Resonance & Q-factor — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Resonance & Q-factor MCQs with step-by-step solutions (21 questions). Part of AC Circuits. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — 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.
Q2 — 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 $\omega_0 = \dfrac{1}{\sqrt{LC}}$.
Q3 — 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 $\dfrac{V}{R}$.
Q4 — 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.
Q5 — Resonance & Q-factor · medium · theory
The quality factor of a series LCR resonant circuit is given by:
A. $\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$ ✓ Correct
B. $\dfrac{1}{R}\sqrt{\dfrac{C}{L}}$
C. $R\sqrt{\dfrac{L}{C}}$
D. $\dfrac{1}{L}\sqrt{\dfrac{R}{C}}$
Solution: Equivalently $Q = \dfrac{\omega_0L}{R}$; a higher $Q$ means a sharper resonance peak.
Q6 — Resonance & Q-factor · medium · theory
The sharpness of resonance of a series LCR circuit increases when:
A. The capacitance is increased
B. The resistance is reduced ✓ Correct
C. The resistance is increased
D. The inductance is reduced
Solution: Lower resistance raises the quality factor, narrowing the band of frequencies to which the circuit responds strongly.
Q7 — Resonance & Q-factor · hard · theory
A series resonant circuit is called an acceptor circuit because at resonance it:
A. Blocks all frequencies equally
B. Offers minimum impedance and accepts the largest current at that frequency ✓ Correct
C. Stores no energy
D. Offers maximum impedance and rejects that frequency
Solution: This selectivity is what allows a radio receiver to pick out one station from many.
Q8 — Resonance & Q-factor · hard · theory
The bandwidth of a series LCR resonant circuit is given by:
A. $\dfrac{L}{R}$
B. $\dfrac{R}{L}$ ✓ Correct
C. $\dfrac{R}{C}$
D. $\dfrac{1}{RC}$
Solution: A smaller resistance gives a narrower bandwidth, which is the same statement as a higher quality factor.
Q9 — Resonance & Q-factor · hard · numerical
A series circuit has $L = 20\text{ mH}$ and $C = 5\,\mu\text{F}$. Its resonant angular frequency is approximately:
A. $5000\text{ rad/s}$
B. $3162\text{ rad/s}$ ✓ Correct
C. $1000\text{ rad/s}$
D. $10^4\text{ rad/s}$
Solution: $\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{20 \times 10^{-3} \times 5 \times 10^{-6}}} = \dfrac{1}{\sqrt{10^{-7}}} \approx 3162\text{ rad/s}$.
Q10 — Resonance & Q-factor · hard · numerical
A series circuit has $L = 2\text{ H}$ and $C = 8\,\mu\text{F}$. Its resonant angular frequency is:
A. $500\text{ rad/s}$
B. $125\text{ rad/s}$
C. $1000\text{ rad/s}$
D. $250\text{ rad/s}$ ✓ Correct
Solution: $\omega_0 = \dfrac{1}{\sqrt{2 \times 8 \times 10^{-6}}} = \dfrac{1}{\sqrt{1.6 \times 10^{-5}}} = \dfrac{1}{4 \times 10^{-3}} = 250\text{ rad/s}$.
Q11 — Resonance & Q-factor · hard · numerical
A series circuit has $L = 1\text{ H}$ and $C = 1\,\mu\text{F}$. Its resonant frequency is approximately:
A. $159\text{ Hz}$ ✓ Correct
B. $50\text{ Hz}$
C. $1000\text{ Hz}$
D. $318\text{ Hz}$
Solution: $f_0 = \dfrac{1}{2\pi\sqrt{LC}} = \dfrac{1}{2\pi \times 10^{-3}} \approx 159\text{ Hz}$.
Q12 — Resonance & Q-factor · medium · numerical
A series circuit has $L = 4\text{ H}$ and $C = 1\,\mu\text{F}$. Its resonant angular frequency is:
A. $250\text{ rad/s}$
B. $2000\text{ rad/s}$
C. $1000\text{ rad/s}$
D. $500\text{ rad/s}$ ✓ Correct
Solution: $\omega_0 = \dfrac{1}{\sqrt{4 \times 10^{-6}}} = \dfrac{1}{2 \times 10^{-3}} = 500\text{ rad/s}$.
