---
title: "Current response and average power, non-sinusoidal source"
subject: "Electrical Circuit Theory"
chapter: 6
key: "semester_3/electrical_circuit_theory:fourier-average-power-nonsinusoidal-r-l"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/fourier-average-power-nonsinusoidal-r-l
site: BCT Engineering
---

# Current response and average power, non-sinusoidal source

Electrical Circuit Theory · Chapter 6: Fourier Series and transform

- Asked in: 2068 Shrawan, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/fourier-average-power-nonsinusoidal-r-l
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-6

## Question

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]

## 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.

## Answer

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)

```text
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)

```text
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

```text
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

```text
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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the response of a circuit with 40 sin ωt + 80 sin 3ωt found?
   A. Superposition: phasor analysis at each harmonic then add the time functions
   B. Use the r.m.s. value of the supply only
   C. Treat the source as a d.c. source
   D. Use the fundamental impedance for both
   Answer: A. Linear circuit, so each harmonic is solved separately. (Key point: Method)

2. Why does the 3rd-harmonic impedance differ?
   A. XL and XC remain the same
   B. Only R changes
   C. Z3 = Z1/3
   D. XL becomes 105 Ω and XC falls to 6.79 Ω, so Z3 = 98.43 ∠86.21°
   Answer: D. XL grows with frequency and XC shrinks. (Key point: Impedance at 3ω)

3. How is total average power obtained?
   A. Multiply total Vrms and Irms
   B. Use only the fundamental power
   C. Add P = (1/2) Im^2 R from each harmonic, since cross terms average to zero
   D. Add voltages first and square
   Answer: C. P1 + P3 = 20.28 + 2.147 = 22.43 W. (Key point: Average power)

4. Which relation confirms the power?
   A. P = V_rms I_rms = 117 W
   B. P = I_rms^2 R = 1.858^2 x 6.5 = 22.43 W
   C. P = I_rms R = 12 W
   D. P = V_rms^2/R
   Answer: B. Only R dissipates power. (Key point: Check)

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