Question:
An LCR circuit as shown in the figure is connected to a voltage source \(V_{ac}\) whose frequency can be varied. 
(Circuit: inductor 24 H, capacitor 2 \(\mu\)F and resistor 15 \(\Omega\) in series with the source \(V_{ac} = V_0 \sin\omega t\); a voltmeter V is connected across the resistor.)
The frequency, at which the voltage across the resistor is maximum, is :
An LCR circuit as shown in the figure is connected to a voltage source \(V_{ac}\) whose frequency can be varied.
(Circuit: inductor 24 H, capacitor 2 \(\mu\)F and resistor 15 \(\Omega\) in series with the source \(V_{ac} = V_0 \sin\omega t\); a voltmeter V is connected across the resistor.)
The frequency, at which the voltage across the resistor is maximum, is :
(Circuit: inductor 24 H, capacitor 2 \(\mu\)F and resistor 15 \(\Omega\) in series with the source \(V_{ac} = V_0 \sin\omega t\); a voltmeter V is connected across the resistor.)
The frequency, at which the voltage across the resistor is maximum, is :
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The voltage across \(R\) is \(IR\); it is maximum when the current is maximum, i.e. at series resonance.
Updated On: Oct 31, 2026
- 902 Hz
- 143 Hz
- 23 Hz
- 345 Hz
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The Correct Option is C
Solution and Explanation
Voltage across \(R\): \(V_R = IR = \frac{V_0 R}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}}\).
\(V_R\) is maximum when \(\omega L = \frac{1}{\omega C}\) (resonance).
\(f = \frac{1}{2\pi\sqrt{LC}}\) with \(L = 24\ \mathrm{H}\), \(C = 2\times10^{-6}\ \mathrm{F}\).
\(LC = 4.8\times10^{-5}\), \(\sqrt{LC} = 6.93\times10^{-3}\ \mathrm{s}\).
\(f = \frac{1}{2\pi \times 6.93\times10^{-3}} = \frac{1}{0.0435} \approx 23\ \mathrm{Hz}\).
Answer: (3) 23 Hz
\(V_R\) is maximum when \(\omega L = \frac{1}{\omega C}\) (resonance).
\(f = \frac{1}{2\pi\sqrt{LC}}\) with \(L = 24\ \mathrm{H}\), \(C = 2\times10^{-6}\ \mathrm{F}\).
\(LC = 4.8\times10^{-5}\), \(\sqrt{LC} = 6.93\times10^{-3}\ \mathrm{s}\).
\(f = \frac{1}{2\pi \times 6.93\times10^{-3}} = \frac{1}{0.0435} \approx 23\ \mathrm{Hz}\).
Answer: (3) 23 Hz
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