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Resistor and capacitor with half voltage

An adjustable resistor R in series with a capacitance of 25μF draws a current of 0.8A, when connected across 50Hz supply. Calculate: i) The value of resistor so that the voltage across the capacitor is 'half' of the supply voltage. ii) Power consumed and iii) The power factor

Given C = 25 μF, I = 0.8 A, f = 50 Hz (series R–C circuit).

X_C = 1/(2πfC) = 1/(314.16 × 25 × 10⁻⁶) = 127.3 Ω

i) Value of R

The voltage across the capacitor is V_C = I X_C = 0.8 × 127.3 = 101.9 V. For V_C = V/2, the supply voltage is V = 203.7 V, so the impedance is

Z = V/I = 203.7/0.8 = 254.6 Ω = 2X_C

Z² = R² + X_C² ⟹ R = √(Z² − X_C²) = √(4X_C² − X_C²) = √3 X_C

R = 1.732 × 127.3 = 220.5 Ω

ii) Power consumed

P = I²R = 0.8² × 220.5 = 141.1 W

iii) Power factor

cos φ = R/Z = 220.5/254.6 = 0.866 leading (φ = 30°)

(Check: sin φ = X_C/Z = 1/2, so φ = 30°.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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