A 50 W, 100 V bulb is to be used on a 200 V, 50 Hz a.c. supply. Calcu…

The correct answer is Option (1) → $\frac{2\sqrt{3}}{\pi}H$

Given:

Bulb rating: $P = 50\ \text{W},\ V_{\text{bulb}} = 100\ \text{V}$

Supply: $V_s = 200\ \text{V},\ f = 50\ \text{Hz}$

Bulb resistance:

$R = \frac{V_{\text{bulb}}^2}{P} = \frac{100^2}{50} = 200\ \Omega$

The inductive reactance $X_L$ of the series inductor satisfies:

$(V_s)^2 = V_R^2 + V_L^2 \Rightarrow V_L = \sqrt{V_s^2 - V_R^2}$

$V_L = \sqrt{200^2 - 100^2} = \sqrt{40000 - 10000} = \sqrt{30000} = 100\sqrt{3}\ \text{V}$

Inductive reactance: $X_L = \frac{V_L}{I}$, where $I = \frac{V_R}{R} = \frac{100}{200} = 0.5\ \text{A}$

$X_L = \frac{100 \sqrt{3}}{0.5} = 200 \sqrt{3}\ \Omega$

Inductance: $X_L = 2 \pi f L \Rightarrow L = \frac{X_L}{2 \pi f} = \frac{200 \sqrt{3}}{2 \pi \cdot 50} = \frac{2 \sqrt{3}}{\pi}\ \text{H}$

Inductance L = $\frac{2 \sqrt{3}}{\pi}$ H