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AC Circuits — MH-CET Physics MCQs with Solutions
Free MH-CET Physics AC Circuits MCQs with step-by-step solutions covering Alternating Current & RMS Values, AC through Resistor, Inductor & Capacitor, Series LCR Circuit & Impedance, Resonance & Q-factor, Power in AC Circuits, Transformers & LC Oscillations. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Alternating Current & RMS Values · easy · theory
An alternating current is one whose:
A. Magnitude and direction are both constant
B. Direction changes but magnitude stays fixed
C. Magnitude changes but direction stays fixed
D. Magnitude and direction both change periodically with time ✓ Correct
Solution: A sinusoidal alternating current completes a full reversal in each cycle, which is why its mean value over a cycle is zero.
Q2 — Alternating Current & RMS Values · easy · theory
For a sinusoidal alternating current of peak value $I_0$, the RMS value is:
A. $I_0\sqrt{2}$
B. $\dfrac{2I_0}{\pi}$
C. $\dfrac{I_0}{\sqrt{2}}$ ✓ Correct
D. $\dfrac{I_0}{2}$
Solution: Averaging $\sin^2$ over a full cycle gives one-half, so the root mean square is $\dfrac{I_0}{\sqrt{2}} \approx 0.707I_0$.
Q3 — Alternating Current & RMS Values · easy · theory
The average value of a sinusoidal alternating current over one complete cycle is:
A. $\dfrac{2I_0}{\pi}$
B. Zero ✓ Correct
C. $I_0$
D. $\dfrac{I_0}{\sqrt{2}}$
Solution: The positive and negative half cycles are identical in shape, so they cancel exactly.
Q4 — Alternating Current & RMS Values · easy · numerical
An alternating voltage is $V = 200\sqrt{2}\sin(100t)\text{ V}$. Its RMS value is:
A. $141.4\text{ V}$
B. $200\text{ V}$ ✓ Correct
C. $100\text{ V}$
D. $282.8\text{ V}$
Solution: $V_{rms} = \dfrac{V_0}{\sqrt{2}} = \dfrac{200\sqrt{2}}{\sqrt{2}} = 200\text{ V}$.
Q5 — Alternating Current & RMS Values · easy · numerical
An alternating current has a peak value of $5\text{ A}$. Its RMS value is approximately:
A. $3.54\text{ A}$ ✓ Correct
B. $7.07\text{ A}$
C. $3.18\text{ A}$
D. $2.5\text{ A}$
Solution: $I_{rms} = \dfrac{5}{\sqrt{2}} \approx 3.54\text{ A}$.
Q6 — Alternating Current & RMS Values · easy · numerical
An alternating voltage has angular frequency $100\pi\text{ rad/s}$. Its frequency is:
A. $314\text{ Hz}$
B. $25\text{ Hz}$
C. $100\text{ Hz}$
D. $50\text{ Hz}$ ✓ Correct
Solution: $f = \dfrac{\omega}{2\pi} = \dfrac{100\pi}{2\pi} = 50\text{ Hz}$.
Q7 — Alternating Current & RMS Values · easy · numerical
An alternating current has angular frequency $314\text{ rad/s}$. Its frequency is approximately:
A. $50\text{ Hz}$ ✓ Correct
B. $100\text{ Hz}$
C. $314\text{ Hz}$
D. $157\text{ Hz}$
Solution: $f = \dfrac{314}{2\pi} \approx 50\text{ Hz}$.
Q8 — Alternating Current & RMS Values · easy · numerical
For an alternating voltage, the ratio of the peak value to the RMS value is:
A. $0.707$
B. $1.11$
C. $2$
D. $1.414$ ✓ Correct
Solution: This ratio, the peak factor, equals $\sqrt{2}$ for a sinusoidal waveform.
Q9 — Alternating Current & RMS Values · easy · numerical
An alternating supply has a frequency of $50\text{ Hz}$. Its time period is:
A. $50\text{ s}$
B. $0.05\text{ s}$
C. $0.1\text{ s}$
D. $0.02\text{ s}$ ✓ Correct
Solution: $T = \dfrac{1}{f} = \dfrac{1}{50} = 0.02\text{ s}$.
