91Ó°ÊÓ

1. The resistance .$R_1=100\text{‰ө}$
2. The resistance$R_2=50\text{‰ө}$.
3. The voltage${\mathrm{Î\mu }}_1=6.0\text{ V}$.
4. The voltage${\mathrm{Î\mu }}_2=5.0\text{ V}$.
5. The voltage${\mathrm{Î\mu }}_3=4.0\text{ V}$ .

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 $R_1$,

$\begin{array}{c}{\mathrm{Î\mu }}_2−i_1{R}_1=0\\ {\mathrm{Î\mu }}_2=i_1{R}_1\\ i_1=\frac{{\mathrm{Î\mu }}_2}{R_1}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}{i}_1=\frac{5.0\text{ V}}{100\text{‰ө}}\\ =0.050\text{‼î}\end{array}$

Hence the current in resistor 1 is, $0.050\text{‼î}$.

The current in resistor 2:

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

$\begin{array}{c}{\mathrm{Î\mu }}_1−{\mathrm{Î\mu }}_2−{\mathrm{Î\mu }}_3−i_2{R}_2=0\\ {\mathrm{Î\mu }}_1−{\mathrm{Î\mu }}_2−{\mathrm{Î\mu }}_3=i_2{R}_2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}{i}_2\left(50\text{‰ө}\right)=6.0\text{ V}−5.0\text{ V}−4.0\text{ V}\\ i_2=\frac{−3.0\text{ V}}{50\text{‰ө}}\\ =−0.06\text{‼î}\end{array}$

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

Hence the current in resistor 2 is, $0.060\text{‼î}$.

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

$V_a−V_b={\mathrm{Î\mu }}_2+{\mathrm{Î\mu }}_3$

Substitute all the value in the above equation.

$\begin{array}{c}{V}_a−V_b=5.0\text{ V}+4.0\text{ V}\\ =9.0\text{ V}\end{array}$

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