Question

A resistance R, an inductance L = 0.01 H and a capacitance C are connected in series. When a voltage v = 400 cos (300 t − 10°) votls is applied to the series combination, the current flowing is 10√2 cos(3000 t - 55°) amperes. Find R and C.

Answer 1

The phase difference between the applied voltage and circuit current is (55° − 10°) = 45° with current lagging. The angular frequency is ω = 3000 radian/second. Since current lags, X_L > X_C. 

Net reactance X = (X_L − X_C). Also X_L = ω_L = 3000 × 0.01 = 30 Ω

tan φ = X/R or tan 45° = X/R 

∴ X = R Now, Z = V_m/I_m = 400/10√2 = 28.3Ω

Z² = R² + X² = 2R²

∴ R = Z/√2 = 28.3/√2 = 20Ω; X = X_L - X_C = 30 - X_C = 20

X_C = 10Ω or 1/ωC = 10 or 1/3000C = 3.3 or C = 33μF