Q 12-07-032JEE MainMedium
A $220$ V, $100$ W lamp is to be run from a $220$ V, $50$ Hz supply through a series choke (pure inductor) so that the lamp gets only $110$ V. Treating the lamp as a pure resistor, find the inductance of the choke in henry, to two decimal places. (take $\pi = 3.14$)
Numerical value type. Enter your answer.
Answer: 2.67
Lamp resistance $R = \dfrac{220^2}{100} = 484\ \Omega$. At $110$ V across it, $I = \dfrac{110}{484}$ A.
$V_L = \sqrt{220^2 - 110^2} = 110\sqrt{3}$ V, so $X_L = \dfrac{V_L}{I} = 484\sqrt{3} \approx 838.3\ \Omega$.
$$L = \frac{X_L}{2\pi f} = \frac{838.3}{314} \approx 2.67\ \text{H}$$
Solution by Sreeraj P, M.Sc Physics