A coil consumes 1 KW power from 50 Hz single phase ac supply and volt-ampere input is 2 KVA. If value of resistance of coil is 10 Ω then find the following parameters with respect to a coil. (i) Value of inductance, (ii) Phase angle, (iii) Power fact…

Given: P = 1000 W, S = 2000 VA, R = 10 Ω, f = 50 Hz.

Current: \( P = I^2R \Rightarrow I = \sqrt{1000/10} = 10 \) A. Voltage: \( V = S/I = 2000/10 = 200 \) V.

(iv) Impedance: \( Z = V/I = 200/10 = \) 20 Ω.

(iii) Power factor: \( \cos\phi = P/S = 1000/2000 = \) 0.5 lagging (also R/Z = 10/20 = 0.5).

(ii) Phase angle: \( \phi = \cos^{-1}0.5 = \) 60° (current lags voltage).

(i) Inductance: \( X_L = \sqrt{Z^2-R^2} = \sqrt{400-100} = 17.32\ \Omega \); \( L = \dfrac{X_L}{2\pi f} = \dfrac{17.32}{314.16} = \) 0.0551 H = 55.1 mH.

(v) Equations: V_m = 200√2 = 282.84 V, I_m = 10√2 = 14.14 A, ω = 314.16 rad/s.

v = 282.84 sin(314.16 t) V; i = 14.14 sin(314.16 t − 60°) A = 14.14 sin(314.16 t − π/3) A.

(vi) Reactive power: \( Q = VI\sin\phi = 2000\times\sin60^\circ = 1732 \) VAR = 1.732 kVAR. Check: \( \sqrt{S^2-P^2} = \sqrt{4-1} = 1.732 \) kVAR; also I²X_L = 100 × 17.32 = 1732 VAR ✓.