Question

Two coils A and B are connected in series across a 240-V, 50-Hz supply. The resistance of A is 5 Ω and the inductance of B is 0.015 H. If the input from the supply is 3 kW and 2 kVAR, find the inductance of A and the resistance of B. Calculate the voltage across each coil.

Answer 1

The kVA triangle is shown in Fig. The circuit kVA is given by,

kVA = √(32 + 22) = 3.606 or VA = 3,606 voltmeters

Circuit current = 3,606/240 = 15.03 A

∴ 15.032 (RA + RB) = 3,000

∴ RA + RB = 3,000/15.032 = 13.3 Ω

∴ RB = 13.3 − 5 = 8.3 Ω

Now, impedance of the whole circuit is given by

Z = 240/15.03 = 15.97 Ω

∴ XA + XB = √(Z2 - (RA + RB)2) = √(15.972 - 13.33) = 8.84 Ω

Now XB = 2π × 50 × 0.015 = 4.713 Ω

∴ XA = 8.843 − 4.713 = 4.13Ω or 2π × 50 × LA = 4.13

∴ LA = 0.0132 H (approx)

Now ZA = √(R2A + X2A) = √(52 + 4.132) = 6.585Ω

P.D. across coil A = I . ZA = 15.03 × 6.485 = 97.5 V; Zb = √(8.32 + 4.7132) = 9.545Ω

∴ p.d. across coil B = I . ZB = 15.03 × 9.545 = 143.5 V