Internal Resistance

Branch note: This page deepens one part of Current Electricity Fundamentals.

Overview

This note deepens Current Electricity Fundamentals by isolating the behaviour of real sources.

The main idea is simple: when a source supplies current, not all the energy it supplies reaches the external circuit because some is dissipated inside the source.

Related topics:

• Electrical Power and Ratings
• Current Electricity Common Exam Traps

Core Ideas

• A real source can be modelled as an ideal emf source in series with internal resistance.
• The terminal p.d. is the energy delivered per unit charge to the external circuit.
• The lost volts $Ir$ are the p.d. across the internal resistance.
• When a source supplies current, terminal p.d. decreases according to $V=\mathcal{E}-Ir$.
• Internal resistance changes both useful output power and source efficiency.

Exam Relevance

Internal-resistance questions often test whether students can distinguish emf, terminal p.d., and lost volts, and whether they can read $\mathcal{E}$ and $r$ correctly from a $V$-against-$I$ graph.

Definition

emf $\mathcal{E}$

The emf of a source is the energy supplied per unit charge by the source.

Unit:

$$\text{V}$$

Terminal Potential Difference $V$

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

It is the voltage available to the external circuit.

Internal Resistance $r$

Internal resistance is the resistance inside the source itself.

Unit:

$$\Omega$$

Why It Matters

Internal resistance explains:

• why terminal p.d. falls when current increases
• why batteries and power supplies heat up under load
• why short circuits are dangerous
• why an unused cell reads approximately its emf
• why source efficiency falls as more energy is lost inside the source

Key Representations

Source Model

A practical source is modelled as:

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

Figure: A real source is modelled as an ideal emf $\mathcal{E}$ in series with internal resistance $r$. When it supplies current, each coulomb receives $\mathcal{E}$ joules, loses $Ir$ joules internally, and delivers $V$ joules externally.

Lost Volts

When the source supplies current $I$, the lost volts across the internal resistance are

$$Ir$$

Figure: Lost volts are the p.d. across the internal resistance only, so the terminal p.d. is lower than the emf when the source supplies current.

Core Equation

For a source supplying current, energy conservation gives

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

so

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

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

$$V=\mathcal{E}$$

For an ideal voltmeter, which draws negligible current, the reading equals the emf in this model. A real meter makes the equality approximate.

Short Circuit

In the idealised short-circuit model, the external resistance is approximately zero and current is limited mainly by internal resistance:

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

This can overheat the source and damage it.

V-against-I Graph

From

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

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

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

Figure: The straight $V$–$I$ graph gives $\mathcal{E}$ from its vertical intercept and $r$ from the magnitude of its gradient. The power curve shows that useful load power is zero at open circuit and in the ideal short-circuit limit, with a maximum between them.

Power

Power supplied by the source:

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

Power delivered to the external circuit:

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

Power lost inside the source:

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

For an external load resistance $R$,

$$I=\frac{\mathcal{E}}{R+r},V=IR=\frac{\mathcal{E}R}{R+r}.$$

The useful load power is

$$P_{\text{load}}=I^{2}R=\frac{{\mathcal{E}}^{2}R}{(R+r{)}^2}.$$

It is maximum when

$$R=r.$$

Equivalently, the maximum occurs at $I=\mathcal{E}/(2r)$ and $V=\mathcal{E}/2$. This is the maximum-power transfer condition.

Efficiency

Source efficiency is useful power output divided by total power supplied:

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

Using $V=IR$ and $\mathcal{E}=I(R+r)$,

$$\eta =\frac{R}{R+r}.$$

At maximum load power, $R=r$ and $\eta =50\%$. Increasing $R/r$ above 1 raises efficiency but eventually reduces the output power. Maximum power and maximum efficiency are therefore different design aims.

Enrichment: charging a source

The relation $V=\mathcal{E}-Ir$ assumes current leaves the positive terminal. When a rechargeable source is charged, current enters its positive terminal and the terminal p.d. is $V=\mathcal{E}+Ir$.

Worked Example

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

The terminal p.d. is

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

The power lost inside the source is

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

Common Exam Traps

emf is not a force

emf is energy per unit charge, measured in volts.

Terminal p.d. is not always emf

For a real source supplying current,

$$V<\mathcal{E}$$

Links

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