Problem — internal resistance of a battery

Problem

A battery has EMF $\varepsilon = 1{,}5\ \text{V}$. At no load its terminals measure $V_1 = 1{,}45\ \text{V}$; when powering a load $R = 6\ \Omega$ the terminal voltage drops to $V_2 = 1{,}20\ \text{V}$. Calculate the internal resistance $r$ of the battery and comment on its state of use.

Real battery ($\varepsilon$ in series with the internal resistance $r$) powering the load $R$.

Solution

Real battery model — a real battery is modelled as an ideal e.m.f. $\varepsilon$ in series with an internal resistance $r$. When it delivers current, part of the e.m.f. is lost across $r$ and less remains at the terminals: $V_\text{terminals} = \varepsilon - r\,i$

Current in the load — with $R = 6\ \Omega$ and terminal voltage $V_2 = 1{,}20\ \text{V}$: $i = \frac{V_2}{R} = \frac{1{,}20\ \text{V}}{6\ \Omega} = 0{,}20\ \text{A}$

Internal resistance — the drop $\varepsilon - V_2$ occurs entirely across $r$: $r = \frac{\varepsilon - V_2}{i} = \frac{1{,}5 - 1{,}20}{0{,}20} = \frac{0{,}30}{0{,}20} = 1{,}5\ \Omega$

Comment on the state of use — an internal resistance of $1{,}5\ \Omega$ is high for an AA battery (a new battery has $r$ of a fraction of an ohm). This explains why even at no load the terminals measure $V_1 = 1{,}45\ \text{V} < \varepsilon$ (the real voltmeter draws a small current) and why under load the voltage collapses to $1{,}20\ \text{V}$. The battery is therefore worn/nearly exhausted: it holds up at no load but “gives way” as soon as current is demanded from it.

$\ev{r = 1{,}5\ \Omega \ (\text{worn battery})}$

Collegamenti

Argomenti: Circuiti elettrici Concetti: Forza elettromotrice