Problem — characteristic curve of a real battery

Problem

Characteristic curve of a real battery. The figure shows the terminal voltage of a battery as a function of the current delivered.

The terminal voltage decreases linearly as the delivered current increases.

(a) Write the equation $V_\text{mors}(i) = \mathcal{E} - r\,i$. (b) Obtain $\mathcal{E}$ from the intercept with the vertical axis. (c) Obtain the internal resistance $r$ from the slope. (d) At what current would the battery short-circuit according to this model? Comment.

Solution

(a) Model. A real battery is an EMF $\mathcal{E}$ in series with an internal resistance $r$. The terminal voltage is the EMF minus the internal drop: $V_\text{mors}(i) = \mathcal{E} - r\,i$ It is a line decreasing in $i$: intercept $\mathcal{E}$, slope $-r$.

(b) EMF from the intercept. At zero current ($i = 0$) there is no internal drop: $V_\text{mors}(0) = \mathcal{E}$. From the graph the intercept with the vertical axis is at $1{,}5\ \text{V}$: $\mathcal{E} = 1{,}5\ \text{V}$

(c) Internal resistance from the slope. The line drops from $1{,}5\ \text{V}$ (at $i=0$) to $0{,}75\ \text{V}$ (at $i = 5\ \text{A}$): $r = -\frac{\Delta V}{\Delta i} = -\frac{0{,}75 - 1{,}5}{5 - 0} = \frac{0{,}75}{5} = 0{,}15\ \Omega$

(d) Short-circuit current. In a short circuit the terminal voltage is zero ($V_\text{mors} = 0$): the model gives $0 = \mathcal{E} - r\,i_\text{cc} \;\Rightarrow\; i_\text{cc} = \frac{\mathcal{E}}{r} = \frac{1{,}5}{0{,}15} = 10\ \text{A}$ Comment: this is the maximum current predicted by the linear model. A real battery cannot sustain it for long (the internal resistance increases with heating and discharge, and the characteristic ceases to be linear at high currents); however, it helps explain why short-circuiting a battery heats it up and damages it.

$\ev{\mathcal{E} = 1{,}5\ \text{V} \qquad r \approx 0{,}15\ \Omega \qquad i_\text{cc} = 10\ \text{A}}$

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