---
title: "Full-bridge square-wave inverter, 400 V, RL load"
subject: "Power Electronics"
chapter: 5
key: "semester_7/power_electronics:full-bridge-inverter-400v-rl"
source: https://www.bctengineer.com/q/semester_7/power_electronics/full-bridge-inverter-400v-rl
site: BCT Engineering
---

# Full-bridge square-wave inverter, 400 V, RL load

Power Electronics · Chapter 5: Inverter

- Asked in: 2073 Shrawan, 8 marks
- Video: [Single phase full bridge inverter with RL-load](https://www.youtube.com/watch?v=wG8ZpdyTwa0) — Anet Jose
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/full-bridge-inverter-400v-rl
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5

## Question

A single phase full bridge inverter with dc input voltage of Vs = 400V and generating output square wave of 50Hz is connected to inductive load having R = 10 Ω and L = 50mH. Calculate the magnitude and phase of fundamental component and third harmonic component of output voltage and load current. [Figure: block 'Bridge inverter' fed from Vs = 400 V dc, output Vo across the RL load]

## Short answer

For square-wave output, V1m = 4×400/π = 509.30 V, V3m = 169.77 V. Load impedance: Z1 = 18.62 Ω, θ1 = 57.52°; Z3 = 48.17 Ω, θ3 = 78.02°. Fundamental current: 27.35 A peak, lagging 57.52°; third harmonic: 3.52 A peak, lagging 78.02°.

## Answer

The square-wave output of a full-bridge inverter has the Fourier series $v_0 = \sum_{n\ \text{odd}}\frac{4V_s}{n\pi}\sin n\omega t$. Taking the voltage as the reference (phase 0°), each harmonic current lags its voltage by the load angle $\theta_n = \tan^{-1}(n\omega L/R)$.

**Given:** $V_s = 400$ V, $f = 50$ Hz ($\omega = 314.16$ rad/s), $R = 10\ \Omega$, $L = 50$ mH.

#### Fundamental component ($n = 1$)
$$
\begin{aligned}
V_{1m} &= \frac{4V_s}{\pi} = \frac{4\times400}{\pi} = 509.30\ \text{V}, \qquad V_{1,rms} = 360.13\ \text{V} \\
X_1 &= \omega L = 314.16\times0.05 = 15.708\ \Omega \\
Z_1 &= \sqrt{10^2 + 15.708^2} = 18.621\ \Omega, \quad \theta_1 = \tan^{-1}\frac{15.708}{10} = 57.52^\circ \\
I_{1m} &= \frac{509.30}{18.621} = 27.35\ \text{A}, \qquad I_{1,rms} = 19.34\ \text{A}
\end{aligned}
$$

#### Third harmonic ($n = 3$)
$$
\begin{aligned}
V_{3m} &= \frac{509.30}{3} = 169.77\ \text{V}, \qquad V_{3,rms} = 120.04\ \text{V} \\
X_3 &= 3\omega L = 47.124\ \Omega \\
Z_3 &= \sqrt{10^2 + 47.124^2} = 48.173\ \Omega, \quad \theta_3 = \tan^{-1}\frac{47.124}{10} = 78.02^\circ \\
I_{3m} &= \frac{169.77}{48.173} = 3.524\ \text{A}, \qquad I_{3,rms} = 2.492\ \text{A}
\end{aligned}
$$

#### Results
| Component | Voltage (peak, phase) | Current (peak, phase) |
|---|---|---|
| Fundamental (50 Hz) | 509.30 V ∠0° | 27.35 A ∠−57.52° |
| 3rd harmonic (150 Hz) | 169.77 V ∠0° | 3.524 A ∠−78.02° |

$$
\begin{aligned}
v_1 &= 509.30\sin\omega t, & i_1 &= 27.35\sin(\omega t - 57.52^\circ) \\
v_3 &= 169.77\sin3\omega t, & i_3 &= 3.524\sin(3\omega t - 78.02^\circ)
\end{aligned}
$$

The third harmonic voltage is one third of the fundamental, but its current is only about 13% of the fundamental current, because the inductive reactance triples at 150 Hz.

## Check yourself

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

1. What is the peak fundamental output voltage of the full-bridge square-wave inverter with Vs = 400 V?
   A. 400 V
   B. 254.65 V
   C. 360.13 V
   D. 509.30 V
   Answer: D. V1m = 4Vs/pi = 509.30 V (360.13 V is the rms). (Key point: Fundamental voltage)

2. How is the phase of the nth harmonic current found relative to the voltage?
   A. It leads by tan^-1(R/(n w L))
   B. It lags by tan^-1(n w L/R)
   C. It lags by tan^-1(w L/R) for every n
   D. It is in phase with the voltage
   Answer: B. The reactance scales with n, so theta_n = tan^-1(n w L/R), with the current lagging. (Key point: Load angle per harmonic)

3. What are the fundamental current magnitude and phase (R = 10 ohm, L = 50 mH, 50 Hz)?
   A. 27.35 A peak, lagging 57.52 degrees
   B. 27.35 A peak, lagging 78.02 degrees
   C. 50.9 A peak, lagging 57.52 degrees
   D. 27.35 A peak, leading 57.52 degrees
   Answer: A. Z1 = 18.62 ohm, I1m = 509.3/18.62 = 27.35 A, theta1 = tan^-1(15.708/10) = 57.52 degrees. (Key point: Fundamental current)

4. Why is the third harmonic current only about 13% of the fundamental current although the voltage is one third?
   A. The third harmonic voltage is in antiphase
   B. The resistance triples at 150 Hz
   C. The inductive reactance triples at 150 Hz, giving a much larger impedance
   D. The supply cannot deliver 150 Hz
   Answer: C. X3 = 3wL = 47.1 ohm so Z3 = 48.17 ohm and I3m = 169.77/48.17 = 3.52 A. (Key point: Third harmonic current)

## Related in Power Electronics

- [Single-phase full-wave inverter with 5 Ω load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-wave-inverter-5ohm) — Power Electronics, chapter 5
- [Half-bridge square-wave inverter at 120 Hz](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asquare-wave-inverter-120hz) — Power Electronics, chapter 5
- [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×
- [Square-wave inverter ±100 V: fundamental load current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ainverter-100v-rl-fundamental) — Power Electronics, chapter 5, asked 2×
- [Full-bridge inverter RLC load, THD](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-bridge-inverter-rlc-112uf) — Power Electronics, chapter 5
- [Single-phase square-wave inverter rms and fundamental](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asquare-wave-inverter-rms-fundamental) — Power Electronics, chapter 5

## Related in other subjects

- [Full bridge inverter; dc-dc output power control](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Afull-bridge-inverter-dc-dc-control) — Advanced Electronics, chapter 7
- [Choppers; single-phase full bridge inverter](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Achoppers-full-bridge-inverter) — Advanced Electronics, chapter 7, asked 2×
- [Full wave bridge rectifier load power](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Abridge-rectifier-power-5ohm) — Electronics Devices and Circuit, chapter 6
- [Full-bridge cantilever load cell output](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Acantilever-full-bridge-output-numerical) — Instrumentation and Measurement, chapter 6