91Ó°ÊÓ

Ohm’s law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant

$V=IR$

According to Kirchhoff’s Lawthe sum of the voltages around the closed loop is equal to null.

$\underset{\mathbf{}}{\mathbf{∑}}V=0$

The internal resistance is the slope of the V-I graph, since resistance R is variable the internal resistance is indeed the slope

$$\begin{gathered} r=-\frac{∆V}{∆I} \\ r=-\frac{30V-24V}{3A-6A}=2\mathrm{Î\copyright } \end{gathered}$$

The internal resistance$r=2\mathrm{Î\copyright }$

We know terminal voltage of the battery is given by

$V=\mathrm{Î\mu }-Ir$

Using the values of $V=30V$and $I=3A$and $r=2\mathrm{Î\copyright }$rearrange

$$\begin{gathered} \mathrm{Î\mu }=V_{ab}+Ir \\ \mathrm{Î\mu }=30V+\left(3A\right)\left(2\mathrm{Î\copyright }\right) \\ \mathrm{Î\mu }=36V \end{gathered}$$

The emf of the battery$\mathrm{Î\mu }=36V$

When the voltage drops to 80% of emf , use krichhoff’s law

$V=\mathrm{Î\mu }-Ir$

Rearranging,

$I=\frac{\mathrm{Î\mu }-V}{r}$

Input $V=0.8\mathrm{Î\mu }$and$r=2\mathrm{Î\copyright }$

$$\begin{gathered} I=\frac{\mathrm{Î\mu }-0.8\mathrm{Î\mu }}{r} \\ I=\frac{0.2\mathrm{Î\mu }}{r} \\ I=\frac{0.2\left(36V\right)}{0.2\mathrm{Î\copyright }} \\ I=3.6A \end{gathered}$$

Use Ohm’s Law to find R

$$\begin{gathered} R=\frac{V}{I}=\frac{0.8\mathrm{Î\mu }}{I} \\ R=\frac{0.8\left(36V\right)}{3.6A}=8\mathrm{Î\copyright } \end{gathered}$$

Hence the external resistance $R=8\mathrm{Î\copyright }$.