# Internal resistance

> CIE A-Level Physics · D.C. circuits
> Source: https://www.owlsprep.com/study/cie-9702-u10-internal-resistance/

This sub-topic explains the origin of internal resistance in voltage sources, its effect on terminal potential difference, and problem-solving methods for CIE A-Level Physics DC circuit questions.

**Prerequisites:** [Ohm's Law](https://www.owlsprep.com/study/cie-9702-u10-ohms-law/); [Electromotive force (e.m.f.)](https://www.owlsprep.com/study/cie-9702-u10-emf/); [Kirchhoff's Laws](https://www.owlsprep.com/study/cie-9702-u10-kirchhoffs-laws/)

## Learning objectives

- Explain the origin of internal resistance in electrical sources
- Derive and use the relationship $\textbackslashvarepsilon = V + Ir$
- Calculate terminal potential difference and lost volts
- Determine e.m.f. and internal resistance from experimental V-I data
- Solve common exam problems involving internal resistance

## What is Internal Resistance?

All sources of e.m.f. (like batteries, generators, and power packs) have some internal resistance. This comes from the resistance of the materials that make up the source itself: for a chemical battery, this is the resistance of the electrolyte between electrodes. When current flows through the source, energy is dissipated as heat in the internal resistance, just like in an external resistor.

**Internal resistance** — The resistance of the materials inside a source of e.m.f. that causes energy loss when current flows.

*Notation:* $r$

*Example:* A new AA battery has an internal resistance of ~0.1 Ω to 1 Ω, which increases as the battery ages.

**Worked example:** A 1.5 V AA battery has internal resistance $r = 0.4 \ \Omega$. It is connected to an external resistor of $2.6 \ \Omega$. Calculate the current flowing in the circuit and the voltage drop across the internal resistance.

1. Total resistance in the series circuit is the sum of external resistance $R$ and internal resistance $r$:
2. $$R_{\text{total}} = R + r = 2.6 + 0.4 = 3.0 \ \Omega$$
3. Use Ohm's law, with e.m.f. $\textbackslashvarepsilon$ equal to the total voltage:
4. $$I = \frac{\textbackslashvarepsilon}{R_{\text{total}}} = \frac{1.5}{3.0} = 0.5 \ \text{A}$$
5. Voltage drop (lost volts) across $r$ is calculated using Ohm's law:
6. $$V_r = Ir = 0.5 \times 0.4 = 0.2 \ \text{V}$$

*Calculator:* allowed

## The E.m.f. Equation and Lost Volts

We can derive the key equation for internal resistance using Kirchhoff's second law: the sum of e.m.f. around a series circuit equals the sum of potential drops. When current $I$ flows through the source, the potential drop across the internal resistance is $Ir$, called lost volts.

**Lost volts** — The potential difference across the internal resistance of a source when current flows, equal to $\textbackslashvarepsilon - V$.

*Example:* A 12 V car battery with 0.1 Ω internal resistance supplying 100 A has lost volts = 10 V, leaving only 2 V terminal pd when starting the engine.

$$\textbackslashvarepsilon = V + Ir$$

**Worked example:** A battery of e.m.f. 12 V has an unknown internal resistance. When it supplies a current of 2 A to a circuit, the terminal potential difference is measured as 11.4 V. Calculate the internal resistance of the battery.

1. First calculate lost volts by rearranging the core e.m.f. equation:
2. $$\text{Lost volts} = \textbackslashvarepsilon - V = 12 - 11.4 = 0.6 \ \text{V}$$
3. Lost volts equal $Ir$, so rearrange to solve for $r$:
4. $$r = \frac{\text{Lost volts}}{I} = \frac{0.6}{2} = 0.3 \ \Omega$$

> **Exam tip:** E.m.f. is only equal to terminal pd when the circuit is open (no current, $I=0$). Always account for lost volts when current flows.

