When we represent a simple circuit, we usually assume that the resistance of the battery is zero or negligible. In real life, of course, this is not the case. A reliable way of thinking about this is that we can represent the battery as a cell (the e.m.f. with no resistance) and a resistor encased in an outer shell:
The resistance of this imaginary resistor is referred to as internal resistance ($r$).
How e.m.f. and internal resistance are related
The terminal potential difference, represented by $V$, is the $1.5V$, $9V$ or other value advertised on the battery. Its relationship with internal resistance, $r$ and the e.m.f., $E$ is:
$$ \begin{equation}\begin{aligned} E&=V+Ir\\ E&=IR+Ir\\ \end{aligned}\end{equation} $$The terminal p.d. is therefore what voltage remains after the e.m.f. has overcome the battery’s own internal resistance: $$ \begin{equation}\begin{aligned} V&=E-Ir\\ \rightarrow IR&=E-Ir\\ \end{aligned}\end{equation} $$
1) What is the current flowing through a circuit where the e.m.f. is 9.5 volts, internal resistance is 1 ohm and the total external resistance is 4 ohms?
- 1.81 A
- 1.9 A
- 2.125 A
- 0.52 A
$I=\frac{E}{R+r}=\frac{9.5 V}{1\Omega+4\Omega}=1.9 A$
2) The e.m.f. of a cell where the internal resistance is $1.5\Omega$, with a terminal voltage of $9V$, and current of flowing $0.5A$ through it, is
- 10 V
- 8 V
- 15 V
- 9.75 V
$E=V+Ir=9V+0.5A\times 1.5\Omega=9.75V$
Power and internal resistance
The total power produced by the battery is given as: $$ \begin{equation}\begin{aligned} P_T=I^2R+I^2r\\ \end{aligned}\end{equation} $$
Some power is dissipated by the internal resistance and the rest by the load in the circuit. The latter can be expressed as a fraction of the total power: $$ \begin{equation}\begin{aligned} \frac{P_R}{P_T}&=\frac{I^2R}{I^2R+I^2r}\\ \therefore \frac{P_R}{P_T}&=\frac{R}{R+r}\\ \end{aligned}\end{equation} $$
This is the ratio of the external resistance to the total resistance.