In a series LCR circuit $C = 5 μF$ and $ω = 1000\, rad\,s^{-1}$. The value of inductance L for which the current is maximum in this circuit will be

200 mH

The correct answer is Option (4) → 200 mH

For a series LCR circuit, the current is maximum at resonance. At resonance, the inductive reactance equals the capacitive reactance:

$\omega L = \frac{1}{\omega C}$

Solving for $L$:

$L = \frac{1}{\omega^2 C}$

Given $C = 5 \mu F = 5 \times 10^{-6} F$, $\omega = 1000\ \text{rad/s}$

$L = \frac{1}{(1000)^2 \cdot 5 \times 10^{-6}} = \frac{1}{10^6 \cdot 5 \times 10^{-6}} = \frac{1}{5} = 0.2\ H$

Final Answer: $L = 0.2\ H$