---
title: "Square-wave inverter ±100 V: fundamental load current"
subject: "Power Electronics"
chapter: 5
key: "semester_7/power_electronics:inverter-100v-rl-fundamental"
source: https://www.bctengineer.com/q/semester_7/power_electronics/inverter-100v-rl-fundamental
site: BCT Engineering
---

# Square-wave inverter ±100 V: fundamental load current

Power Electronics · Chapter 5: Inverter

- Asked in: 2070 Asar, 8 marks
- Video: [Deriving an Equation for the Inductive Load Current Fed from a Square Wave Inverter, 17/11/2016](https://www.youtube.com/watch?v=PS7BU8cw1-Y) — Lutfi Al-Sharif
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/inverter-100v-rl-fundamental
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5

## Question

The following Fig.4b shows the waveform of the output voltage of a single phase inverter with an inductive load of R = 10 ohm and L = 20mH connected in series. Calculate the magnitude and phase of fundamental component of the output voltage and write down the time domain equation of fundamental component of load current. [Figure: square wave, +100 V for the first half period and −100 V for the second half, period T = 0.02 sec]

## Short answer

Fundamental voltage of the ±100 V square wave is 127.32 V peak at 0° (90.03 V rms). With R=10 Ω, L=20 mH, fundamental current is i1(t)=10.78 sin(314.16t - 32.14°) A.

## Answer

The output is a square wave of $\pm100$ V with $T = 0.02$ s, so $f = 50$ Hz and $\omega = 2\pi f = 314.16$ rad/s. Its fundamental is found from the Fourier series $v_0 = \sum_{n\ \text{odd}}\frac{4V}{n\pi}\sin n\omega t$.

**Load:** $R = 10\ \Omega$, $L = 20$ mH in series.

#### Fundamental component of output voltage
$$
\begin{aligned}
V_{1m} &= \frac{4V}{\pi} = \frac{4\times100}{\pi} = 127.32\ \text{V} \\
V_{1,rms} &= \frac{127.32}{\sqrt2} = 90.03\ \text{V}
\end{aligned}
$$

Phase: the square wave starts its positive half at $t = 0$, so the fundamental is in phase with it (0°):

$$
v_1(t) = 127.32\sin(314.16t)\ \text{V}
$$

#### Fundamental load current
$$
\begin{aligned}
X_1 &= \omega L = 314.16\times0.02 = 6.283\ \Omega \\
Z_1 &= \sqrt{10^2 + 6.283^2} = 11.810\ \Omega \\
\theta_1 &= \tan^{-1}\frac{6.283}{10} = 32.14^\circ \ (\text{lagging}) \\
I_{1m} &= \frac{127.32}{11.810} = 10.78\ \text{A}, \qquad I_{1,rms} = 7.62\ \text{A}
\end{aligned}
$$

$$
i_1(t) = 10.78\sin(314.16t - 32.14^\circ)\ \text{A}
$$

**Answer:** fundamental voltage $= 127.32$ V peak (90.03 V rms) at 0°; $i_1(t) = 10.78\sin(314.16t - 32.14^\circ)$ A. Fundamental power delivered $= 7.62^2\times10 = 581$ W.

## Check yourself

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

1. What is the peak fundamental of the +/-100 V square wave?
   A. 100 V
   B. 90.03 V
   C. 157.1 V
   D. 127.32 V
   Answer: D. V1m = 4V/pi = 127.32 V peak. (Key point: Fundamental voltage)

2. Why is the fundamental voltage at phase 0 degrees?
   A. Because the square wave starts its positive half at t = 0, so the fundamental is in phase with it
   B. Because the load is resistive
   C. Because the inductor sets it to zero
   D. Because the fundamental is always leading
   Answer: A. The wave begins positive at t = 0, so its fundamental is sin(wt) with 0 degree phase. (Key point: Phase of voltage)

3. For R = 10 ohm, L = 20 mH at 50 Hz, what are Z1 and the load angle?
   A. 11.81 ohm and 32.14 degrees
   B. 16.28 ohm and 57.5 degrees
   C. 11.81 ohm and 57.86 degrees
   D. 10 ohm and 0 degrees
   Answer: A. X1 = 6.283 ohm, Z1 = 11.81 ohm, theta = tan^-1(0.6283) = 32.14 degrees. (Key point: Load impedance and angle)

4. Which is the fundamental load current?
   A. 12.73 sin(314.16t - 32.14 degrees) A
   B. 10.78 sin(314.16t + 32.14 degrees) A
   C. 10.78 sin(314.16t - 32.14 degrees) A
   D. 10.78 sin(314.16t - 57.86 degrees) A
   Answer: C. I1m = 127.32/11.81 = 10.78 A, lagging by 32.14 degrees. (Key point: Fundamental current equation)

## Close variants set in other exams

- [Square-wave inverter RL load current up to 5th order](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ainverter-square-wave-rl-fifth-order) — Power Electronics, chapter 5, asked 2×

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