91Ó°ÊÓ

We have given,

For open circuit $\mathrm{I}=1.636\mathrm{A}$

For closed circuit $\mathrm{I}=1.565\mathrm{A}$

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

For open circuit,

Apply Kirchhoff voltage law in the left loop,

$$\begin{gathered} -\mathrm{Î\mu }+\mathrm{Ir}+5\mathrm{I}=0 \\ \mathrm{Î\mu }=\mathrm{I}(\mathrm{r}+5) \\ \mathrm{Î\mu }=(1.636\mathrm{A})(\mathrm{r}+5)....................................\left(1\right) \end{gathered}$$

For closed circuit, let us consider that the circuit flowing through the$10Ω$is${\mathrm{I}}_{2.}$

Voltage drop across localid="1649414542659" $5Ω$= voltage drop across the $10Ω$

localid="1649414568230" $$\begin{gathered} 5Ω×1.565\mathrm{A}={\mathrm{I}}_2×10Ω \\ {\mathrm{i}}_2=0.7825\mathrm{A} \end{gathered}$$

Then total current through the battery will be

localid="1649414580361" $$\begin{gathered} \mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2 \\ \mathrm{I}=1.565\mathrm{A}+0.7825\mathrm{A} \\ \mathrm{I}=2.3475\mathrm{A} \end{gathered}$$

Now apply the Kirchhoff voltage law in left loop.

localid="1649414662726" $$\begin{gathered} \mathrm{Î\mu }=\mathrm{r}(2.4575\mathrm{A})+5Ω×1.565\mathrm{A} \\ \mathrm{Î\mu }=7.825+\mathrm{r}(2.4575)...........................\left(2\right) \end{gathered}$$

Using both above equations (1) and (2)

localid="1649212926983" $$\begin{gathered} 1.636\mathrm{r}+8.18=7.825+2.4575\mathrm{r} \\ 0.355=0.8215\mathrm{r} \\ \mathrm{r}=0.43Ω \end{gathered}$$

put this value in above equation (1)

$\mathrm{Î\mu }=8.88\mathrm{V}$