Question

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?

Short answer

(a)The outcomes are

$$\begin{gathered} {\mathrm{I}}_{\mathrm{bat}}=8\mathrm{A} \\ ∆{\mathrm{V}}_{\mathrm{ab}}=8\mathrm{V} \end{gathered}$$

(b) The outcomes are

$$\begin{gathered} {\mathrm{I}}_{\mathrm{bat}}=9\mathrm{A} \\ ∆{\mathrm{V}}_{\mathrm{ab}}=0\mathrm{V}. \end{gathered}$$

Step-by-step solution

Part (a) Step 1: Given information

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.

Simplify

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.

$∆\mathrm{V}=20-12=8\mathrm{V}$

Part (b) Step 1: Given information

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.

Simplify

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.

$∆\mathrm{V}=0$