Question

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).

Hint:

At resonance, the impedance in a series RLC circuit is equal to the resistance only. This simplifies the calculation of the inductor.

Answer 1

Correct Answer: 2.59

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