Q10 — AC through Resistor, Inductor & Capacitor · easy · theory
In a purely resistive AC circuit, the current and the applied voltage are:
A. Out of phase by $\pi$
B. Out of phase by $\dfrac{\pi}{2}$
C. Out of phase by $\dfrac{\pi}{4}$
D. In phase with each other ✓ Correct
Solution: A resistor offers no reactance, so the current follows the voltage instantaneously.
Q11 — AC through Resistor, Inductor & Capacitor · easy · theory
In a purely inductive AC circuit, the current:
A. Leads the applied voltage by $\dfrac{\pi}{2}$
B. Lags behind the applied voltage by $\dfrac{\pi}{2}$ ✓ Correct
C. Is in phase with the voltage
D. Lags behind the voltage by $\pi$
Solution: The back EMF of the inductor opposes the rise of current, delaying it by a quarter cycle.
Q12 — AC through Resistor, Inductor & Capacitor · easy · theory
In a purely capacitive AC circuit, the current:
A. Leads the applied voltage by $\dfrac{\pi}{2}$ ✓ Correct
B. Is in phase with the voltage
C. Leads the voltage by $\pi$
D. Lags behind the applied voltage by $\dfrac{\pi}{2}$
Solution: Charge must flow before the voltage across the plates can build up, so current leads by a quarter cycle.
Q13 — AC through Resistor, Inductor & Capacitor · easy · theory
The inductive reactance of a coil of inductance $L$ at angular frequency $\omega$ is:
A. $\omega^2L$
B. $\omega L$ ✓ Correct
C. $\dfrac{\omega}{L}$
D. $\dfrac{1}{\omega L}$
Solution: It is measured in ohm and grows in proportion to the frequency.
Q14 — AC through Resistor, Inductor & Capacitor · easy · theory
The capacitive reactance of a capacitor $C$ at angular frequency $\omega$ is:
A. $\dfrac{\omega}{C}$
B. $\dfrac{1}{\omega^2C}$
C. $\dfrac{1}{\omega C}$ ✓ Correct
D. $\omega C$
Solution: Capacitive reactance falls as the frequency rises, which is why a capacitor passes high frequencies readily.
Q15 — AC through Resistor, Inductor & Capacitor · easy · numerical
If the frequency of the supply is doubled, the inductive reactance of a coil:
A. Remains unchanged
B. Halves
C. Doubles ✓ Correct
D. Becomes four times
Solution: $X_L = 2\pi fL \propto f$.
Q16 — AC through Resistor, Inductor & Capacitor · easy · numerical
If the frequency of the supply is doubled, the capacitive reactance of a capacitor:
A. Halves ✓ Correct
B. Remains unchanged
C. Becomes four times
D. Doubles
Solution: $X_C = \dfrac{1}{2\pi fC} \propto \dfrac{1}{f}$.
Q17 — AC through Resistor, Inductor & Capacitor · easy · numerical
An inductor of $0.5\text{ H}$ is connected to a supply of angular frequency $100\text{ rad/s}$. Its reactance is:
A. $50\,\Omega$ ✓ Correct
B. $200\,\Omega$
C. $0.005\,\Omega$
D. $100\,\Omega$
Solution: $X_L = \omega L = 100 \times 0.5 = 50\,\Omega$.
Q18 — AC through Resistor, Inductor & Capacitor · easy · numerical
An RMS voltage of $100\text{ V}$ is applied across a $50\,\Omega$ resistor. The RMS current is:
A. $0.5\text{ A}$
B. $5000\text{ A}$
C. $1.41\text{ A}$
D. $2\text{ A}$ ✓ Correct
Solution: $I_{rms} = \dfrac{V_{rms}}{R} = \dfrac{100}{50} = 2\text{ A}$.
Q19 — Series LCR Circuit & Impedance · easy · theory
The impedance of a series LCR circuit is given by:
A. $Z = R + X_L + X_C$
B. $Z = \sqrt{R^2 - (X_L - X_C)^2}$
C. $Z = \sqrt{R^2 + (X_L - X_C)^2}$ ✓ Correct
D. $Z = \sqrt{R^2 + X_L^2 + X_C^2}$
Solution: Resistance and net reactance add as perpendicular sides of the impedance triangle.
