The resistances in Figures (a) and (b) are all $\mathbf{}\mathbf{6}\mathbf{.0}\mathit{\Omega }$ , and the batteries are ideal $\mathbf{\text{12V}}$batteries. (a) When switch S in Figure (a) is closed, what is the change in the electric potential ${\mathit{V}}_{\mathbf{1}}$ across resistor 1, or does ${\mathit{V}}_{\mathbf{1}}$ remain the same?

(a) When switch S in Figure (b) is closed, what is the change in $V_1$across resistor 1, or does $V_1$ remain the same?

Resistances,$R_1=R_2=R_3=6.0\Omega$

Battery voltage,$V=12\text{V}$

Use the loop rule to find if the change in the electrical potential ${\mathit{V}}_{\mathbf{1}}$across resistor1when the switch S in figure$\mathbf{27}\mathbf{-}\mathbf{45}\mathit{a}$is closed or if${\mathit{V}}_{\mathbf{1}}$remains the same. In addition to the loop rule, use the equivalent resistance formula for parallel and series resistance and Ohm’s law to find if the change in the electrical potential ${\mathit{V}}_{\mathbf{1}}$across resistor1when the switch S in figure$\mathbf{27}\mathbf{-}\mathbf{45}\mathit{b}$is closed or if${\mathit{V}}_{\mathbf{1}}$remains the same.

Formulae:

The current is define by,

$I=\frac{V}{R}$

Equivalent resistance for series combination,$R_{eq}=\underset{J=1}{\overset{n}{}}{R}_J$

Equivalent resistance for parallel combination,$R_{eq}=\underset{J=1}{\overset{n}{}}\frac{1}{R_J}$

The battery is connected across$R_1$. Therefore, according to the loop rule, when the switch is closed, the voltage across$R_1$is$12\text{V}$and when the switch is open, the voltage across$R_1$is still$12\text{V}$.

Hence, the voltage across the resistance$R_1$remains the same.

For the open switch, resistances$R_1$and$R_3$are in series. Therefore, by using the formula for equivalent resistance for series combination, we get

$R_{eq}=R_1+R_2$

Therefore, current through them is,

$I_1=\frac{V}{R_{eq}}=\frac{V}{R_1+R_2}$

The voltage across the resistanceis,

$\begin{array}{c}{V}_1=I_1{R}_1\\ =\frac{V{R}_1}{R_1+R_2}\end{array}$

Substitute known values in the above equation.

$\begin{array}{c}{V}_1=\frac{\left(12\text{V}\right)\left(6.0\Omega \right)}{(6.0\Omega +6.0\Omega )}\\ =6\text{V}\end{array}$

For closed switch, the resistances$R_1$and$R_2$are parallel and they are in series with resistance$R_3$.

Therefore, by using the formula for the equivalent resistance for series and parallel combinations, you get

$\begin{array}{c}{R}_{eq}=R_3+\frac{R_1{R}_2}{R_1+R_2}\\ =\left(6.0\Omega \right)+\frac{\left(6.0\Omega \right)\left(6.0\Omega \right)}{\left(6.0\Omega \right)+\left(6.0\Omega \right)}\\ =9.0\Omega \end{array}$

Therefore, the current in the circuit will be

$\begin{array}{c}{I}_1^{\text{'}}=\frac{V}{R_{eq}}\\ =\frac{12\text{V}}{9.0\Omega }\\ =1.33\text{A}\end{array}$

So, voltage across$R_1$is,

$\begin{array}{c}{V}_1^{\text{'}}=V-I_1^{\text{'}}{R}_3\\ =12\text{V}-\left(1.33\text{A}\right)\left(6.0\Omega \right)\\ =4.0\text{V}\end{array}$

Therefore, the change in voltage is

$\Delta V=V_1-V_1^{\text{'}}$

$\begin{array}{c}\Delta V=6\text{V}-4\text{V}\\ =2\text{V}\end{array}$

Hence,the change in the electrical potential$V_1$ across the resistor $1$ when the switch S in figure$27-45b$ is closed is $\text{2.0V}$.