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Coil on 240 V, power, current, inductance

Short answer

Power dissipated is 500 W. Current is 5.42 A. Inductance of the coil is 0.130 H.

A coil takes 1.3 kVA and 1.2 kVAR when connected to a 240 V, 50 Hz sinusoidal supply. Calculate: (i) Power dissipated (ii) Current and (iii) Inductance of the coil.

Given S = 1.3 kVA = 1300 VA, Q = 1.2 kVAR = 1200 VAR, V = 240 V, f = 50 Hz.

(i) Power dissipated (active power)

From the power triangle S² = P² + Q²:

P = √(1300² − 1200²) = √(1 690 000 − 1 440 000) = √250 000 = 500 W

(Power factor = P/S = 500/1300 = 0.385 lagging.)

(ii) Current

I = S/V = 1300/240 = 5.42 A

(iii) Inductance of the coil

Q = I² X_L, so X_L = 1200 / 5.417² = 1200/29.34 = 40.9 Ω

L = X_L/(2πf) = 40.9/314.16 = 0.130 H (≈ 130 mH)

The coil resistance is R = P/I² = 500/29.34 = 17.0 Ω. Check: Z = √(17.0² + 40.9²) = 44.3 Ω = 240/5.417.

✓ Answer checked by Claude Opus 5.5 (high effort)

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