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Coil in series with non-inductive resistance

Short answer

Current I=4.333 A from resistor voltage. Coil resistance R_c=7.37 Ω. Inductive reactance 40.88 Ω yields L=0.130 H. Power absorbed by coil 138.3 W, total circuit power 701.7 W.

A coil is connected in series with a non-inductive resistance of 30Ω across 240V, 50Hz, 1-φ supply. The reading of voltmeters across the coil is 180 V and across the resistance is 130 V. Calculate, i) Inductance of coil ii) Resistance of coil iii) Power absorbed by coil iv) Power absorbed by whole circuit

Let the coil have resistance R_c and reactance X_L. The 30 Ω resistor is non-inductive. Supply V = 240 V, f = 50 Hz.

Current

I = V_R/R = 130/30 = 4.333 A (same current in the series circuit)

Impedances

  • Coil: Z_c = 180/4.333 = 41.54 Ω, so R_c² + X_L² = 41.54² = 1725.6 …(1)
  • Whole circuit: Z = 240/4.333 = 55.38 Ω, so (R_c + 30)² + X_L² = 55.38² = 3067.1 …(2)

(2) − (1): 60R_c + 900 = 1341.5, so R_c = 7.37 Ω

ii) Resistance of coil

R_c = 7.37 Ω

i) Inductance of coil

X_L = √(1725.6 − 54.3) = √1671.3 = 40.88 Ω

L = X_L/(2πf) = 40.88/314.16 = 0.130 H

iii) Power absorbed by coil

P_c = I²R_c = 4.333² × 7.37 = 138.3 W

iv) Power absorbed by whole circuit

P = I²(R_c + 30) = 18.78 × 37.37 = 701.7 W

(Power factor of circuit = 37.37/55.38 = 0.675 lagging; P = 240 × 4.333 × 0.675 = 702 W.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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