An emf $E= 200 \sin 300t$ volt is applied across an inductor coil having a resistance of 1.0 Ω. If the maximum current is found to be 10 A, then the value of inductance is approximately:

67 mH

The correct answer is Option (1) → 67 mH

Given:

EMF: $E = 200 \sin 300 t \, \text{V}$

Resistance: $R = 1 \, \Omega$

Maximum current: $I_{\max} = 10 \, \text{A}$

Angular frequency: $\omega = 300 \, \text{rad/s}$

For an RL circuit: $I_{\max} = \frac{E_0}{\sqrt{R^2 + (\omega L)^2}}$

Peak EMF: $E_0 = 200 \, \text{V}$

Substitute values:

$10 = \frac{200}{\sqrt{1^2 + (300 L)^2}}$

$\sqrt{1 + (300 L)^2} = \frac{200}{10} = 20$

$1 + (300 L)^2 = 400$

$(300 L)^2 = 399 \approx 400$

$300 L \approx 20 \Rightarrow L \approx \frac{20}{300} \approx 0.0667 \, \text{H}$

Answer: Inductance $L \approx 0.067 \, \text{H}$