---
title: "Ideal and practical sources, conversion"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:ideal-practical-sources-conversion"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/ideal-practical-sources-conversion
site: BCT Engineering
---

# Ideal and practical sources, conversion

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2074 Asoj, 5 marks
- Video: [Ideal vs. Practical Sources: Voltage and Current Sources](https://www.youtube.com/watch?v=-YWwL6mgeLg) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/ideal-practical-sources-conversion
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

What do you mean by ideal and practical voltage and current source? Explain the method for converting practical voltage source into current source and vice versa.

## Short answer

Ideal voltage source: constant voltage, zero internal resistance; practical: series resistance, V = E−IRs. Ideal current source: constant current, infinite internal resistance; practical: parallel resistance, I = Is−V/Rp. Convert voltage to current: Is = E/Rs, Rp = Rs. Convert current to voltage: E = Is×Rp, Rs = Rp.

## Answer

#### Ideal and practical sources
- **Ideal voltage source:** gives constant voltage *E* at any current; internal resistance zero.
- **Practical voltage source:** an ideal source *E* in series with an internal resistance *Rs*; V = E − I Rs, so it falls as current rises.
- **Ideal current source:** gives constant current *Is* at any voltage; internal resistance infinite.
- **Practical current source:** an ideal source *Is* in parallel with a resistance *Rp*; I = Is − V/Rp.

```text
 Practical V source        Practical I source
  o--[Rs]--o a              o--+------+--o a
    (E)                       (Is)   [Rp]
  o--------o b              o--+------+--o b
```

#### Converting voltage source to current source
Short-circuit current of the voltage source is E/Rs. So:
Is = E/Rs, and Rp = Rs (connected in parallel).

#### Converting current source to voltage source
Open-circuit voltage of the current source is Is × Rp. So:
E = Is × Rp, and Rs = Rp (connected in series).

#### Example
A 20 V source with Rs = 5 Ω is equivalent to a 4 A source with Rp = 5 Ω. A 2 A source with 10 Ω is equivalent to a 20 V source in series with 10 Ω.

The equivalence is valid only for the terminals seen from the load, not for the internal power loss.

## Check yourself

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

1. An ideal current source has internal resistance:
   A. Zero
   B. Infinite
   C. Equal to the load
   D. Finite and small
   Answer: B. Ideal current source gives constant current at any voltage, so Rp is infinite. (Key point: Ideal vs practical)

2. Converting a voltage source (E, Rs) to a current source uses:
   A. Is = E x Rs, Rp = Rs in series
   B. Is = Rs/E, Rp = E
   C. Is = E/Rs, Rp = Rs in parallel
   D. Is = E, Rp = 0
   Answer: C. Is is the short-circuit current E/Rs, and Rs goes in parallel. (Key point: Voltage to current)

3. A 2 A source with Rp = 10 ohm is equivalent to:
   A. A 5 V source with 10 ohm in series
   B. A 0.2 V source with 10 ohm in series
   C. A 20 V source with 10 ohm in parallel
   D. A 20 V source with 10 ohm in series
   Answer: D. E = Is x Rp = 20 V with Rs = Rp = 10 ohm in series. (Key point: Current to voltage)

4. The equivalence is valid:
   A. For the terminals seen by the load, not for internal power loss
   B. For the internal power loss as well
   C. Only for ideal sources
   D. Only without a load
   Answer: A. Both forms give the same load voltage and current but differ internally. (Key point: Validity)

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