91Ó°ÊÓ

Consider the expression for the voltage at the battery terminals.

$\mathbf{V}\mathbf{=}\mathbf{Î\mu }\mathbf{-}\mathbf{iR}$

Here, the voltage is the V, the emf of the battery is , the current is and the resistance is Ris the internal resistance.

From the concept of internal-resistance the relation is$V_{ab}=\mathrm{Î\mu }-Ir$whereelectric current is which moving through the battery.

Forwhich state that circuit is open and no current flow through the circuit. Now the equation of internal resistance reduces to$V_{ab}=\mathrm{Î\mu }$which state that open circuit voltage equal to the emf.

Now when current is flow which state that circuit is short circuit and there is an external resistors

$$\begin{gathered} V_{sc}=I_{sc}R \\ {V}_{sc}=0 \end{gathered}$$

In that case use Ohm’s law, which state that

Now for short circuit

$$\begin{gathered} V_{sc}=\mathrm{Î\mu }-I_{sc}r \\ \mathrm{Î\mu }-I_{sc}r=0 \\ r=\frac{\mathrm{Î\mu }}{I_{sc}} \end{gathered}$$

Since,then.

$r=\frac{V_{oc}}{I_{sc}}$

Hence, yes it is correct the internal resistance of a battery the open circuit voltage divide by the short circuit current.