(B) In an $LCR$ circuit, the current is maximum at resonance.At resonance, the angular frequency $\omega$ is given by $\omega = \frac{1}{\sqrt{LC}}$.F

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(B) In an $LCR$ circuit, the current is maximum at resonance.At resonance, the angular frequency $\omega$ is given by $\omega = \frac{1}{\sqrt{LC}}$.From the given voltage equation $V = 5 \sin(100t)$, we have $\omega = 100 \text{ rad/s}$.Given $L = 2 \text{ H}$, we can substitute these values into the resonance formula:$100 = \frac{1}{\sqrt{2 \times C}}$Squaring both sides:$10000 = \frac{1}{2C}$$C = \frac{1}{2 \times 10000} = \frac{1}{20000} \text{ F}$$C = 0.5 \times 10^{-4} \text{ F} = 50 \times 10^{-6} \text{ F} = 50 \mu\text{F}$.

Class 12 Physics Alternating Current Half Power Frequency , Quality Factor ,Resonance in AC Circuit

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Using a variable-frequency a.c. voltage source, the maximum current measured in the given $LCR$ circuit is $50 \text{ mA}$ for $V = 5 \sin(100t)$. The values of $L$ and $R$ are shown in the figure. The capacitance of the capacitor $(C)$ used is . . . . . . $\mu\text{F}$.

JEE MAIN 2026 All JEE MAIN

A
$25$
B
$50$
C
$75$
D
$100$

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Half Power Frequency , Quality Factor ,Resonance in AC Circuit

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Using a variable-frequency a.c. voltage source, the maximum current measured in the given $LCR$ circuit is $50 \text{ mA}$ for $V = 5 \sin(100t)$. The values of $L$ and $R$ are shown in the figure. The capacitance of the capacitor $(C)$ used is . . . . . . $\mu\text{F}$.

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$A$ resonance circuit having inductance and resistance $2 \times 10^{-4} \ H$ and $6.28 \ \Omega$ respectively oscillates at $10 \ MHz$ frequency. The value of the quality factor of this resonator is .........
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