---
title: "Single-phase full-wave inverter with 5 Ω load"
subject: "Power Electronics"
chapter: 5
key: "semester_7/power_electronics:full-wave-inverter-5ohm"
source: https://www.bctengineer.com/q/semester_7/power_electronics/full-wave-inverter-5ohm
site: BCT Engineering
---

# Single-phase full-wave inverter with 5 Ω load

Power Electronics · Chapter 5: Inverter

- Asked in: 2079 Baishakh, 8 marks
- Video: [Single Phase Full Bridge Inverter | Power Electronics | Lecture 77](https://www.youtube.com/watch?v=Nl3saO5AG4A) — Electronics Maddy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/full-wave-inverter-5ohm
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5

## Question

A single phase full wave inverter has a resistive load of 5 ohm. The dc input voltage is 30. Find (i) rms value of output voltage (ii) rms value of fundamental component of output voltage (iii) output power (iv) peak current in each thyristors.

## Short answer

For Vs=30 V, R=5 Ω: V0,rms=30 V, fundamental rms=27.01 V, output power=180 W, and peak thyristor current=6 A.

## Answer

A single-phase full-wave (full-bridge) inverter with square-wave switching gives $v_0 = \pm V_s$ across the load; with a resistive load the current is a square wave of $\pm V_s/R$.

**Given:** $R = 5\ \Omega$, $V_s = 30$ V (unit taken as volts).

#### (i) RMS output voltage
$$
V_{0,rms} = \left[\frac{2}{T}\int_0^{T/2}V_s^2\,dt\right]^{1/2} = V_s = 30\ \text{V}
$$

#### (ii) RMS of fundamental component
From $v_0 = \sum_{n\ \text{odd}}\frac{4V_s}{n\pi}\sin n\omega t$:

$$
V_{1,rms} = \frac{4V_s}{\sqrt2\,\pi} = 0.9003\times30 = 27.01\ \text{V}
$$

#### (iii) Output power
$$
P_0 = \frac{V_{0,rms}^2}{R} = \frac{30^2}{5} = 180\ \text{W}
$$

(Fundamental power alone is $27.01^2/5 = 145.9$ W.)

#### (iv) Peak current in each thyristor
Each pair (T1, T2 or T3, T4) carries the full load current for its half cycle:

$$
I_{T,peak} = \frac{V_s}{R} = \frac{30}{5} = 6\ \text{A}
$$

Also, average thyristor current $= 6\times0.5 = 3$ A and rms thyristor current $= 6/\sqrt2 = 4.24$ A.

**Answer:** $V_{0,rms} = 30$ V; $V_{1,rms} = 27.01$ V; $P_0 = 180$ W; $I_{T,peak} = 6$ A.

## Check yourself

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

1. For a full-bridge square wave of ±30 V, what is the rms output?
   A. 27.01 V
   B. 21.2 V
   C. 30 V
   D. 42.4 V
   Answer: C. A ±Vs square wave has rms equal to Vs = 30 V. (Key point: RMS output)

2. What is the rms of the fundamental?
   A. 30 V
   B. 27.01 V
   C. 38.2 V
   D. 13.5 V
   Answer: B. V1,rms = 4Vs/(sqrt2 pi) = 0.9003 x 30 = 27.01 V. (Key point: Fundamental)

3. What is the output power (R = 5 Ω)?
   A. 180 W
   B. 145.9 W
   C. 90 W
   D. 360 W
   Answer: A. P = V0,rms^2/R = 900/5 = 180 W; 145.9 W would be the fundamental alone. (Key point: Power)

4. What is the peak current in each thyristor?
   A. 3 A
   B. 4.24 A
   C. 12 A
   D. 6 A
   Answer: D. The pair carries the full load current Vs/R = 30/5 = 6 A. (Key point: Thyristor peak current)

## Related in Power Electronics

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