*Calculator:* allowed

## Experimental Measurement of E.m.f. and Internal Resistance

A common exam practical experiment involves varying the external load resistance connected to a source, and recording pairs of terminal potential difference $V$ and current $I$. Rearranging the core equation gives a straight-line relationship:

$$V = -rI + \textbackslashvarepsilon$$

This matches the form $y = mx + c$, so plotting $V$ on the y-axis against $I$ on the x-axis gives: a y-intercept equal to e.m.f. $\textbackslashvarepsilon$, and a gradient equal to $-r$. The magnitude of the gradient is $r$.

**Worked example:** Two pairs of data from an experiment: $I=0.2 \ \text{A}, V=2.18 \ \text{V}$ and $I=0.5 \ \text{A}, V=1.55 \ \text{V}$. Calculate e.m.f. and internal resistance of the cell.

1. Substitute both pairs into $\textbackslashvarepsilon = V + Ir$ to get two simultaneous equations:
2. $$\textbackslashvarepsilon = 2.18 + 0.2r \\ \textbackslashvarepsilon = 1.55 + 0.5r$$
3. Equate the two expressions for $\textbackslashvarepsilon$ and solve for $r$:
4. $$2.18 + 0.2r = 1.55 + 0.5r \\ 0.63 = 0.3r \\ r = 2.1 \ \Omega$$
5. Substitute $r$ back to find $\textbackslashvarepsilon$:
6. $$\textbackslashvarepsilon = 2.18 + (0.2 \times 2.1) = 2.6 \ \text{V}$$

> **Exam tip:** Always write the equation in $y = mx + c$ form before interpreting the gradient to avoid sign errors.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Assuming terminal pd is always equal to e.m.f.
  - Why it fails: This is only true when no current flows (open circuit). Voltage is lost across internal resistance when current flows.
  - Correct: Always use $\textbackslashvarepsilon = V + Ir$ to account for lost volts for non-zero current.
- **Wrong:** Taking the gradient of a $V$-$I$ graph as $+r$ and using the negative value for $r$.
  - Why it fails: The equation $V = -rI + \textbackslashvarepsilon$ gives a negative gradient by definition.
  - Correct: Internal resistance is always positive, so take the magnitude of the gradient.
- **Wrong:** Writing total resistance as $R - r$ instead of $R + r$ in $\textbackslashvarepsilon = I(R + r)$.
  - Why it fails: Confusion about whether internal resistance adds to total circuit resistance.
  - Correct: Internal resistance is in series with the external circuit, so total resistance is the sum of $R$ and $r$.
- **Wrong:** Claiming internal resistance is caused by connecting wires outside the source.
  - Why it fails: Internal resistance is a property of the voltage source itself, not external components.
  - Correct: Internal resistance originates from materials inside the source (e.g. electrolyte in a battery).
- **Wrong:** Calculating power available to the external circuit as $\textbackslashvarepsilon I$.
  - Why it fails: This includes power lost as heat in the internal resistance.
  - Correct: Available power is $VI = \textbackslashvarepsilon I - I^2 r$, where $I^2 r$ is power lost in $r$.

## Cheatsheet

| Concept | Formula/Property | Key Exam Notes |
| --- | --- | --- |
| Internal resistance | $r$ | Resistance inside the voltage source |
| Electromotive force | $\textbackslashvarepsilon$ | Y-intercept of $V$-$I$ graph, open-circuit terminal pd |
| Core equation | $\textbackslashvarepsilon = V + Ir = I(R + r)$ | $V$ = terminal pd, $Ir$ = lost volts |
| Lost volts | $\textbackslashvarepsilon - V = Ir$ | Voltage drop across internal resistance |
| V-I graph relationship | $V = -rI + \textbackslashvarepsilon$ | $r = \|\text{gradient}\|$, y-intercept = $\textbackslashvarepsilon$ |
| Power lost in $r$ | $P = I^2 r$ | Dissipated as heat inside the source |

## What's next

Internal resistance is a foundational concept for all practical DC circuit analysis, and is regularly tested in both multiple choice and structured questions in CIE A-Level 9702 papers. It forms the basis for understanding power transfer between sources and loads, including the maximum power theorem, and is a core topic for practical assessment questions requiring analysis of experimental data. Mastery of this topic also helps you spot errors in circuit calculations caused by unaccounted source resistance.

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