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Three coils in delta, 415 V

Short answer

Each coil impedance Z = sqrt(20² + (2π×50×0.08)²) = 32.12 Ω. In delta, phase voltage = 415 V; phase current = 415/32.12 = 12.92 A. Line current = √3 × 12.92 = 22.38 A. Total power = 3 × 12.92² × 20 = 10.02 kW. Power factor = R/Z = 20/32.12 = 0.623 lagging.

Three identical coils, each having resistance of 20 Ω and inductance of 0.08 H are connected in delta across a three-phase, 415 V, 50 Hz supply. Calculate: i) The phase current, ii) The line current, iii) The total power consumed, and iv) power factor.

Given: each coil R=20 ΩR=20\ \Omega, L=0.08L=0.08 H, delta connection, VL=415V_L=415 V, f=50f=50 Hz. In delta, Vph=VL=415V_{ph}=V_L=415 V.

Coil impedance:

XL=2π×50×0.08=25.13 Ω,Zph=202+25.132=32.12 ΩX_L=2\pi\times50\times0.08=25.13\ \Omega, \qquad Z_{ph}=\sqrt{20^2+25.13^2}=32.12\ \Omega

(i) Phase current:

Iph=41532.12=12.92 AI_{ph}=\frac{415}{32.12}=12.92\ \text{A}

(ii) Line current:

IL=3 Iph=3×12.92=22.38 AI_L=\sqrt3\,I_{ph}=\sqrt3\times12.92=22.38\ \text{A}

(iii) Total power:

P=3Iph2R=3×(12.92)2×20=10016 WP=3I_{ph}^2R=3\times(12.92)^2\times20=10016\ \text{W}

(iv) Power factor:

cos⁡ϕ=RZ=2032.12=0.623 lagging (ϕ=51.5∘)\cos\phi=\frac{R}{Z}=\frac{20}{32.12}=0.623\ \text{lagging}\ (\phi=51.5^\circ)

(Check: P=3VLILcos⁡ϕ=3×415×22.38×0.6227=10016P=\sqrt3V_LI_L\cos\phi=\sqrt3\times415\times22.38\times0.6227=10016 W.)

Answer: (i) 12.92 A, (ii) 22.38 A, (iii) 10.02 kW, (iv) 0.623 lagging.

✓ Answer checked by Claude Opus 5.5 (high effort)

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