Problem — internal resistance of an alkaline cell

Problem

An alkaline cell has nominal EMF $\varepsilon = 1{,}5\ \text{V}$. Connected to a light bulb of resistance $R = 4{,}5\ \Omega$, its terminals read $V = 1{,}35\ \text{V}$. Determine the internal resistance $r$ of the cell and the fraction of power it dissipates internally.

Solution

Current in the circuit: the terminal voltage is the one applied to the bulb, so $i = \frac{V}{R} = \frac{1{,}35\ \text{V}}{4{,}5\ \Omega} = 0{,}30\ \text{A}$

Internal resistance: the voltage drop missing relative to the EMF occurs across the internal resistance, $\varepsilon = V + r\,i$, hence $r = \frac{\varepsilon - V}{i} = \frac{1{,}5 - 1{,}35}{0{,}30} = \frac{0{,}15}{0{,}30} = 0{,}5\ \Omega$

Fraction of power dissipated internally: the internal power is $P_r = i^2 r$ and the total power delivered by the EMF is $P_\text{tot} = i^2 (R + r)$, so the ratio does not depend on the current: $\frac{P_r}{P_\text{tot}} = \frac{r}{R + r} = \frac{0{,}5}{4{,}5 + 0{,}5} = \frac{0{,}5}{5{,}0} = 0{,}10$

Only $10\%$ of the power is wasted inside the cell.

$\ev{r = 0{,}5\ \Omega \qquad P_r/P_\text{tot} = 10\%}$

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Topics: Electric circuits Concepts: Electromotive force