Series RL circuit with 283 sin(300t+90°)

Given R = 4 Ω, L = 0.01 H, v = 283 sin(300t + 90°) V. So Vm = 283 V, ω = 300 rad/s.

• X_L = ωL = 300 × 0.01 = 3 Ω
• Z = 4 + j3 = 5∠36.87° Ω

i) Power factor

cos φ = R/Z = 4/5 = 0.8 lagging

ii) Expression for i(t)

Im = Vm/Z = 283/5 = 56.6 A. The current lags the voltage by 36.87°, so its phase is 90° − 36.87° = 53.13°.

i(t) = 56.6 sin(300t + 53.13°) A

iii) Power dissipated

I_rms = 56.6/√2 = 40.02 A

P = I_rms² R = 40.02² × 4 = 6407 W (≈ 6.4 kW)

Check: V_rms I cos φ = 200.1 × 40.02 × 0.8 = 6407 W.

iv) Voltage drops (rms)

• V_R = I R = 40.02 × 4 = 160.1 V (peak 226.4 V)
• V_L = I X_L = 40.02 × 3 = 120.1 V (peak 169.8 V)

Check: √(160.1² + 120.1²) = 200.1 V = 283/√2.

v) Phasor diagram (I drawn as reference)

                         V = 200.1 V
                       ↗ ↑
                     ╱   │
                   ╱     │ V_L = 120.1 V (leads I by 90°)
                 ╱       │
               ╱ 36.87°  │
   ───────────●──────────┴──────► I = 40.02 A
               V_R = 160.1 V (in phase with I)

V_R is in phase with I, V_L leads I by 90°, and their sum V leads I by 36.87°. In absolute terms I is at +53.13° and V at +90°, so the whole diagram is simply rotated by 53.13°.