In a series resonant LCR circuit,the voltage | vedclass

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(D) In a series $LCR$ circuit,the current $I$ flowing through the circuit is given by $I = \frac{V_R}{R}$.Given $V_R = 100 \ V$ and $R = 1 \ k\Omega =

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$250 \ V$

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(D) In a series $LCR$ circuit,the current $I$ flowing through 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 $X_L = X_C = \frac{1}{200 	imes 2 	imes 10^{-6}} = \frac{1}{400 	imes 10^{-6}} = \frac{10^6}{400} = 2500 \ \Omega$.The voltage across the inductor $L$ is given by $V_L = I X_L$.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:

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