Q 12-07-107JEE MainJEE Main 2022 (27 Jul, Shift 1)Medium
To light, a $50$ W, $100$ V lamp is connected, in series with a capacitor of capacitance $\dfrac{50}{\pi\sqrt x}\ \mu$F, with $200$ V, $50$ Hz AC source. The value of $x$ will be ______ .
Numerical value type. Enter your answer.
Answer: 3
Lamp: $R = \dfrac{V^2}{P} = \dfrac{100^2}{50} = 200\ \Omega$, rated current $I = \dfrac{50}{100} = 0.5$ A.
$$Z = \frac{200}{0.5} = 400\ \Omega \Rightarrow X_C = \sqrt{400^2 - 200^2} = 200\sqrt3\ \Omega$$
$$C = \frac{1}{2\pi fX_C} = \frac{1}{2\pi(50)(200\sqrt3)} = \frac{10^{-4}}{2\pi\sqrt3}\ \text{F} = \frac{50}{\pi\sqrt3}\ \mu\text{F}$$
$x = 3$.
Solution by Sreeraj P, M.Sc Physics