91Ó°ÊÓ

We have to find the emf and internal resistance of the battery .

Real batteries always have an internal resistance $r$due to their components while ideal batteries have zero internal resistance. When an ideal wire with zero resistance is connected to the two terminals of the battery, it makes a short circuit where the current is maximum. For a battery that is connected to a circuit, the current that flows through the circuit is given by equation in the form

localid="1648906626706" $I=\frac{\mathrm{Î\mu }}{R+r}\left(1\right)$

Solve equation $\left(1\right)$for to be in the form

localid="1648906022667" $\mathrm{Î\mu }=IR+Ir\left(2\right)$

The current is maximum $I=6A$when the resistance is zero $R=0$. So, these two values into equation $\left(2\right)$to get $\mathrm{Î\mu }$by

localid="1648906236806" $\mathrm{Î\mu }=I\left(0\right)+\left(6\right)r=6r(*)$

Again, for any two points, when $R=20\mathrm{Î\copyright }$, the current is $I=2A$. So, we use equation $\left(2\right)$to find $\mathrm{Î\mu }$by

localid="1648906396994" $\mathrm{Î\mu }=\left(2A\right)\left(20\mathrm{Î\copyright }\right)+(2A)r=40+2r(**)$

In equation $(*)$we get $\mathrm{Î\mu }=6r$, so we use this expression into equation $(**)$to get $r$by

$$\begin{gathered} 6r=40+2r \\ r=10\mathrm{Î\copyright } \end{gathered}$$

Putting the values of $r$into equation $(*)$to get $\mathrm{Î\mu }$by

$\mathrm{Î\mu }=6r=6\left(10\right)=60V$