A 50 W, 100 V bulb is to be used on a 200 V, 50 Hz a.c. supply. Calculate the inductance of inductor used so that the bulb glows with normal brightness
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{2\sqrt{3}}{\pi}\,\mathrm{H} \)
For normal brightness the bulb must carry its rated current. An inductor in series takes up the extra voltage; an ideal inductor consumes no power. The bulb's rated current and resistance are
\( \displaystyle I = \frac{P}{V} = \frac{50\,\mathrm{W}}{100\,\mathrm{V}} = 0.5\,\mathrm{A} \)
\( \displaystyle R = \frac{V^{2}}{P} = \frac{(100)^{2}}{50} = 200\,\Omega \)
The same current must flow from the 200 V supply, so the impedance of the circuit is
\( \displaystyle Z = \frac{200\,\mathrm{V}}{0.5\,\mathrm{A}} = 400\,\Omega \)
\( \displaystyle Z = \sqrt{R^{2} + X_L^{2}} \;\Rightarrow\; X_L = \sqrt{400^{2} - 200^{2}} = 200\sqrt{3}\,\Omega \)
\( \displaystyle L = \frac{X_L}{2\pi f} = \frac{200\sqrt{3}}{2\pi \times 50} = \frac{2\sqrt{3}}{\pi}\,\mathrm{H} \)