Question: A rule of thumb used to determine the internal resistance of a source is that it is the open circuit voltage divide by the short circuit current. Is this correct? Why or why not?

Consider the expression for the voltage at the battery terminals.

$\mathbf{V}\mathbf{=}\mathbf{\epsilon }\mathbf{-}\mathbf{iR}$

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$V_{ab}=\epsilon -Ir$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$V_{ab}=\epsilon$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

$$\begin{gathered} V_{sc}=\epsilon -I_{sc}r \\ \epsilon -I_{sc}r=0 \\ r=\frac{\epsilon }{I_{sc}} \end{gathered}$$

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.