---
title: "Full-bridge inverter RLC load to fifth harmonic"
subject: "Power Electronics"
chapter: 5
key: "semester_7/power_electronics:full-bridge-inverter-rlc-155uf"
source: https://www.bctengineer.com/q/semester_7/power_electronics/full-bridge-inverter-rlc-155uf
site: BCT Engineering
---

# Full-bridge inverter RLC load to fifth harmonic

Power Electronics · Chapter 5: Inverter

- Asked in: 2081 Bhadra, 8 marks
- Video: [Lec 38(e) Full Bridge Inverter | RLC load | Power Electronics](https://www.youtube.com/watch?v=cc2p3yFM7Vc) — Gatematic by Sohail Sir
- 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-rlc-155uf
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5

## Question

A single-phase full-bridge inverter has RLC load of R = 4 Ω, L = 35 mH and C = 155 µF. The dc input voltage is 230 V and the output frequency is 50 Hz. i) Find an expression for load current up to fifth harmonic. Also, calculate ii) Rms value of fundamental load current. iii) The power absorbed by load and the fundamental power.

## Short answer

Harmonic voltages Vnm=4×230/(nπ) V; load impedances Zn=√[R²+(nωL-1/(nωC))²] give currents. Up to 5th: i0(t)=28.31 sin(314.16t+67.25°)+3.69 sin(942.48t-81.30°)+1.15 sin(1570.8t-85.50°) A. Fundamental RMS=20.02 A; total power=1632.5 W, fundamental power=1602.6 W.

## Answer

For a square-wave full-bridge inverter, the output voltage contains only odd harmonics, $v_0 = \sum \frac{4V_s}{n\pi}\sin n\omega t$. Each harmonic current is found by dividing by the load impedance at that harmonic, and the currents are added (superposition).

**Given:** $R = 4\ \Omega$, $L = 35$ mH, $C = 155\ \mu$F, $V_s = 230$ V, $f = 50$ Hz, $\omega = 314.16$ rad/s.

#### i) Load current up to the fifth harmonic
$$
\begin{aligned}
V_{nm} &= \frac{4V_s}{n\pi} = \frac{292.85}{n}\ \text{V} \\
Z_n &= \sqrt{R^2 + \Big(n\omega L - \frac{1}{n\omega C}\Big)^2}, \qquad \theta_n = \tan^{-1}\frac{n\omega L - 1/(n\omega C)}{R}
\end{aligned}
$$

| $n$ | $V_{nm}$ (V) | $X_L$ (Ω) | $X_C$ (Ω) | $X$ (Ω) | $Z_n$ (Ω) | $\theta_n$ | $I_{nm}$ (A) |
|---|---|---|---|---|---|---|---|
| 1 | 292.85 | 11.00 | 20.54 | −9.54 | 10.35 | −67.25° | 28.31 |
| 3 | 97.62 | 32.99 | 6.85 | 26.14 | 26.45 | 81.30° | 3.69 |
| 5 | 58.57 | 54.98 | 4.11 | 50.87 | 51.03 | 85.50° | 1.15 |

The current is $i_n = (V_{nm}/Z_n)\sin(n\omega t - \theta_n)$. The fundamental sees a capacitive load, so it leads:

$$
i_0(t) = 28.31\sin(314.16t + 67.25^\circ) + 3.69\sin(942.48t - 81.30^\circ) + 1.15\sin(1570.8t - 85.50^\circ)\ \text{A}
$$

#### ii) RMS value of fundamental load current
$$
I_{1} = \frac{28.31}{\sqrt2} = 20.02\ \text{A}
$$

#### iii) Power absorbed by the load and fundamental power
RMS load current (harmonics up to the 5th):

$$
\begin{aligned}
I_3 &= \frac{3.691}{\sqrt2} = 2.61\ \text{A}, \qquad I_5 = \frac{1.148}{\sqrt2} = 0.81\ \text{A} \\
I_0 &= \sqrt{20.02^2 + 2.61^2 + 0.81^2} = 20.20\ \text{A}
\end{aligned}
$$

$$
\begin{aligned}
P_0 &= I_0^2R = 20.20^2\times4 = 1632.5\ \text{W} \\
P_1 &= I_1^2R = 20.02^2\times4 = 1602.6\ \text{W}
\end{aligned}
$$

(Including all higher harmonics changes $P_0$ only slightly, to about 1633.6 W.)

**Answer:** $I_{1,rms} = 20.02$ A; load power $\approx 1632.5$ W; fundamental power $= 1602.6$ W.

## Check yourself

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

1. How is the load current of a square-wave inverter feeding an RLC load found?
   A. Divide the dc voltage by R
   B. Use only the fundamental impedance
   C. Use the rms voltage divided by the average impedance
   D. Divide each harmonic voltage by the impedance at that harmonic and add the currents
   Answer: D. Superposition: each odd harmonic 4Vs/(n pi) is divided by Z_n at n*omega. (Key point: Method)

2. At 50 Hz, X_L = 11.0 Ω and X_C = 20.54 Ω. What is the load nature at the fundamental?
   A. Inductive, current lags
   B. Resistive, current in phase
   C. Capacitive, current leads
   D. Series resonant
   Answer: C. X = X_L - X_C = -9.54 Ω is negative, so the load is capacitive and the current leads. (Key point: Nature of fundamental)

3. What is the rms fundamental load current?
   A. 28.31 A
   B. 20.02 A
   C. 14.15 A
   D. 41.4 A
   Answer: B. I_1m = 292.85/10.35 = 28.31 A, so I_1 = 28.31/sqrt2 = 20.02 A. (Key point: Fundamental current)

4. Why is the load power (about 1632 W) slightly larger than the fundamental power (1602.6 W)?
   A. Harmonic currents also dissipate I^2R in R
   B. The capacitor consumes power
   C. The inductor consumes real power
   D. The dc source adds power
   Answer: A. Total power = I_0^2 R, with I_0 including the rms harmonic currents; only R dissipates power. (Key point: Power)

## Related in Power Electronics

- [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 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
- [Single-phase diode bridge with constant load current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Adiode-bridge-rl-load-numerical) — Power Electronics, chapter 2
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- [Single-phase inverter fundamental output voltage](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asingle-phase-inverter-r-fundamental) — Power Electronics, chapter 5

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