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RMS, phase difference and power from v and i

Short answer

RMS: Vrms=353.55 V, Irms=3.54 A; average half-cycle: Vav=318.3 V, Iav=3.18 A. At t=0.001 s, ωt=0.4 rad. Phase difference = 90° (voltage leads). Power consumed = 0 W (purely reactive).

The impressed voltage and the current flowing through a circuit are: v = 500 sin (400t + π/3) and i = 5 sin (400t – π/6). Determine the i) rms and average value of v and i, ii) Radian through which it's vector has gone when t = 0.001 sec, iii) Phase difference, and iv) Power consumed in the circuit.

Given v = 500 sin(400t + π/3) V and i = 5 sin(400t − π/6) A. So Vm = 500 V, Im = 5 A, ω = 400 rad/s.

i) RMS and average values

  • V_rms = Vm/√2 = 500/1.414 = 353.55 V
  • I_rms = 5/1.414 = 3.54 A
  • V_av (half cycle) = 2Vm/π = 2×500/π = 318.3 V
  • I_av (half cycle) = 2×5/π = 3.18 A

(Over a full cycle the average of a sine wave is zero.)

ii) Radian passed by the vector at t = 0.001 s

Angle swept = ωt = 400 × 0.001 = 0.4 rad (= 22.9°).

The actual phase angles at that instant are: voltage 0.4 + π/3 = 1.447 rad (82.9°), current 0.4 − π/6 = −0.124 rad (−7.1°).

iii) Phase difference

φ = π/3 − (−π/6) = π/2 rad = 90°, with the voltage leading the current.

iv) Power consumed

P = V_rms I_rms cos φ = 353.55 × 3.536 × cos 90° = 0 W

The circuit is purely reactive (an ideal inductor) so no real power is consumed. The reactive power is Q = 353.55 × 3.536 = 1250 VAR.

✓ Answer checked by Claude Opus 5.5 (high effort)

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