In a series resonant LCR circuit,the voltage | vedclass

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(B) In a series $LCR$ circuit at resonance,the inductive reactance $X_L$ equals the capacitive reactance $X_C$,i.e.,$X_L = X_C = \frac{1}{\omega C}$.T

Vedclass · Accepted Answer

$250\,V$

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(B) In a series $LCR$ circuit at resonance,the inductive reactance $X_L$ equals the capacitive reactance $X_C$,i.e.,$X_L = X_C = \frac{1}{\omega C}$.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 voltage across the inductor $V_L$ is given by $V_L = I 	imes X_L$.Since $X_L = X_C = \frac{1}{\omega C}$,we have $V_L = I 	imes \frac{1}{\omega C}$.Substituting the given values: $V_L = 0.1 	imes \frac{1}{200 	imes 2 	imes 10^{-6}}$.$V_L = 0.1 	imes \frac{1}{400 	imes 10^{-6}} = 0.1 	imes \frac{10^6}{400} = 0.1 	imes 2500 = 250\,V$.Thus,the voltage across $L$ is $250\,V$.

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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