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RL series circuit current, pf, power

Short answer

XL = 2π×50×0.03 = 9.425 Ω, Z = √(5²+9.425²) = 10.669 Ω. Current I = 230/10.669 = 21.56 A. Power factor cosφ = 5/10.669 = 0.469 lagging, power consumed P = I²×5 = 2.32 kW.

A coil having a resistance of 5 Ω and inductance of 30 mH in series are connected across a 230 V, 50 Hz supply. Calculate current, power factor, and power consumed.

Given: R=5 ΩR=5\ \Omega, L=30L=30 mH, V=230V=230 V, f=50f=50 Hz.

Inductive reactance:

XL=2πfL=2π×50×0.03=9.425 ΩX_L=2\pi fL=2\pi\times50\times0.03=9.425\ \Omega

Impedance:

Z=R2+XL2=52+9.4252=10.669 ΩZ=\sqrt{R^2+X_L^2}=\sqrt{5^2+9.425^2}=10.669\ \Omega

Current:

I=VZ=23010.669=21.56 AI=\frac{V}{Z}=\frac{230}{10.669}=21.56\ \text{A}

Power factor:

cos⁡ϕ=RZ=510.669=0.469 (lagging)\cos\phi=\frac{R}{Z}=\frac{5}{10.669}=0.469\ \text{(lagging)}

Power consumed:

P=I2R=(21.56)2×5=2323.7 WP=I^2R=(21.56)^2\times5=2323.7\ \text{W}

(Check: VIcos⁡ϕ=230×21.56×0.4686=2323.7VI\cos\phi=230\times21.56\times0.4686=2323.7 W.)

Answer: I=21.56I=21.56 A, pf =0.469=0.469 lagging, P=2.32P=2.32 kW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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