Internal Resistance

Overview

Real sources of emf such as cells, batteries, and power supplies are not ideal. Besides supplying energy to the external circuit, some energy is dissipated inside the source itself due to its internal resistance.

This explains why the terminal potential difference of a battery can fall when current is drawn.

This page deepens ideas introduced in Current Electricity Fundamentals.

Related topics:

• DC Circuits
• Electrical Power and Ratings
• Current Electricity Common Exam Traps

Definition

emf $\mathcal{E}$

The emf of a source is:

energy supplied per unit charge by the source.

Unit:

• volt (V)

Terminal Potential Difference $V$

The terminal p.d. is the voltage across the external terminals of the source.

It is the voltage available to the external circuit.

Internal Resistance $r$

Internal resistance is the resistance within the source itself.

Unit:

$$\Omega$$

Why It Matters

Internal resistance explains:

• why terminal p.d. decreases when current increases
• why batteries heat up under load
• why short circuits are dangerous
• why a voltmeter across an unused cell reads approximately emf
• why source efficiency depends on load resistance

Key Representations

Core Model

A practical cell can be modeled as:

• ideal emf source $\mathcal{E}$
• internal resistance $r$ in series with external load

Lost Volts

When current $I$ flows through internal resistance $r$, the p.d. lost inside the source is:

$$Ir$$

This is called the lost volts or internal p.d. drop.

Core Equation

Applying energy conservation around the circuit:

$$\mathcal{E}=V+Ir$$

Hence:

$$V=\mathcal{E}-Ir$$

Meaning:

• $\mathcal{E}$ is energy supplied per unit charge
• $Ir$ is energy dissipated per unit charge inside the source
• $V$ is energy delivered per unit charge to the external circuit

Open Circuit

When no current flows:

$$I=0$$

so:

$$V=\mathcal{E}$$

An ideal voltmeter across an unused battery therefore reads approximately its emf.

Short Circuit

If external resistance is approximately zero, current is limited mainly by internal resistance:

$$I_{\text{sc}}=\frac{\mathcal{E}}{r}$$

This can cause overheating, battery damage, and fire risk.

Terminal p.d. Against Current Graph

From:

$$V=\mathcal{E}-Ir$$

a graph of $V$ against $I$ is a straight line:

• vertical intercept = $\mathcal{E}$
• gradient = $-r$
• horizontal intercept = $\mathcal{E}/r$

Power

Power delivered to the external load:

$$P_{\text{load}}=IV$$

Using $V=IR$ for load resistance $R$:

$$P_{\text{load}}=I^{2}R$$

Power lost inside the source:

$$P_{\text{lost}}=I^{2}r$$

Power supplied by the source:

$$P_{\text{supplied}}=I\mathcal{E}$$

Efficiency of a Source

Efficiency is useful power output divided by total power supplied:

$$\eta =\frac{IV}{I\mathcal{E}}=\frac{V}{\mathcal{E}}$$

Maximum Power Transfer

For a source with internal resistance $r$ connected to a load resistance $R$, maximum power is delivered to the load when:

$$R=r$$

At this condition, efficiency is only $50\%$, because equal power is dissipated in the load and in the internal resistance.

This result is useful for optimisation questions, but it should be treated as enrichment unless explicitly included in the assessed syllabus.

Worked Example

A cell has emf $1.5\text{V}$ and internal resistance $0.50\Omega$. It supplies a current of $0.80\text{A}$.

Terminal p.d.:

$$V=\mathcal{E}-Ir$$ $$V=1.5-(0.80)(0.50)=1.1\text{V}$$

Power lost internally:

$$P_{\text{lost}}=I^{2}r=(0.80{)}^2(0.50)=0.32\text{W}$$

Common Exam Traps

Treating emf as a force

emf is energy per unit charge, measured in volts.

Forgetting lost volts

For a real source supplying current:

$$V<\mathcal{E}$$

Wrong Graph Gradient

For a $V$ against $I$ graph:

$$\text{gradient}=-r$$

not $r$.

Assuming Open Circuit and Loaded Voltage Are Equal

They are equal only when $I=0$ or $r$ is negligible.

Thinking Maximum Power Means Maximum Efficiency

At maximum power transfer, efficiency is $50\%$, not maximum.

For a compact revision warning sheet, see:

Current Electricity Common Exam Traps

Summary Formula Table

Quantity	Formula
emf	$\mathcal{E}=\frac{W}{Q}$
Lost volts	$Ir$
Terminal p.d.	$V=\mathcal{E}-Ir$
Short-circuit current	$I_{\text{sc}}=\frac{\mathcal{E}}{r}$
External power	$P_{\text{load}}=IV=I^{2}R$
Internal power loss	$P_{\text{lost}}=I^{2}r$
Source efficiency	$\eta =\frac{V}{\mathcal{E}}$
Maximum power transfer	$R=r$

Links

• Related: Current Electricity Fundamentals
• Related: Electrical Power and Ratings
• Related: DC Circuits
• Related: Work, Energy and Power
• Related: Current Electricity Common Exam Traps
• Misconception: emf vs current