Explanation:
C Given,
emf of each battery $\left(\mathrm{E}_{1}=\mathrm{E}_{2}\right)=1.5 \mathrm{~V}$
External resistance $\left(\mathrm{R}_{1}=\mathrm{R}_{2}\right)=20 \Omega$
Voltmeter reading $(\mathrm{V})=1.2 \mathrm{~V}$
Let, internal resistance $=r$

Hence, equivalent external resistance $(R)=10 \Omega$
Equivalent emf of battery $(\mathrm{E})=1.5 \mathrm{~V}$
Equivalent internal resistance $(r)=r / 2$
We know that,
Voltage drop
$\mathrm{V}=\mathrm{E}-\mathrm{i} \frac{\mathrm{r}}{2}$
$1.2=1.5-\mathrm{i}\left(\frac{\mathrm{r}}{2}\right)$
$\mathrm{i} \frac{\mathrm{r}}{2}=0.3$
And $\quad i=\frac{E}{R+r}$
$i=\frac{1.5}{10+r / 2}$
$10 i+i r / 2=1.5$
Putting the value of ir/2 from equation (i) to equation (ii) we get -
$10 \mathrm{i}+0.3=1.5$
$10 \mathrm{i}=1.2$
$\mathrm{i}=0.12 \mathrm{~A}$
Then, from equation (i),
$0.12 \frac{\mathrm{r}}{2}=0.3$
$\mathrm{r}=\frac{0.6}{0.12}$
$\mathrm{r}=5 \Omega$