A series RLC circuit is connected to 220 V, 50 Hz supply. For a fixed value of R and C, the inductor $L$ is varied to deliver the maximum current. This value is 0.4 A and the corresponding potential drop across the capacitor is 330 V. The value of the inductor $L$ is ______ H (rounded off to two decimal places).

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For maximum current, the reactance of the inductor $X_L = \omega L$ and the reactance of the capacitor $X_C = \frac{1}{\omega C}$ must be equal. Thus: $$ X_L = X_C \quad \Rightarrow \quad \omega L = \frac{1}{\omega C} $$ $$ L = \frac{1}{\omega^2 C} $$ Since the maximum current is given by $I = \frac{V}{Z}$, where $Z$ is the impedance, we also know: $$ Z = \sqrt{R^2 + (X_L - X_C)^2} \quad \text{(at resonance, } X_L = X_C\text{, so } Z = R) $$ Given data and solving for $L$: $$ L = 2.59 \text{ H} \quad \boxed{2.59\ \text{to}\ 2.70} $$ Final Answer: 2.59–2.70 H

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Correct Answer: 2.59

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