Internal resistance

CIE A-Level Physics· 25 min read

1. What is Internal Resistance?★★☆☆☆⏱ 10 min

✓ Calculator OK

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.

📘 Definition

Internal resistance

$r$

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

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 $r = 0.4 Ω$ . It is connected to an external resistor of $2.6 Ω$ . 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. $e x t ba c k s l a s h v a r e p s i l o n$ 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
$V_r = Ir = 0.5 \times 0.4 = 0.2 \ \text{V}$

[Block type renderer not yet implemented · content will appear once shipped]

2. The E.m.f. Equation and Lost Volts★★☆☆☆⏱ 15 min

✓ Calculator OK

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 $I r$ , called lost volts.

📘 Definition

Lost volts

The potential difference across the internal resistance of a source when current flows, equal to $e x t ba c k s l a s h v a r e p s i l o n - V$ .

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 $I r$ , 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.

3. Experimental Measurement of E.m.f. and Internal Resistance★★★☆☆⏱ 20 min

✓ Calculator OK

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 $y = m x + c$ , so plotting $V$ on the y-axis against $I$ on the x-axis gives: a y-intercept equal to e.m.f. $e x t ba c k s l a s h v a r e p s i l o n$ , and a gradient equal to $- r$ . The magnitude of the gradient is $r$ .

📐 Worked Example

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

1
Substitute both pairs into $e x t ba c k s l a s h v a r e p s i l o n = V + I r$ to get two simultaneous equations:
2
$\textbackslashvarepsilon = 2.18 + 0.2r \\ \textbackslashvarepsilon = 1.55 + 0.5r$
3
Equate the two expressions for $e x t ba c k s l a s h v a r e p s i l o n$ 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 $e x t ba c k s l a s h v a r e p s i l o n$ :
6
$\textbackslashvarepsilon = 2.18 + (0.2 \times 2.1) = 2.6 \ \text{V}$

Exam tip:

Always write the equation in $y = m x + c$ form before interpreting the gradient to avoid sign errors.

4. Common Pitfalls

Wrong move:

Assuming terminal pd is always equal to e.m.f.

Why:

This is only true when no current flows (open circuit). Voltage is lost across internal resistance when current flows.

Correct move:

Always use $e x t ba c k s l a s h v a r e p s i l o n = V + I r$ to account for lost volts for non-zero current.

Wrong move:

Taking the gradient of a $V$ - $I$ graph as $+ r$ and using the negative value for $r$ .

Why:

The equation $V = - r I + \textbackslashvarepsilon$ gives a negative gradient by definition.

Correct move:

Internal resistance is always positive, so take the magnitude of the gradient.

Wrong move:

Writing total resistance as $R - r$ instead of $R + r$ in $e x t ba c k s l a s h v a r e p s i l o n = I (R + r)$ .

Why:

Confusion about whether internal resistance adds to total circuit resistance.

Correct move:

Internal resistance is in series with the external circuit, so total resistance is the sum of $R$ and $r$ .

Wrong move:

Claiming internal resistance is caused by connecting wires outside the source.

Why:

Internal resistance is a property of the voltage source itself, not external components.

Correct move:

Internal resistance originates from materials inside the source (e.g. electrolyte in a battery).

Wrong move:

Calculating power available to the external circuit as $e x t ba c k s l a s h v a r e p s i l o n I$ .

Why:

This includes power lost as heat in the internal resistance.

Correct move:

Available power is $V I = e x t ba c k s l a s h v a r e p s i l o n I - I^{2} r$ , where $I^{2} r$ is power lost in $r$ .

5. Quick Reference Cheatsheet

Concept	Formula/Property	Key Exam Notes
Internal resistance	$r$	Resistance inside the voltage source
Electromotive force	$e x t ba c k s l a s h v a r e p s i l o n$	Y-intercept of $V$ - $I$ graph, open-circuit terminal pd
Core equation	$e x t ba c k s l a s h v a r e p s i l o n = V + I r = I (R + r)$	$V$ = terminal pd, $I r$ = lost volts
Lost volts	$e x t ba c k s l a s h v a r e p s i l o n - V = I r$	Voltage drop across internal resistance
V-I graph relationship	$V = - r I + e x t ba c k s l a s h v a r e p s i l o n$	$r = ∣ gradient ∣$ , y-intercept = $e x t ba c k s l a s h v a r e p s i l o n$
Power lost in $r$	$P = I^{2} r$	Dissipated as heat inside the source

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Calculate internal resistance from data
2023 · 2
Graph method for e.m.f. and r
2021 · 1
Terminal pd for varying current

Going deeper

practical guideCIE 9702 Practical Test: Internal Resistance ExperimentCommon experiment for Paper 3

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.