91Ó°ÊÓ

Consider the real source of emf has internal energy then its terminal potential difference$V_{ab}$ is as follows:

${\mathit{V}}_{\mathbf{a}\mathbf{b}}\mathbf{=}\mathit{Î\mu }\mathbf{-}\mathbf{IR}$ ….. (1)

(a)

Consider the given expression for the current and the voltage.

$$\begin{gathered} V_{ba}=8.40V \\ {I}_{ba}=1.5A \end{gathered}$$

Substitute the values in the equation (1).

$$\begin{gathered} 8.40=\mathrm{Î\mu }-(1.5)\left(R\right) \\ \mathrm{Î\mu }=8.40+1.5r \end{gathered}$$

Consider the current is 3.50 A in the reverse direction, the potential difference becomes 10.20 V.

Substitute the values in the equation $V=\mathrm{Î\mu }+lr$and solve.

$$\begin{gathered} 10.20=(8.30+1.5\mathrm{r})+3.5\mathrm{r} \\ 5\mathrm{r}=1.8 \\ \mathrm{r}=0.36\mathrm{Î\copyright } \end{gathered}$$

Hence, the internal resistance of the battery is$0.36\mathrm{Î\copyright }$

(b)

Given: the value of the internal resistance is $0.36\mathrm{Î\copyright }$.

Substitute $0.36\mathrm{Î\copyright }$for r in the equation $\mathrm{Î\mu }=8.40+1.5r$.

$$\begin{gathered} \mathrm{Î\mu }=8.40+1.5×0.36 \\ =8.94V \end{gathered}$$

Hence, the emf of the battery is 8.94 V.