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RL coil on sinusoidal source

Short answer

Reactance X_L=31.4 Ω, impedance 37.23 Ω. Instantaneous current is 5.372 sin(314t−57.5°) A, power factor 0.537 lagging. Active power 288.6 W, reactive 453 VAR, apparent 537 VA. Voltage drops: resistor 76 V, inductor 119.3 V.

An alternating source of emf v =200sin (314t) volts is applied to a practical coil with resistance 20Ω and inductance 0.1 H respectively. Determine (i) expression for instantaneous current and power factor (ii) active, reactive and apparent power of circuit (iii) voltage drop on resistor and inductor and (iv) construct phasor diagram for above circuit.

Given v = 200 sin 314t, so Vm = 200 V, ω = 314 rad/s. R = 20 Ω, L = 0.1 H.

  • X_L = ωL = 314 × 0.1 = 31.4 Ω
  • Z = R + jX_L = 20 + j31.4 = 37.23∠57.5° Ω

(i) Instantaneous current and power factor

  • Im = Vm/Z = 200/37.23 = 5.372 A
  • φ = tan⁻¹(31.4/20) = 57.5° (current lags)

i = 5.372 sin(314t − 57.5°) A

Power factor = cos 57.5° = R/Z = 20/37.23 = 0.537 lagging

(ii) Powers

V_rms = 200/√2 = 141.4 V, I_rms = 5.372/√2 = 3.80 A

  • Apparent: S = VI = 141.4 × 3.80 = 537 VA
  • Active: P = I²R = 3.80² × 20 = 288.6 W (= VI cos φ)
  • Reactive: Q = I²X_L = 3.80² × 31.4 = 453 VAR (lagging)

(iii) Voltage drops (rms)

  • V_R = I R = 3.80 × 20 = 76.0 V (peak 107.5 V)
  • V_L = I X_L = 3.80 × 31.4 = 119.3 V (peak 168.7 V)

Check: √(76.0² + 119.3²) = 141.4 V.

(iv) Phasor diagram

  V_L = 119.3 V ▲          ↗ V = 141.4 V
                │        ↗
                │      ↗
                │    ↗
                │  ↗  φ = 57.5°
                O──────────────────► I (reference)
                └── V_R = 76 V ──┘

Take I as reference along the horizontal axis. V_R is in phase with I, V_L is 90° ahead of I, and the supply voltage V is their phasor sum, leading I by 57.5°.

✓ Answer checked by Claude Opus 5.5 (high effort)

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