To light, a 50 W, 100 V lamp is connected, in series with a capacitor of capacitance (50)/(π√(x)) μ F, with 200 V, 50 Hz AC source. The value of x will be ___________. Numerical answer type. Work it out, then open the answer.

Answer: 3. Lamp: R = V²/P = 100²/50 = 200 Ω, I = P/V = 0.5 A. Z = 200/0.5 = 400 Ω XC = √(400²-200²) = 200√(3) Ω C = (1)/(2π f XC) = (1)/(100π× 200√(3)) F = (50)/(π√(3)) μ F So x = 3.

To light, a 50 W, 100 V lamp is connected, in series with a capacitor… | JEE Main 2022 (27 Jul, Shift 1)

Q54JEE Main 2022 · 27 JUL · SHIFT 1Medium

Power, Power Factor & Wattless Current

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

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 3

Lamp: $R = V^2/P = 100^2/50 = 200\ \Omega$ , $I = P/V = 0.5$ A.
$Z = 200/0.5 = 400\ \Omega$
$X_C = \sqrt{400^2-200^2} = 200\sqrt{3}\ \Omega$
$C = \frac{1}{2\pi f X_C} = \frac{1}{100\pi\times 200\sqrt{3}}$ F $= \frac{50}{\pi\sqrt{3}}\ \mu$ F
So $x = 3$ .

Report a mistake

Power, Power Factor & Wattless Current question from JEE Main 2022 (27 Jul, Shift 1)

More questions on Power, Power Factor & Wattless Current