Question

In Figure, the ideal batteries have emfs ${\mathit{\epsilon }}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.0}\mathbf{\text{\hspace{0.17em}V}}$and ${\mathit{\epsilon }}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.0}\mathbf{\text{\hspace{0.17em}V}}$, the resistances are each $\mathbf{2}\mathbf{.0}\mathbf{\text{\hspace{0.17em}Ω}}$, and the potential is defined to be zero at the grounded point of the circuit. What are potentials

(a) What are potential $V_1$at the indicated points?

(b) What are potential ${\mathit{V}}_{\mathbf{2}}$at the indicated points?

Short answer

1. The potential $V_1$ is, $-11.0\text{\hspace{0.17em}V}$.
2. The potential $V_2$ is, $-9.0\text{\hspace{0.17em}V}$.

Step-by-step solution

Given

1. The EMF${\epsilon }_1=5.0\text{\hspace{0.17em}V}$ .
2. The EMF${\epsilon }_2=12\text{\hspace{0.17em}V}$.
3. The resistances are, $R_1=R_2=R_3=R_4=R_5=2.00\text{\hspace{0.17em}Ω}$

Understanding the concept

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" ${\mathit{R}}_{\mathbf{e}\mathbf{q}}\mathbf{=}\sum _{\mathbf{J}\mathbf{=}\mathbf{1}}^{\mathbf{n}}{\mathit{R}}_{\mathbf{J}}$
2. Equivalent resistance for parallel combination,
   $R_{eq}=\underset{J=1}{\overset{n}{}}\frac{1}{R_J}$

(a) Calculate potential V1  at the indicated points

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.

$\begin{array}{c}R\text{'}=\frac{\left(2.00\text{\hspace{0.17em}Ω}\right)\left(2.00\text{\hspace{0.17em}Ω}\right)}{\left(2.00\text{\hspace{0.17em}Ω}\right)+\left(2.00\text{\hspace{0.17em}Ω}\right)}\\ =1.0\text{\hspace{0.17em}Ω}\end{array}$

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.

$\begin{array}{c}{R}_{eq}=\left(2.00\text{\hspace{0.17em}Ω}\right)+\left(2.00\text{\hspace{0.17em}Ω}\right)+\left(2.00\text{\hspace{0.17em}Ω}\right)+\left(1.0\text{\hspace{0.17em}Ω}\right)\\ =7.00\text{\hspace{0.17em}Ω}\end{array}$

Now, voltage in the loop is

$V={\epsilon }_2-{\epsilon }_1$

Substitute all the value in the above equation

$\begin{array}{c}V=12.00\text{\hspace{0.17em}V}-5.00\text{\hspace{0.17em}V}\\ =7.00\text{\hspace{0.17em}V}\end{array}$

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

$\begin{array}{c}I=\frac{7.00\text{\hspace{0.17em}V}}{7.00\text{\hspace{0.17em}Ω}}\\ =1\text{\hspace{0.17em}A}\end{array}$

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

$\begin{array}{c}{V}^{\text{'}}=I{R}^{\text{'}}\\ =\left(1.0\text{\hspace{0.17em}A}\right)\left(1.0\text{\hspace{0.17em}Ω}\right)\\ =1.0\text{\hspace{0.17em}V}\end{array}$

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

$12.0\text{\hspace{0.17em}V}-1.0\text{\hspace{0.17em}V}=11.0\text{\hspace{0.17em}V}$

Noting the orientation of the battery, we can say that

$V_1=-11.0\text{\hspace{0.17em}V}$

Therefore, the potential $V_1$is$-11.0\text{\hspace{0.17em}V}$.

(b) Calculate potential  at the indicated points

Now, the voltage drop between the two voltage points is

$V"=IR$

$\begin{array}{c}V"=\left(1.0\text{\hspace{0.17em}A}\right)\left(2.00\text{\hspace{0.17em}Ω}\right)\\ =2\text{\hspace{0.17em}V}\end{array}$

So, the potential is

$V_2=V_1+V"$

Substitute all the value in the above equation.

$\begin{array}{c}{V}_2=-11.0\text{\hspace{0.17em}V}+2\text{\hspace{0.17em}V}\\ =-9.0\text{\hspace{0.17em}V}\end{array}$

Therefore, the potential $V_2$ is $-9.0\text{\hspace{0.17em}V}$.