- 2068 Shrawan · 8 marks
Current response and average power, non-sinusoidal source
Short answer
Use superposition: at ω=500 rad/s, Z₁=6.5+j14.63 Ω gives I₁=2.498 A ∠−66.05°; at 3ω, Z₃=6.5+j98.21 Ω gives I₃=0.813 A ∠−86.21°. Total current i(t) = 2.498 sin(500t−66.05°) + 0.813 sin(1500t−86.21°) A. Average power P = ½(2.498²+0.813²)×6.5 = 22.43 W.
The network of figure shown below has an applied voltage of v(t) = (40 sin ωt + 80 sin 3ωt) volts where ω = 500 rad/s. Find the current response and hence the average power. [Figure: source v(t) in series with a 6.5 Ω resistor, a 0.07 H inductor and a 98.2 μF capacitor]
The source has two harmonics (ω and 3ω), so use superposition: find the current for each harmonic by phasor analysis, add the time functions, and add the average powers (powers of different frequencies add, because cross-terms average to zero).
Circuit (assumption). The 6.5 Ω resistor, 0.07 H inductor and 98.2 μF capacitor are in series with the source; ω = 500 rad/s.
Fundamental (ω = 500 rad/s, Vm = 40 V)
XL = ωL = 500 × 0.07 = 35 Ω
XC = 1/(ωC) = 1/(500 × 98.2×10⁻⁶) = 20.37 Ω
Z1 = 6.5 + j(35 − 20.37) = 6.5 + j14.63 = 16.01 ∠66.05° Ω
I1m = 40/16.01 = 2.498 A, i1(t) = 2.498 sin(500t − 66.05°) A
Third harmonic (3ω = 1500 rad/s, Vm = 80 V)
XL = 1500 × 0.07 = 105 Ω
XC = 1/(1500 × 98.2×10⁻⁶) = 6.79 Ω
Z3 = 6.5 + j(105 − 6.79) = 6.5 + j98.21 = 98.43 ∠86.21° Ω
I3m = 80/98.43 = 0.8128 A, i3(t) = 0.8128 sin(1500t − 86.21°) A
Current response
i(t) = 2.498 sin(500t − 66.05°) + 0.813 sin(1500t − 86.21°) A
I_rms = √[(2.498² + 0.8128²)/2] = 1.858 A
Average power
P1 = ½ I1m² R = ½ (2.498)² (6.5) = 20.28 W
P3 = ½ I3m² R = ½ (0.8128)² (6.5) = 2.147 W
P = P1 + P3 = 22.43 W
Check: P = I_rms² R = (1.858)² × 6.5 = 22.43 W. The source r.m.s. voltage is √[(40² + 80²)/2] = 63.25 V, so the power factor is P/(V_rms I_rms) = 0.191.
Answer: i(t) = 2.498 sin(500t − 66.05°) + 0.813 sin(1500t − 86.21°) A; average power P = 22.43 W.
✓ Answer checked by Claude Opus 5.5 (high effort)
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