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(D) In a series $LCR$ circuit,the current $I$ is given by $I = \frac{E}{Z} = \frac{E}{\sqrt{R^2 + (X_L - X_C)^2}}$.Here,$X_L = 2 \pi f L$ and $X_C = \

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both halved simultaneously.

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(D) In a series $LCR$ circuit,the current $I$ is given by $I = \frac{E}{Z} = \frac{E}{\sqrt{R^2 + (X_L - X_C)^2}}$.Here,$X_L = 2 \pi f L$ and $X_C = \frac{1}{2 \pi f C}$.To keep the current $I$ constant for a given voltage $E$ and resistance $R$,the impedance $Z$ must remain constant.This requires the reactances $X_L$ and $X_C$ to remain unchanged.If the frequency $f$ is doubled $(f' = 2f)$,then for $X_L$ to remain constant: $2 \pi (2f) L' = 2 \pi f L \implies L' = \frac{L}{2}$.Similarly,for $X_C$ to remain constant: $\frac{1}{2 \pi (2f) C'} = \frac{1}{2 \pi f C} \implies C' = \frac{C}{2}$.Therefore,both $L$ and $C$ should be halved simultaneously.

In a series $LCR$ circuit,$R$ represents the resistance of an electric bulb. If the frequency of the $A.C.$ supply is doubled,what should be the values of inductance $L$ and capacitance $C$ to maintain the same current?

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