---
title: "Chopper numerical: 220 V, 10 Ω, 50% duty"
subject: "Advanced Electronics"
chapter: 7
key: "semester_4/advanced_electronics:chopper-220v-10ohm"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/chopper-220v-10ohm
site: BCT Engineering
---

# Chopper numerical: 220 V, 10 Ω, 50% duty

Advanced Electronics · Chapter 7: Introduction to Power Electronics

- Asked in: 2073 Shrawan, 2+2+2 marks; 2069 Asar, 2+2+3 marks (asked 2 times)
- 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_4/advanced_electronics/chopper-220v-10ohm
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-7

## Question

For the chopper shown in figure below with a resistive load R = 10Ω, the input voltage is Vs = 220V, when the chopper remains ON its voltage drop Vch = 2V and the chopping frequency is f = 1kHz with duty cycle of 50%. Determine: a) The average output voltage b) The RMS value of output voltage c) The chopper efficiency. [Figure: dc source Vs (+ at top) connected through the chopper switch SW to a resistive load R; the output voltage vo appears across R and the load current io flows through it.]

## Short answer

For Vs=220V, Vch=2V, duty cycle 0.5, ON voltage is 218V. Average output Va = 0.5×218 = 109V. RMS output = √0.5×218 = 154.15V. Efficiency = (218/220)×100% = 99.09%.

## Answer

A **chopper** is a DC-to-DC converter that switches a fixed DC supply ON and OFF to give a variable average DC output.

Given: Vs = 220 V, R = 10 Ω, Vch = 2 V, f = 1 kHz, duty cycle k = 0.5.

During ON time the load voltage is Vs − Vch = 218 V; during OFF time it is 0.

```text
vo
218 |----+    +----+    +---
    |    |    |    |    |
  0 +----+----+----+----+----> t
    |<kT>|
    |<--T=1ms->|
```

#### a) Average output voltage

```text
Va = (1/T)∫₀^(kT) (Vs − Vch) dt = k(Vs − Vch)
   = 0.5 × (220 − 2) = 0.5 × 218
Va = 109 V
```

#### b) RMS output voltage

```text
Vo,rms = [ (1/T)∫₀^(kT) (Vs − Vch)² dt ]^(1/2)
       = √k · (Vs − Vch)
       = √0.5 × 218
Vo,rms = 154.15 V
```

#### c) Chopper efficiency

Output power:
```text
Po = (1/T)∫₀^(kT) vo²/R dt = k(Vs − Vch)²/R
   = 0.5 × 218² / 10 = 2376.2 W
```

Input power (source current = (Vs − Vch)/R during ON):
```text
Pi = (1/T)∫₀^(kT) Vs·(Vs − Vch)/R dt
   = k·Vs(Vs − Vch)/R
   = 0.5 × 220 × 218 / 10 = 2398 W
```

Efficiency:
```text
η = Po/Pi = (Vs − Vch)/Vs = 218/220 = 0.9909
```

**Answer: Va = 109 V, Vo,rms = 154.15 V, η = 99.09 %**

## Check yourself

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

1. With Vs = 220 V and Vch = 2 V, what is the load voltage while the chopper is ON?
   A. 220 V
   B. 218 V
   C. 222 V
   D. 109 V
   Answer: B. The ON-state drop is subtracted: Vs − Vch = 218 V. (Key point: Output during ON)

2. What is the average output voltage at k = 0.5?
   A. 218 V
   B. 110 V
   C. 109 V
   D. 154.15 V
   Answer: C. Va = k(Vs − Vch) = 0.5 × 218 = 109 V. (Key point: Average voltage)

3. Which formula gives the RMS output voltage?
   A. k × (Vs − Vch)
   B. k^2 × (Vs − Vch)
   C. (Vs − Vch)/sqrt(k)
   D. sqrt(k) × (Vs − Vch)
   Answer: D. RMS of a rectangular pulse of height 218 V and duty k is sqrt(k) × 218 = 154.15 V. (Key point: RMS voltage)

4. Why is the chopper efficiency 99.09 %?
   A. Only the 2 V switch drop is lost, so η = (Vs − Vch)/Vs = 218/220
   B. Because the duty cycle is 50 %
   C. Because the load is resistive, so no power is lost
   D. Because RMS equals the average value
   Answer: A. The ratio Po/Pi simplifies to (Vs − Vch)/Vs. (Key point: Efficiency)

## Close variants set in other exams

- [Inverters; chopper numerical 220 V, 10 Ω](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ainverters-chopper-220v-10ohm) — Advanced Electronics, chapter 7, asked 4×

## Related in Advanced Electronics

- [Buck-boost regulator numerical (Vs = 10 V)](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Abuck-boost-10v-45-percent) — Advanced Electronics, chapter 8, asked 3×

## Related in other subjects

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