--- title: "Square-wave inverter RL load current up to 5th order" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:inverter-square-wave-rl-fifth-order" source: https://www.bctengineer.com/q/semester_7/power_electronics/inverter-square-wave-rl-fifth-order site: BCT Engineering --- # Square-wave inverter RL load current up to 5th order Power Electronics · Chapter 5: Inverter - Asked in: 2073 Chaitra, 8 marks - Video: [single phase full bridge inverter with rl load | single phase full bridge inverter power electronics](https://www.youtube.com/watch?v=frBioU4Jalo) — Mind of Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/inverter-square-wave-rl-fifth-order - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question The following figure below shows the waveform of the output voltage of a single phase inverter with an inductive load of R = 5 ohm and L = 10mH connected in series. Calculate the magnitude and phase of fundamental component of the output voltage and write down the time domain equation of the load current considering Fourier series up to 5th Order. [Figure: square wave, +100 V for the first half period and −100 V for the second half, period T = 0.02 sec (the top level is misprinted as −100V in the figure)] ## Short answer Fundamental voltage: 127.32 V peak (90.03 V rms) at 0°. Load current up to 5th order: i₀(t) = 21.56 sin(314.16t – 32.14°) + 3.98 sin(942.48t – 62.05°) + 1.54 sin(1570.8t – 72.34°) A. ## Answer The output is a square wave of $\pm100$ V with period 0.02 s, so $f = 50$ Hz and $\omega = 314.16$ rad/s (the top level is taken as +100 V; the figure's sign is a misprint). Its Fourier series is $v_0 = \sum_{n\ \text{odd}}\frac{4V}{n\pi}\sin n\omega t$ with $V = 100$ V. **Load:** $R = 5\ \Omega$, $L = 10$ mH in series. #### Fundamental component of output voltage $$ \begin{aligned} V_{1m} &= \frac{4\times100}{\pi} = 127.32\ \text{V (peak)}, \qquad V_{1,rms} = 90.03\ \text{V} \\ v_1 &= 127.32\sin(314.16t)\ \text{V} \quad (\text{phase } 0^\circ) \end{aligned} $$ #### Harmonic currents $$ Z_n = \sqrt{R^2 + (n\omega L)^2}, \quad \theta_n = \tan^{-1}\frac{n\omega L}{R}, \quad I_{nm} = \frac{4V}{n\pi Z_n} $$ | $n$ | $V_{nm}$ (V) | $n\omega L$ (Ω) | $Z_n$ (Ω) | $\theta_n$ | $I_{nm}$ (A) | |---|---|---|---|---|---| | 1 | 127.32 | 3.142 | 5.905 | 32.14° | 21.56 | | 3 | 42.44 | 9.425 | 10.669 | 62.05° | 3.98 | | 5 | 25.46 | 15.708 | 16.485 | 72.34° | 1.54 | #### Time-domain load current (up to 5th order) $$ i_0(t) = 21.56\sin(314.16t - 32.14^\circ) + 3.98\sin(942.48t - 62.05^\circ) + 1.54\sin(1570.8t - 72.34^\circ)\ \text{A} $$ RMS of these components: $I_1 = 15.25$ A, $I_3 = 2.81$ A, $I_5 = 1.09$ A, so $I_0 \approx \sqrt{15.25^2 + 2.81^2 + 1.09^2} = 15.54$ A. **Answer:** fundamental voltage $= 127.32$ V peak (90.03 V rms) at 0°; $i_0$ as above, with the fundamental current $21.56$ A peak lagging by $32.14^\circ$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A square wave of period 0.02 s has what fundamental angular frequency? A. 50 rad/s B. 628.3 rad/s C. 157.1 rad/s D. 314.16 rad/s Answer: D. f = 1/0.02 = 50 Hz, w = 2 pi f = 314.16 rad/s. (Key point: Frequency from period) 2. For a +/-100 V square wave, what is the fundamental voltage? A. 127.32 V peak at 0 degrees B. 100 V peak at 0 degrees C. 127.32 V peak at 32.14 degrees D. 90.03 V peak at 0 degrees Answer: A. V1m = 4 x 100/pi = 127.32 V peak (90.03 V is rms), in phase with the square wave. (Key point: Fundamental voltage) 3. How is the nth harmonic current amplitude found? A. I_nm = (4V/(n pi)) / sqrt(R^2 + (n w L)^2) B. I_nm = (4V/pi) / sqrt(R^2 + (n w L)^2) C. I_nm = (4V/(n pi)) / (R + n w L) D. I_nm = (4V/(n pi)) / sqrt(R^2 + (w L)^2) Answer: A. Each harmonic sees Zn = sqrt(R^2 + (n w L)^2) and has voltage 4V/(n pi). (Key point: Harmonic impedance and current) 4. Which is the correct 3rd-harmonic term of the load current? A. 3.98 sin(942.48t + 62.05 degrees) B. 3.98 sin(314.16t - 62.05 degrees) C. 21.56 sin(942.48t - 62.05 degrees) D. 3.98 sin(942.48t - 62.05 degrees) Answer: D. I3m = 42.44/10.669 = 3.98 A, frequency 3w = 942.48 rad/s, lagging by tan^-1(9.425/5) = 62.05 degrees. (Key point: Time-domain current) ## Close variants set in other exams - [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× ## Related in Power Electronics - [Full-bridge square-wave inverter, 400 V, RL load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-bridge-inverter-400v-rl) — Power Electronics, chapter 5 - [Square-wave inverter ±48 V: fundamental and third harmonic](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ainverter-48v-rl-fundamental-third) — 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 - [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 - [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 ## Related in other subjects - [Voltage follower square-wave output with slew rate](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Avoltage-follower-square-wave) — Advanced Electronics, chapter 2 - [Voltage build-up; series generator at no load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Avoltage-buildup-series-generator-no-load) — Electrical Machines, chapter 3