In the circuit shown in the figure, the voltm | vedclass

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(D) From the circuit diagram, the source voltage is $V_{source} = 75 \, V$ and the frequency is $f = 500 \, Hz$. The voltmeter is connected across the

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$10$

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(D) From the circuit diagram, the source voltage is $V_{source} = 75 \, V$ and the frequency is $f = 500 \, Hz$. The voltmeter is connected across the resistor $R$, and it reads $V_R = 75 \, V$.Since $V_R = V_{source}$, the voltage drop across the series combination of the inductor and capacitor must be zero, i.e., $V_L - V_C = 0$, which implies $V_L = V_C$.This condition corresponds to electrical resonance in an $LCR$ circuit, where the inductive reactance $X_L$ equals the capacitive reactance $X_C$.The resonance frequency is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.Rearranging for $C$, we get $C = \frac{1}{4 \pi^2 f^2 L}$.Given $L = 10 \, mH = 10 	imes 10^{-3} \, H = 0.01 \, H$ and $f = 500 \, Hz$.Substituting the values: $C = \frac{1}{4 	imes \pi^2 	imes (500)^2 	imes 0.01} = \frac{1}{4 	imes 10 	imes 250000 	imes 0.01} \approx \frac{1}{10000} \, F = 10^{-4} \, F = 100 \, \mu F$.Wait, re-evaluating the calculation: $C = \frac{1}{4 	imes \pi^2 	imes 250000 	imes 0.01} = \frac{1}{4 	imes 9.8696 	imes 2500} = \frac{1}{98696} \approx 1.013 	imes 10^{-5} \, F = 10.13 \, \mu F$.Rounding to the nearest option, the value is $10 \, \mu F$.

In the circuit shown in the figure, the voltmeter reads $75 \, V$. The value of $C$ is ....... $\mu F$.

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