Q20 — Series LCR Circuit & Impedance · easy · theory
In a series LCR circuit, the phase angle between the applied voltage and the current is zero when:
A. $X_L = X_C$ ✓ Correct
B. $X_L < X_C$
C. $X_L > X_C$
D. $R = 0$
Solution: The reactances cancel, the impedance reduces to $R$, and the circuit is purely resistive.
Q21 — Series LCR Circuit & Impedance · easy · theory
The SI unit of impedance is the:
A. Volt
B. Henry
C. Farad
D. Ohm ✓ Correct
Solution: Impedance is the AC generalisation of resistance and carries the same unit.
Q22 — Series LCR Circuit & Impedance · easy · numerical
A series circuit has $R = 3\,\Omega$ and $X_L = 4\,\Omega$ with no capacitor. Its impedance is:
A. $5\,\Omega$ ✓ Correct
B. $1\,\Omega$
C. $12\,\Omega$
D. $7\,\Omega$
Solution: $Z = \sqrt{3^2 + 4^2} = 5\,\Omega$.
Q23 — Series LCR Circuit & Impedance · easy · numerical
A series circuit has $R = 6\,\Omega$ and net reactance $8\,\Omega$. Its impedance is:
A. $2\,\Omega$
B. $48\,\Omega$
C. $14\,\Omega$
D. $10\,\Omega$ ✓ Correct
Solution: $Z = \sqrt{36 + 64} = 10\,\Omega$.
Q24 — Series LCR Circuit & Impedance · easy · numerical
An RMS voltage of $100\text{ V}$ is applied to a series circuit of impedance $50\,\Omega$. The RMS current is:
A. $5000\text{ A}$
B. $0.5\text{ A}$
C. $2\text{ A}$ ✓ Correct
D. $1.41\text{ A}$
Solution: $I_{rms} = \dfrac{V_{rms}}{Z} = \dfrac{100}{50} = 2\text{ A}$.
Q25 — Series LCR Circuit & Impedance · easy · numerical
In a series LCR circuit a current of $2\text{ A}$ flows through a resistance of $30\,\Omega$. The voltage across the resistor is:
A. $15\text{ V}$
B. $120\text{ V}$
C. $30\text{ V}$
D. $60\text{ V}$ ✓ Correct
Solution: $V_R = IR = 2 \times 30 = 60\text{ V}$.
Q26 — Resonance & Q-factor · easy · theory
Electrical resonance in a series LCR circuit occurs when:
A. $X_L < X_C$
B. $X_L = X_C$ ✓ Correct
C. $X_L > X_C$
D. $R = 0$
Solution: At this frequency the reactances cancel and the circuit behaves as a pure resistance.
Q27 — Resonance & Q-factor · easy · theory
The resonant angular frequency of a series LCR circuit is:
A. $\sqrt{LC}$
B. $\dfrac{1}{2\pi\sqrt{LC}}$
C. $\dfrac{1}{\sqrt{LC}}$ ✓ Correct
D. $\dfrac{L}{C}$
Solution: Setting $\omega L = \dfrac{1}{\omega C}$ gives $\omega_0 = \dfrac{1}{\sqrt{LC}}$.
Q28 — Resonance & Q-factor · easy · theory
At resonance in a series LCR circuit, the impedance is:
A. Zero
B. Infinite
C. Maximum
D. Minimum and equal to $R$ ✓ Correct
Solution: With the reactances cancelling, the current reaches its largest possible value $\dfrac{V}{R}$.
Q29 — Resonance & Q-factor · easy · theory
At electrical resonance, the power factor of a series LCR circuit is:
A. Zero
B. $1.0$ ✓ Correct
C. $\dfrac{1}{\sqrt{2}}$
D. $0.5$
Solution: Since $Z = R$, $\cos\phi = \dfrac{R}{Z} = 1$ and the circuit consumes maximum power.
Q30 — Resonance & Q-factor · easy · numerical
At resonance, the current in a series LCR circuit is:
A. Leading the applied voltage by $\dfrac{\pi}{2}$
B. Zero
C. In phase with the applied voltage ✓ Correct
D. Lagging behind the voltage by $\dfrac{\pi}{2}$
Solution: With the reactances cancelling, the phase angle is zero and the circuit is purely resistive.