values of $L$ and $C$ are not needed, because only their equality matters here.

   ✦ P = 2000 W, the maximum possible, since at resonance Z reduces to R = 20 ohm

8. 85 marksNCERT Exercises, Chapter 7

   A series LCR circuit is connected to a variable-frequency 230 V source with $L=5.0$ H, $C=80\mu$F, $R=40\Omega$. (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance and the amplitude of current at resonance. (c) Determine the rms potential drops across the three elements, and show that the drop across the LC combination is zero at resonance.

   Hint. Find the resonant frequency, then evaluate each reactance at that frequency.

   (a) At resonance the reactances are equal, giving:

   ${\omega }_r=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5.0\times 80\times 10^{-6}}}=\frac{1}{\sqrt{4\times 10^{-4}}}=50{\text{rad s}}^{-1}$

   so $f_r=\frac{50}{2\pi }\approx 7.96$ Hz.

   (b) At resonance the impedance is just the resistance, $Z=R=40\Omega$.

   The current amplitude is:

   $I_0=\frac{V_0}{Z}=\frac{230\sqrt{2}}{40}\approx 8.13\text{A}$

   (c) The rms current is $I_{\text{rms}}=\frac{230}{40}=5.75$ A. Evaluating the reactances at ${\omega }_r=50$:

   $X_L={\omega }_{r}L=50\times 5.0=250\Omega ,X_C=\frac{1}{{\omega }_{r}C}=\frac{1}{50\times 80\times 10^{-6}}=250\Omega$

   Element	rms potential drop
   $V_R$	$5.75\times 40=230$ V
   $V_L$	$5.75\times 250=1437.5$ V
   $V_C$	$5.75\times 250=1437.5$ V

   The drops across $L$ and $C$ are each more than six times the supply voltage, which is characteristic of resonance and is why component voltage ratings matter.

   However, they are exactly antiphase, so across the LC combination together:

   $V_{LC}=V_L-V_C=1437.5-1437.5=0$

   The whole supply voltage therefore appears across the resistor, consistent with $V_R=230$ V.

   ✦ (a) omega_r = 50 rad/s, f_r = 7.96 Hz (b) Z = 40 ohm, current amplitude 8.13 A (c) V_R = 230 V, V_L = V_C = 1437.5 V, and V_LC = 0 since they are antiphase

Solutions written by the tuition.in editorial team and checked against leph107.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.