91Ó°ÊÓ

We have to find the internal resistance of the battery.

The resistor $20\mathrm{Î\copyright }$is connected to the battery, a voltage drop occurs which is $8.5V$. This is the potential difference across the resistor role="math" localid="1648884553842" $â–³V_R=8.5V$. So, determine the current through the circuit by using Ohm's law by

$I=\frac{â–³V_R}{R}=\frac{8.5V}{20\mathrm{Î\copyright }}=0.425A$

A battery that is connected to a circuit, the current that flows through the circuit is given by equation in the form

role="math" localid="1648888679002" $I=\frac{\mathrm{Î\mu }}{R+r}\left(1\right)$

Where $\mathrm{Î\mu }$is the emf of the battery which is $9V$. Solve the equation $\left(1\right)$for $r$to be

$$\begin{gathered} I=\frac{\mathrm{Î\mu }}{R+r} \\ IR+Ir=\mathrm{Î\mu }\left(2\right) \\ r=\frac{\mathrm{Î\mu }-IR}{I} \end{gathered}$$

role="math" localid="1648888910241" $$\begin{gathered} r=\frac{\mathrm{Î\mu }-IR}{I} \\ =\frac{9V-(0.425A)\left(20\mathrm{Î\copyright }\right)}{0.425A} \\ =1.18\mathrm{Î\copyright } \end{gathered}$$