A voltage e(t) = 100 sin 314 t is applied to a series circuit consisting of 10 Ω resistance, 0.0318 H inductance and a capacitor of 63.6 μF. Calculate (i) impedance of a series circuit, (ii) expression for i(t), (iii) power factor (iv) active power,…

Given: V_m = 100 V, ω = 314 rad/s (f = 50 Hz), R = 10 Ω, L = 0.0318 H, C = 63.6 μF.

\( X_L = \omega L = 314\times0.0318 = 9.985\ \Omega \approx 10\ \Omega \); \( X_C = \dfrac{1}{\omega C} = \dfrac{1}{314\times63.6\times10^{-6}} = 50.07\ \Omega \approx 50\ \Omega \).

(i) Impedance: \( Z = R + j(X_L - X_C) = 10 + j(9.985 - 50.07) = 10 - j40.09\ \Omega \)

\( |Z| = \sqrt{10^2 + 40.09^2} = \) 41.32 Ω, φ = tan⁻¹(−40.09/10) = −75.99° (capacitive). (With rounded reactances 10 − j40: 41.23∠−75.96° Ω.)

(ii) Current: \( I_m = \dfrac{V_m}{|Z|} = \dfrac{100}{41.32} = 2.42 \) A; current leads voltage by 75.99° (1.326 rad):

i(t) = 2.42 sin(314 t + 75.99°) A = 2.42 sin(314 t + 1.326) A.

(iii) Power factor: \( \cos\phi = R/|Z| = 10/41.32 = \) 0.242 leading.

(iv) Powers: V_rms = 100/√2 = 70.71 V, I_rms = 2.42/√2 = 1.711 A.

• Active power \( P = I^2R = 1.711^2\times10 = \) 29.29 W (= V I cos φ = 70.71 × 1.711 × 0.242 ✓).
• Reactive power \( Q = I^2(X_C - X_L) = 1.711^2\times40.09 = \) 117.4 VAR (capacitive / leading).
• Apparent power \( S = VI = 70.71\times1.711 = \) 121.0 VA (check √(29.29² + 117.4²) = 121.0 ✓).