A coil takes a current of 6 A when connected to a 24-V d.c. supply. To obtain the same current with a 50-Hz a.c. supply, the voltage required was 30 V. Calculate (i) the inductance of the coil (ii) the power factor of the coil.
It should be kept in mind the coil offers only resistance to direct voltage whereas it offers impedance to an alternating voltage.
∴ R = 24/6 = 4 Ω; Z = 30/6 = 5Ω
(i) ∴ XL = √(Z2 - R2) = √(52 - 42) = 3Ω
(ii) p.f. = cos φ = R/Z = 4/5 = 0.8 (lag)