Q13 — Resonance & Q-factor · hard · numerical
A series LCR circuit has $R = 10\,\Omega$, $L = 0.1\text{ H}$ and $C = 1\,\mu\text{F}$. Its quality factor is approximately:
A. $31.6$ ✓ Correct
B. $316$
C. $3.16$
D. $10$
Solution: $Q = \dfrac{1}{R}\sqrt{\dfrac{L}{C}} = \dfrac{1}{10}\sqrt{\dfrac{0.1}{10^{-6}}} = \dfrac{1}{10}\sqrt{10^5} \approx 31.6$.
Q14 — Resonance & Q-factor · medium · numerical
If the capacitance of a series resonant circuit is made four times as large, the resonant frequency:
A. Becomes one-fourth
B. Doubles
C. Halves ✓ Correct
D. Remains unchanged
Solution: $\omega_0 \propto \dfrac{1}{\sqrt{C}}$, so quadrupling $C$ halves the resonant frequency.
Q15 — Resonance & Q-factor · medium · numerical
If the inductance of a series resonant circuit is made four times as large, the resonant frequency:
A. Remains unchanged
B. Becomes four times
C. Doubles
D. Halves ✓ Correct
Solution: $\omega_0 \propto \dfrac{1}{\sqrt{L}}$, so quadrupling $L$ halves the resonant frequency.
Q16 — Resonance & Q-factor · hard · numerical
A series LCR circuit has $R = 20\,\Omega$ and $L = 0.1\text{ H}$. Its bandwidth is:
A. $2\text{ rad/s}$
B. $0.005\text{ rad/s}$
C. $200\text{ rad/s}$ ✓ Correct
D. $20\text{ rad/s}$
Solution: Bandwidth $= \dfrac{R}{L} = \dfrac{20}{0.1} = 200\text{ rad/s}$.
Q17 — Resonance & Q-factor · hard · numerical
At resonance in a series LCR circuit, the voltages across the inductor and the capacitor are:
A. Unequal in magnitude
B. Equal in magnitude and in phase
C. Equal in magnitude and opposite in phase ✓ Correct
D. Both zero
Solution: Since $X_L = X_C$ and the current is common, $V_L = V_C$; being $180^\circ$ apart they cancel exactly.
Q18 — Resonance & Q-factor · medium · numerical
A series LCR circuit at resonance has $R = 50\,\Omega$ across a supply of $100\text{ V}$ RMS. The current is:
A. $5000\text{ A}$
B. $2\text{ A}$ ✓ Correct
C. $1.41\text{ A}$
D. $0.5\text{ A}$
Solution: At resonance $Z = R$, so $I = \dfrac{100}{50} = 2\text{ A}$ — the largest current the circuit can carry.
Q19 — Resonance & Q-factor · hard · numerical
A circuit has $L = 1\text{ mH}$ and $C = 1\text{ nF}$. Its resonant angular frequency is:
A. $10^3\text{ rad/s}$
B. $10^9\text{ rad/s}$
C. $10^{12}\text{ rad/s}$
D. $10^6\text{ rad/s}$ ✓ Correct
Solution: $\omega_0 = \dfrac{1}{\sqrt{10^{-3} \times 10^{-9}}} = \dfrac{1}{\sqrt{10^{-12}}} = 10^6\text{ rad/s}$.
Q20 — 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.
Q21 — Resonance & Q-factor · medium · numerical
Two series resonant circuits have quality factors $10$ and $50$. The one with the sharper resonance is:
A. Both are equally sharp
B. The circuit of $Q = 10$
C. The circuit of $Q = 50$ ✓ Correct
D. Neither shows resonance
Solution: Sharpness rises with the quality factor, since the bandwidth is $\dfrac{\omega_0}{Q}$.