Question

What is the internal resistance of an automobile battery that has an emf of $\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}$and a terminal voltage of $\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}$while a current of $\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{A}$is charging it?

Short answer

The internal resistance is $0.375\Omega$.

Step-by-step solution

Terminal voltage

The terminal voltage of the cell is

$\mathbf{V}\mathbf{=}\mathbf{emf}\mathbf{-}\mathbf{lR}$

Here, emf is the output voltage of the battery, $\mathbf{I}$is the current and $\mathbf{R}$is the resistance.

Calculation of the internal resistance

Given value of the EMF is E=12.0 V

Current is$\mathrm{l}=8.00\mathrm{A}$

Terminal voltage,$\mathrm{V}=15.0\mathrm{V}$.

When charging , $\mathrm{E}=\mathrm{V}-\mathrm{lr}$.

So internal, resistance

role="math" localid="1656141444437" $$\begin{gathered} \mathrm{l}=\frac{\mathrm{V}-\mathrm{E}}{\mathrm{r}} \\ =\frac{15.0\mathrm{V}-12.0\mathrm{V}}{8.00\mathrm{A}} \\ =0.375\mathrm{\Omega } \end{gathered}$$

Therefore, the internal resistance is $0.375\mathrm{\Omega }$.