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RLC series choke coil and capacitor

Short answer

Net capacitive reactance 16.12 Ω, impedance 18.97 Ω. Current 10.54 A leading 58.2°, power factor 0.527 leading. Active power 1111 W, apparent 2108 VA, reactive 1792 VAR.

A choke coil having a resistance of 10Ω and inductance of 0.05H is connected in series with a condenser of 100 μF. The whole circuit has been connected to 200V, 50 Hz supply. Calculate (i) impedance (ii) current (iii) power factor (iv) power input (v) Apparent and reactive power of circuit.

Given R = 10 Ω, L = 0.05 H, C = 100 μF in series; V = 200 V, 50 Hz. ω = 314.16 rad/s.

  • X_L = ωL = 15.71 Ω
  • X_C = 1/(ωC) = 31.83 Ω
  • Net reactance X = X_L − X_C = −16.12 Ω (capacitive)

(i) Impedance

Z = R + jX = 10 − j16.12 Ω

|Z| = √(10² + 16.12²) = 18.97 Ω, angle −58.2°

(ii) Current

I = V/Z = 200/18.97 = 10.54 A (leading the voltage by 58.2°)

(iii) Power factor

cos φ = R/Z = 10/18.97 = 0.527 leading

(iv) Power input

P = I²R = 10.54² × 10 = 1111 W (= VI cos φ = 200 × 10.54 × 0.527)

(v) Apparent and reactive power

  • Apparent power S = VI = 200 × 10.54 = 2108 VA
  • Reactive power Q = VI sin φ = I²|X| = 10.54² × 16.12 = 1792 VAR (capacitive)

Check: √(1111² + 1792²) = 2108 VA.

✓ Answer checked by Claude Opus 5.5 (high effort)

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