In a series resonant LCR circuit,the voltage | vedclass

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(C) At resonance,the impedance of the circuit is purely resistive,so the current $I$ in the circuit is given by $I = \frac{V_R}{R}$.Given $V_R = 100 \

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$250$

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(C) At resonance,the impedance of the circuit is purely resistive,so the current $I$ in the circuit is given by $I = \frac{V_R}{R}$.Given $V_R = 100 \, V$ and $R = 1 \, k\Omega = 1000 \, \Omega$,we have $I = \frac{100}{1000} = 0.1 \, A$.At resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$,where $X_C = \frac{1}{\omega C}$.Given $\omega = 200 \, rad/s$ and $C = 2 \, \mu F = 2 	imes 10^{-6} \, F$,we calculate $X_C = \frac{1}{200 	imes 2 	imes 10^{-6}} = \frac{1}{400 	imes 10^{-6}} = \frac{10^6}{400} = 2500 \, \Omega$.Since $X_L = X_C$ at resonance,$X_L = 2500 \, \Omega$.The voltage across the inductor $V_L$ is given by $V_L = I 	imes X_L$.Substituting the values,$V_L = 0.1 	imes 2500 = 250 \, V$.

In a series resonant $LCR$ circuit,the voltage across $R$ is $100 \, V$ and $R = 1 \, k\Omega$ with $C = 2 \, \mu F$. The resonant frequency $\omega$ is $200 \, rad/s$. At resonance,the voltage across $L$ is......$V$.

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