91Ó°ÊÓ

1. The EMF${\mathrm{Î\mu }}_1=5.0\text{ V}$ .
2. The EMF${\mathrm{Î\mu }}_2=12\text{ V}$.
3. The resistances are, $R_1=R_2=R_3=R_4=R_5=2.00\text{‰ө}$

We use the formula forequivalentresistancefor theparallel and series combinations to find theequivalentresistance. We use it in Ohm’s law to findthe current and then the potentials at the indicated points.

Formula:

$\mathit{V}\mathbf{=}\mathit{I}\mathit{R}$

1. Equivalent resistance for series combination,
   role="math" localid="1662707670118" ${\mathit{R}}_{\mathbf{e}\mathbf{q}}\mathbf{=}\underset{\mathbf{J}\mathbf{=}\mathbf{1}}{\overset{\mathbf{n}}{∑}}{\mathit{R}}_{\mathbf{J}}$
2. Equivalent resistance for parallel combination,
   $R_{eq}=\underset{J=1}{\overset{n}{}}\frac{1}{R_J}$

Resistances on the right side are parallel.

Since the equivalent resistance for parallel combination is
$R_{eq}=\underset{J=1}{\overset{n}{}}\frac{1}{R_J}$

Therefore,

$R\text{'}=\frac{R_1{R}_2}{R_1+R_2}$

Substitute all the value in the above equation.

$\begin{array}{c}R\text{'}=\frac{\left(2.00\text{‰ө}\right)\left(2.00\text{‰ө}\right)}{\left(2.00\text{‰ө}\right)+\left(2.00\text{‰ө}\right)}\\ =1.0\text{‰ө}\end{array}$

This resistance R is in series with the remaining three resistors.

Since equivalent resistance for series combination is
$R_{eq}=\underset{J=1}{\overset{n}{}}{R}_J$

Therefore,

$R_{eq}=R_3+R_4+R_5+R^{\text{'}}$

Substitute all the value in the above equation.

$\begin{array}{c}{R}_{eq}=\left(2.00\text{‰ө}\right)+\left(2.00\text{‰ө}\right)+\left(2.00\text{‰ө}\right)+\left(1.0\text{‰ө}\right)\\ =7.00\text{‰ө}\end{array}$

Now, voltage in the loop is

$V={\mathrm{Î\mu }}_2−{\mathrm{Î\mu }}_1$

Substitute all the value in the above equation

$\begin{array}{c}V=12.00\text{ V}−5.00\text{ V}\\ =7.00\text{ V}\end{array}$

So, by using Ohm’s law, $V=IR$ ,the current is

$\begin{array}{c}I=\frac{7.00\text{ V}}{7.00\text{‰ө}}\\ =1\text{‼î}\end{array}$

The voltage across $R^{\text{'}}$ is

$\begin{array}{c}{V}^{\text{'}}=I{R}^{\text{'}}\\ =\left(1.0\text{‼î}\right)\left(1.0\text{‰ө}\right)\\ =1.0\text{ V}\end{array}$

So, from the right side of the circuit, the voltage difference between the ground and is

$12.0\text{ V}−1.0\text{ V}=11.0\text{ V}$

Noting the orientation of the battery, we can say that

$V_1=−11.0\text{ V}$

Therefore, the potential $V_1$is$−11.0\text{ V}$.

Now, the voltage drop between the two voltage points is

$V"=IR$

$\begin{array}{c}V"=\left(1.0\text{‼î}\right)\left(2.00\text{‰ө}\right)\\ =2\text{ V}\end{array}$

So, the potential is

$V_2=V_1+V"$

Substitute all the value in the above equation.

$\begin{array}{c}{V}_2=−11.0\text{ V}+2\text{ V}\\ =−9.0\text{ V}\end{array}$

Therefore, the potential $V_2$ is $−9.0\text{ V}$.