In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance CμF across a 200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drops across it is 100 V. Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude of the phase-angle (in degrees) between the current and the supply voltage is φ. Assume, $\mathrm{\pi }\sqrt{3}\approx 5\text{. }$

Answer: The phase angle between the current and the 200 V, 50 Hz supply is φ = 60° (current leads the supply).

Phase angle: φ = 60° (lead). The circuit is a series R–C load, so current leads voltage by φ.

Given Data

• Supply voltage, V_s = 200 V (rms), frequency f = 50 Hz
• Lamp (purely resistive): Power P = 500 W, voltage across lamp V_R = 100 V (rms)
• Capacitor in series with the lamp, no inductance

Step 1: Lamp resistance and circuit current

The metal filament lamp acts as a resistor R.

R = V_R² / P = (100)² / 500 = 10,000 / 500 = 20 Ω.

Series current I (same through R and C): I = P / V_R = 500 / 100 = 5 A (rms).

Step 2: Capacitor voltage from the phasor (right-triangle) relation

In a series R–C circuit, V_R is in phase with I, and V_C lags I by 90°. The supply phasor is the vector sum:

V_s² = V_R² + V_C².

So V_C = √(V_s² − V_R²) = √(200² − 100²) = √(40,000 − 10,000) = √30,000 ≈ 173.2 V.

Step 3: Capacitive reactance

X_C = V_C / I = 173.2 / 5 = 34.64 Ω.

Step 4: Phase angle φ between current and supply

For series R–C, the impedance is Z = R − jX_C. The current leads the supply by angle φ where

tan φ = X_C / R = 34.64 / 20 = 1.732 ⇒ φ = arctan(1.732) = 60°.

Optional: Capacitance value (to connect with C μF)

X_C = 1 / (2π f C) ⇒ C = 1 / (2π f X_C) = 1 / (2π × 50 × 34.64) ≈ 9.19 × 10⁻⁵ F = ≈ 92 μF.

Cross-check and interpretation

• Power in R: P = I²R = 5² × 20 = 500 W ✔
• Voltage triangle: 100–173.2–200 gives tan φ = 173.2 / 100 = 1.732 ⇒ φ = 60° ✔
• Current leads because the net reactance is capacitive (no inductance), so φ is a lead angle.