In a series LCR resonant circuit,R = 800 Ome | vedclass

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(D) At resonance,the current $I$ in the circuit is given by $I = V_R / R$. Given $V_R = 200 \ V$ and $R = 800 \ \Omega$,we have $I = 200 / 800 = 0.25

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$500$

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(D) At resonance,the current $I$ in the circuit is given by $I = V_R / R$. Given $V_R = 200 \ V$ and $R = 800 \ \Omega$,we have $I = 200 / 800 = 0.25 \ A$.At resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$,where $X_C = 1 / (\omega C)$.Given $\omega = 250 \ rad/s$ and $C = 2 	imes 10^{-6} \ F$,we calculate $X_C = 1 / (250 	imes 2 	imes 10^{-6}) = 1 / (500 	imes 10^{-6}) = 10^6 / 500 = 2000 \ \Omega$.Since $X_L = X_C$ at resonance,$X_L = 2000 \ \Omega$.The voltage across the inductance $V_L$ is given by $V_L = I 	imes X_L$.Substituting the values,$V_L = 0.25 	imes 2000 = 500 \ V$.

In a series $LCR$ resonant circuit,$R = 800 \ \Omega$,$C = 2 \ \mu F$,and the voltage across the resistance is $200 \ V$. The angular frequency is $250 \ rad/s$. At resonance,the voltage across the inductance is: (in $V$)

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