Series RLC circuit analysis

Given R = 100 Ω, L = 0.12 H, C = 28.27 μF, V = 100 V, f = 50 Hz. ω = 314.16 rad/s.

Reactances

• X_L = ωL = 37.70 Ω
• X_C = 1/(ωC) = 1/(314.16 × 28.27 × 10⁻⁶) = 112.60 Ω
• X = X_L − X_C = −74.90 Ω (capacitive)

Impedance

Z = 100 − j74.90 = 124.94∠−36.83° Ω

Current

I = V/Z = 100/124.94 = 0.80 A (leading V by 36.83°)

Power factor

cos φ = R/Z = 100/124.94 = 0.80 leading

Powers

• Active: P = I²R = 0.80² × 100 = 64.1 W
• Reactive: Q = I²X = 0.80² × 74.9 = 48.0 VAR (capacitive)
• Apparent: S = VI = 100 × 0.80 = 80 VA

Voltages across elements (rms)

• V_R = I R = 80.0 V
• V_L = I X_L = 0.80 × 37.70 = 30.2 V
• V_C = I X_C = 0.80 × 112.60 = 90.1 V

Check: √(80² + (90.1 − 30.2)²) = √(6400 + 3588) = 100 V.

Phasor diagram (I reference)

          ↑ V_L = 30.2 V
          │
          ●──────────────►  I (reference); V_R = 80 V along I
          │  ╲
          │    ╲ 36.8°
          │      ↘ V = 100 V (lags I)
          ↓
          V_C = 90.1 V

V_R is along I, V_L is 90° ahead of I and V_C is 90° behind I. Since V_C > V_L, the resultant V lags I by 36.8°, so the circuit is capacitive.