Question

A series RLC circuit with \(R=10\,\Omega\), \(L=50\text{ mH}\), \(C=100\ \mu\text{F}\) connected to \(200\ \text{V}\), \(50\ \text{Hz}\) supply consumes power \(P\). The value of \(L\) is changed such that the circuit consumes the same power but operates with lagging power factor. The new value of \(L\) is _____ mH (rounded off to two decimal places).

Hint:

Maintaining the same real power in an RLC circuit requires keeping current magnitude constant.

Answer 1

Correct Answer: 152.01

Original inductive reactance: \[ X_L = 2\pi (50)(0.05) = 15.708~\Omega \] Capacitive reactance: \[ X_C = \frac{1}{2\pi (50)(100\times 10^{-6})} = 31.83~\Omega \] Net reactance: \[ X = X_L - X_C = -16.12\ \Omega \quad (\text{leading power factor}) \] Circuit power: \[ P = V I \cos\phi = I^2 R \] To keep same power, the magnitude of current must remain same \(\Rightarrow |X_{\text{new}}| = |X| = 16.12\ \Omega\). Since new PF is lagging: \[ X_{\text{new}} = +16.12 \] So: \[ X_L^{\text{new}} = X_{\text{new}} + X_C = 16.12 + 31.83 = 47.95\ \Omega \] \[ L_{\text{new}} = \frac{X_L^{\text{new}}}{2\pi f} = \frac{47.95}{2\pi(50)} = 0.1529\ \text{H} \] \[ L_{\text{new}} \approx 152.90\ \text{mH} \] Thus the answer lies in: \[ \boxed{152.01\ \text{to}\ 152.99} \] Final Answer: 152.90 mH