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
Concept. The lamp (resistor) and choke voltages are at right angles: $V^2 = V_R^2 + V_L^2$. Then $X_L = \dfrac{V_L}{I}$ and $L = \dfrac{X_L}{2\pi f}$.
Step 1. Lamp. $R = \dfrac{220^2}{100} = 484\ \Omega$; at $110$ V, $I = \dfrac{110}{484}$ A.
Step 2. Choke voltage. $V_L = \sqrt{220^2 - 110^2} = 110\sqrt3$ V.
Step 3. Reactance. $X_L = \dfrac{110\sqrt3}{110/484} = 484\sqrt3 \approx 838.3\ \Omega$.
Step 4. Inductance. $L = \dfrac{838.3}{314} \approx 2.67$ H.
Answer. $2.67$.
Solution by Sreeraj P, M.Sc Physics