In Figure,${\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{100}\mathbf{\text{\hspace{0.17em}Ω}}$,${\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{50}\mathbf{\text{\hspace{0.17em}Ω}}$, and the ideal batteries have emfs$\mathbf{}{\mathit{\epsilon }}_{\mathbf{1}}\mathbf{=}\mathbf{6}\mathbf{.0}\mathbf{\text{\hspace{0.17em}V}}$,$\mathbf{}{\mathit{\epsilon }}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{.0}\mathbf{\text{\hspace{0.17em}V}}$, and.${\mathit{\epsilon }}_{\mathbf{3}}\mathbf{=}\mathbf{4}\mathbf{.0}\mathbf{\text{\hspace{0.17em}V}}$Find

(a) The current in resistor 1,

(b) The current in resistor 2, and

(c) The potential difference between points aand b.

1. The resistance .$R_1=100\text{\hspace{0.17em}Ω}$
2. The resistance$R_2=50\text{\hspace{0.17em}Ω}$.
3. The voltage${\epsilon }_1=6.0\text{\hspace{0.17em}V}$.
4. The voltage${\epsilon }_2=5.0\text{\hspace{0.17em}V}$.
5. The voltage${\epsilon }_3=4.0\text{\hspace{0.17em}V}$ .

We use the concept of Kirchhoff’s voltage law. We write the equation of Kirchhoff’s voltage for the loops to find the currents and the voltage.

${\mathit{V}}_{\mathbf{1}}\mathbf{+}{\mathit{V}}_{\mathbf{2}}\mathbf{+}{\mathit{V}}_{\mathbf{3}}\mathbf{=}\mathbf{0}$

The current in resistor 1:

We consider the lower loop to find the current through $R_1$,

$\begin{array}{c}{\epsilon }_2-i_1{R}_1=0\\ {\epsilon }_2=i_1{R}_1\\ i_1=\frac{{\epsilon }_2}{R_1}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}{i}_1=\frac{5.0\text{\hspace{0.17em}V}}{100\text{\hspace{0.17em}Ω}}\\ =0.050\text{\hspace{0.17em}A}\end{array}$

Hence the current in resistor 1 is, $0.050\text{\hspace{0.17em}A}$.

The current in resistor 2:

Now, we consider the upper loop to find the current through $R_{2,}$we get

$\begin{array}{c}{\epsilon }_1-{\epsilon }_2-{\epsilon }_3-i_2{R}_2=0\\ {\epsilon }_1-{\epsilon }_2-{\epsilon }_3=i_2{R}_2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}{i}_2\left(50\text{\hspace{0.17em}Ω}\right)=6.0\text{\hspace{0.17em}V}-5.0\text{\hspace{0.17em}V}-4.0\text{\hspace{0.17em}V}\\ i_2=\frac{-3.0\text{\hspace{0.17em}V}}{50\text{\hspace{0.17em}Ω}}\\ =-0.06\text{\hspace{0.17em}A}\end{array}$

The negative sign indicates that the current direction is downward. We take$\left|i_2\right|=0.060\text{\hspace{0.17em}A}$.

Hence the current in resistor 2 is, $0.060\text{\hspace{0.17em}A}$.

The potential difference between the points a and b:

The potential difference between the points a and b is the sum of the potential between them, we can write

$V_a-V_b={\epsilon }_2+{\epsilon }_3$

Substitute all the value in the above equation.

$\begin{array}{c}{V}_a-V_b=5.0\text{\hspace{0.17em}V}+4.0\text{\hspace{0.17em}V}\\ =9.0\text{\hspace{0.17em}V}\end{array}$

Hence the potential difference between point a and b is, $9.0\text{\hspace{0.17em}V}$.