To light, a $50 \mathrm{~W}, 100 \mathrm{~V}$ lamp is connected, in series with a capacitor of capacitance $\frac{50}{\pi \sqrt{\mathrm{x}}} \mu \mathrm{F}$, with $200 \mathrm{~V}, 50 \mathrm{Hz}$ AC source. The value of x will be ______.

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Alternating Current

To light, a

50 W, 100 V

lamp is connected, in series with a capacitor of capacitance

\frac{50}{π x} μ F

, with

200 V, 50 Hz

AC source. The value of x will be ______.

Answer: 3

Lamp resistance

R = V^{2} / P = 10 0^{2} /50 = 200Ω

. In series with capacitor across 200 V:

V_{R}^{2} + V_{C}^{2} = V^{2}

, so

10 0^{2} + V_{C}^{2} = 20 0^{2}

, giving

V_{C} = 1003

V. Current

i = V_{R} / R = 100/200 = 0.5

A. Capacitive reactance

X_{C} = V_{C} / i = 1003 /0.5 = 2003 Ω

. Since

X_{C} = 1/ (ω C) = 1/ (2 π f C)

, we have

C = 1/ (2 π \times 50 \times 2003) = 1/ (20000 π 3)

F. Given

C = 50/ (π x) \times 1 0^{- 6}

F, equating:

50 \times 1 0^{- 6} / (π x) = 1/ (20000 π 3)

, solving gives

x = 3

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Integer answer