A lamp rated 8 V, 800 W is connected to a 100 V, 50 Hz a.c. source. The inductance of the choke required in the circuit is: (Take \( \pi \simeq 3 \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3.3 mH
The choke is joined in series with the lamp so that the lamp gets its rated voltage and current. The lamp is a pure resistance; the choke (an ideal inductor) adds reactance, and the impedance must limit the current to the rated value at 100 V.
\( \displaystyle I = \frac{P}{V_{\text{lamp}}} = \frac{800}{8} = 100\,\mathrm{A} \)
\( \displaystyle R = \frac{V_{\text{lamp}}}{I} = \frac{8}{100} = 0.08\,\Omega \)
\( \displaystyle Z = \frac{V}{I} = \frac{100}{100} = 1\,\Omega \)
\( \displaystyle X_L = \sqrt{Z^{2} - R^{2}} = \sqrt{1 - 0.0064} = 0.997\,\Omega \)
\( \displaystyle \omega = 2\pi f = 2 \times 3 \times 50 = 300\,\mathrm{rad\,s^{-1}} \)
\( \displaystyle L = \frac{X_L}{\omega} = \frac{0.997}{300} = 3.3\times10^{-3}\,\mathrm{H} = 3.3\,\mathrm{mH} \)