Internal resistance

In a nutshell

The internal resistance is the resistance of the power sources themselves which results in the circuit being provided a terminal voltage instead of their actual e.m.f..

Equations

Description	equation
Electromotive force	$\varepsilon=I(R+r)$
Electromotive force	$\varepsilon=V+Ir$

Variable definitions

quantity name	symbol	derived units	alternate units	si base units
$resistance$	$R$	$\Omega$	$V\ A^{-1}$	$kg\ m^2\ s^{-3}\ A^{-2}$
$internal\ resistance$	$r$	$\Omega$	$V\ A^{-1}$	$kg\ m^2 \ s^{-3}\ A^{-2}$
$electromotive\ force$	$\varepsilon$	$V$	$J\ C^{-1}$	$kg\ m^2 \ s^{-3}\ A^{-1}$
$potential\ difference$	$V$	$V$	$J\ C^{-1}$	$kg\ m^2 \ s^{-3}\ A^{-1}$
$current$	$I$	$A$		$A$

Lost volts

When current flows through a power source the charges still need to do work to get through it. This means that some energy gets lost or transferred into heat while inside the cells and batteries themselves.

If you measure the potential difference at the terminals of a power source, known as terminal potential difference, you would find that it is in fact less than the e.m.f of the power source itself; this lost p.d. is called lost volts $v$ .

Due to Kirchhoff's second law one can write:

$\varepsilon=V+v$

Internal resistance

Lost volts are used on the internal resistance $r$ of the power source, which is the resistance of the power sources themselves. You can imagine the power sources as having a small resistor inside them and it can be represented as follows:

Physics; Electrical circuits; KS5 Year 12; Internal resistance

The current is the same everywhere since these "components" are in series. Since $V=IR$ it means that $v=Ir$ and it can be substituted into the equation above:

$\varepsilon=V+Ir$

This equation can be modified further by substituting the terminal p.d. with the resistance of the rest of the circuit and current:

$\varepsilon=IR+Ir$

Simplifying this you get:

$\varepsilon=I(R+r)$

This last equation is essentially another version of $V=IR$ but taking internal resistance into account.

Exam tip: unless it has been specified that the power source has an internal resistance you should always assume it doesn't.

Example

A circuit is powered by a cell with e.m.f. of $5\ V$ . The total resistance of the circuit is $1.8\ \Omega$ and the current is $2.5\ A$ . What is the internal resistance of the cell?

Firstly write down what you know:

$\begin{aligned}\varepsilon&=5\ V\\R&=1.8\ \Omega \\I&=2.5\ A\end{aligned}$ 

Next, write down the correct version of the equation needed. In this case it is the one without the terminal p.d.:

$\varepsilon=I(R+r)$ 

Rearrange this equation to make $r$ the subject:

$\dfrac{\varepsilon}{I}=R+r$ 

$r=\dfrac{\varepsilon}{I}-R$ 

Substitute all the values into the equation and calculate the internal resistance:

$r=\dfrac{5}{2.5}-1.8=0.2\ \Omega$ 

The internal resistance of the cell is $\underline{0.2\ \Omega}$ .