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(D) Given that in the series $LCR$ circuit,$V_R = V_L = V_C = 10 \, V$. Since $V = IR$,this implies $R = X_L = X_C$.The source voltage $E$ is equal to

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$\frac{10}{\sqrt{2}} \, V$

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(D) Given that in the series $LCR$ circuit,$V_R = V_L = V_C = 10 \, V$. Since $V = IR$,this implies $R = X_L = X_C$.The source voltage $E$ is equal to $V_R$ because $V_L$ and $V_C$ cancel each other out in resonance $(X_L = X_C)$. Thus,$E = 10 \, V$.When the inductor is short-circuited,the circuit becomes an $RC$ series circuit with the same source voltage $E = 10 \, V$.The new impedance of the circuit is $Z' = \sqrt{R^2 + X_C^2}$. Since $R = X_C$,$Z' = \sqrt{R^2 + R^2} = R\sqrt{2}$.The new current in the circuit is $i' = \frac{E}{Z'} = \frac{10}{R\sqrt{2}}$.The voltage across the capacitor is $V_C' = i' X_C = \left( \frac{10}{R\sqrt{2}} ight) 	imes R = \frac{10}{\sqrt{2}} \, V$.

In a series $LCR$ circuit,the voltage across the resistance,capacitance,and inductance is $10 \, V$ each. If the inductor is short-circuited,what will be the voltage across the capacitor?

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