---
title: "Step-down chopper with 2 V drop: average and rms"
subject: "Power Electronics"
chapter: 4
key: "semester_7/power_electronics:step-down-chopper-2v-drop-rms"
source: https://www.bctengineer.com/q/semester_7/power_electronics/step-down-chopper-2v-drop-rms
site: BCT Engineering
---

# Step-down chopper with 2 V drop: average and rms

Power Electronics · Chapter 4: DC chopper

- Asked in: 2070 Chaitra, 8 marks
- Video: [SOLVED PROBLEMS OF CHOPPERS](https://www.youtube.com/watch?v=9U2TWnzduhA) — Prof. Barapate's Tutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/step-down-chopper-2v-drop-rms
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-4

## Question

The supply voltage of a step down chopper is 230V dc, load resistance is 10Ω. Take the voltage drop of 2V across chopper when it is on. For a duty cycle of 0.4, calculate average and rms value of output voltage. [Figure: Vdc source, series chopper switch and 10 Ω load]

## Short answer

Account for 2 V on-state drop; effective ON voltage is 228 V. Average Vo = k × 228 = 0.4 × 228 = 91.2 V. RMS Vo = √k × 228 = √0.4 × 228 ≈ 144.2 V.

## Answer

With an ON-state drop of $V_{ch}$, the load receives $(V_s - V_{ch})$ for $kT$ and zero for the rest of the period.

**Given:** $V_s = 230$ V, $R = 10\ \Omega$, $V_{ch} = 2$ V, $k = 0.4$.

```text
 v0   228|_____       _____
         |     |     |     |
       0 |_____|_____|_____|____ t
          kT    T
```

#### Average output voltage
$$
\begin{aligned}
V_a &= \frac{1}{T}\int_0^{kT}(V_s - V_{ch})\,dt = k\,(V_s - V_{ch}) \\
&= 0.4\times(230 - 2) = 0.4\times228 = 91.2\ \text{V}
\end{aligned}
$$

#### RMS output voltage
$$
\begin{aligned}
V_{rms} &= \left[\frac{1}{T}\int_0^{kT}(V_s - V_{ch})^2\,dt\right]^{1/2} = \sqrt{k}\,(V_s - V_{ch}) \\
&= \sqrt{0.4}\times228 = 0.63246\times228 = 144.20\ \text{V}
\end{aligned}
$$

Corresponding load currents: $I_a = 91.2/10 = 9.12$ A, $I_{rms} = 14.42$ A. Chopper efficiency would be $(V_s - V_{ch})/V_s = 228/230 = 99.13\%$.

**Answer:** $V_a = 91.2$ V, $V_{rms} = 144.20$ V.

## Check yourself

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

1. With a 2 V drop, the load voltage during ON is:
   A. 228 V
   B. 230 V
   C. 232 V
   D. 2 V
   Answer: A. Vs - Vch = 230 - 2. (Key point: Pulse height)

2. The average output voltage is:
   A. 92 V
   B. 91.2 V
   C. 228 V
   D. 144.2 V
   Answer: B. 0.4 x 228 = 91.2 V. (Key point: Average)

3. The rms output voltage is:
   A. 91.2 V
   B. 228 V
   C. 144.20 V
   D. 182.4 V
   Answer: C. sqrt(0.4) x 228 = 144.2 V. (Key point: RMS)

4. Chopper efficiency in this case is:
   A. Vch/Vs = 0.87%
   B. k = 40%
   C. 100%
   D. (Vs - Vch)/Vs = 99.13%
   Answer: D. 228/230 = 0.9913. (Key point: Efficiency)

## Close variants set in other exams

- [Step-down chopper with 1 V drop: average and rms](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-down-chopper-1v-drop-rms) — Power Electronics, chapter 4, asked 2×

## Related in Power Electronics

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- [Step-down chopper with DC motor load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-down-chopper-dc-motor) — Power Electronics, chapter 4

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