What are the battery current ${\mathrm{I}}_{\mathrm{bat}}$and the potential difference ${\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{b}}$between points a and b when the switch in FIGURE P28.55 is (a) open and (b) closed?

We have given, the circuit diagram as shown in the above figure.

We have to find the current and potential difference when switch will be open.

The total resistance in the circuit of series resistances,

$$\begin{gathered} {\mathrm{R}}_1=3\Omega +3\Omega =6\Omega \\ {\mathrm{R}}_2=5\Omega +1\Omega =6\Omega \\ \mathrm{Then}\mathrm{for}\mathrm{parallel}, \\ \frac{1}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2} \\ \frac{1}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{1}{6\Omega }+\frac{1}{6\Omega }=\frac{1}{3\Omega } \\ {\mathrm{R}}_{\mathrm{eq}}=3\Omega \end{gathered}$$

then current through the battery will be using ohm's law.

${\mathrm{I}}_{\mathrm{bat}}=\frac{\mathrm{E}}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{24\mathrm{V}}{3\Omega }=8\mathrm{A}$

Since both braches in the circuit have same value of resistance, then amount of current will be in both should be same.

then in each branch current will be $4\mathrm{A}.$since the equal amount of current will divide.

Then potential drop across $3\Omega$resistance =$3\Omega \times 4\mathrm{A}=12\mathrm{V}$

potential difference across $5\Omega$resistance =localid="1649417786411" $5\Omega \times 4\mathrm{A}=20\mathrm{V}$

Then potential difference between the point a and b will be difference of potential drop across $3\Omega$and $5\Omega$resistances.

$∆\mathrm{V}=20-12=8\mathrm{V}$

We have given, the circuit diagram as shown in the above figure.

We have to find the current and potential difference when switch will be closed.

Total resistance in the circuit will be

$$\begin{gathered} \frac{1}{{\mathrm{R}}_1}=\frac{1}{3\Omega }+\frac{1}{5\Omega }=\frac{8}{15} \\ {R}_1=1.9\Omega \\ \frac{1}{{\mathrm{R}}_2}=\frac{1}{3\Omega }+\frac{1}{1\Omega }=\frac{4}{3} \\ {R}_{\mathit{2}}=0.75\Omega \\ \mathrm{Then}, \\ {\mathrm{R}}_{\mathrm{eq}}={\mathrm{R}}_1+{\mathrm{R}}_2 \\ {\mathrm{R}}_{\mathrm{eq}}=1.9\Omega +0.75\Omega =2.65\Omega \end{gathered}$$

Then current through the battery will be

$I_{\mathrm{bat}}=\frac{\mathrm{E}}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{24\mathrm{V}}{2.65\Omega }=9\mathrm{A}$

Since both the point are joined together then potential difference will be zero.

$∆\mathrm{V}=0$