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 ___________.

Solution

Physics sir · 4 steps

The lamp tells you the current; the leftover voltage, added at right angles, sits across the capacitor.

1. Read the lamp. To glow normally, the lamp must carry its rated current I = P/V = 50/100 = 0.5 A with 100 V across it. That makes it a resistor of R = 100/0.5 = 200 Ω.
2. Voltage on capacitor. In a series RC circuit the resistor voltage and the capacitor voltage are 90° out of phase. So they add like perpendicular vectors, not as plain numbers.

   V_C = √(V² - V_R²) = √(200² - 100²) = 100√3 V

3. Find the capacitance. X_C = V_C/I = 100√3/0.5 = 200√3 Ω. Then C = 1/(2π f X_C) = 1/(2π × 50 × 200√3) = 1/20000π√3 F = 50/π√3 μF.
4. So. Comparing with 50/π√x μF gives √x = √3, so x = 3.

The common slip: 1. Subtracting the voltages directly (200 - 100 = 100 V on the capacitor) gives X_C = 200 Ω and C = 50/π μF, so x = 1.

The idea to keep. A rated lamp in an AC circuit fixes the current and its own voltage. In series R with L or C, voltages combine as V² = V_R² + V_X², never by simple subtraction.

Answer from the NTA final answer key. Solution by PrepWiser.