In Figure, $\mathbf{}{\mathit{\epsilon }}_{\mathbf{1}}\mathbf{=}\mathbf{6}\mathbf{.00}\mathbf{\text{V}}$, ${\mathit{\epsilon }}_{\mathbf{2}}\mathbf{=}\mathbf{12}\mathbf{.0}\mathbf{\text{V}}$ ,${\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{100}\mathit{\Omega }$ , ${\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{200}\mathit{\Omega }$ ,and${\mathit{R}}_{\mathbf{3}}\mathbf{=}\mathbf{300}\mathit{\Omega }$ . One point of the circuit is grounded$\mathbf{(}\mathbf{V}\mathbf{=}\mathbf{0}\mathbf{)}$ .(a)What is the size of the current through resistance 1? (b) What is the direction (up or down) of the current through resistance 1? (c) What is the size of the current through resistance 2?(d) What is the direction (left or right) of the current through resistance 2? (e) What is the size of the current through resistance 3? (f) What is the direction of the current through resistance 3? (g) What is the electric potential at point A?

The emf voltage, ${\epsilon }_1=6.00\text{V}$

The emf voltage, ${\epsilon }_2=12.0\text{V}$

The resistor, $R_1=100\Omega$

The resistor, $R_2=200\Omega$

The resistor, $R_3=300\Omega$

Use the loop rule (Kirchhoff’s Voltage law) and the junction rule (Kirchhoff’s Current law) to find the current flowing through each resistance. Once you get the value of current,you can tell that in which direction it is moving.

Formula:

For any loop,$\sum V=0$

At any junction$\sum I_i=\sum I_o$

The Ohm’s law is, $V=IR$

Using the junction rule,you can say that

$I_1=I_2+I_3$ ….. (1)

Using the loop rule,you can say that

${\epsilon }_1-I_2{R}_2-I_1{R}_1=0$ ….. (2)

${\epsilon }_2-I_3{R}_3-I_1{R}_1=0$ ….. (3)

Substituting the value of$I_1$ into equation (1).

$\begin{array}{l}{\epsilon }_1-I_2{R}_2-(I_2+I_3)R_1=0\\ {\epsilon }_1-(R_2+R_1)I_2-I_3{R}_1=0\\ (R_2+R_1)I_2+I_3{R}_1={\epsilon }_1\end{array}$

Substitute known values in the above equation.

$300I_2+100I_3=6.0$ ….. (4)

Now substitute equation (1) into equation (2).

$\begin{array}{l}{\epsilon }_2-I_3{R}_3-(I_2+I_3)R_1=0\\ {\epsilon }_2-I_3(R_3+R_1)-I_2{R}_1=0\\ I_3(R_3+R_1)+I_2{R}_1={\epsilon }_2\end{array}$

Substitute known values in the above equation.

$100I_2+400I_3=12.0$ ….. (5)

Solving equations (4) and (5), and multiply equation (5) with , you get,

$\begin{array}{c}300I_2+100I_3-300I_2-1200I_3=6.0-36.0\\ -1100I_3=-30.0\\ I_3=0.0273\text{A}\end{array}$

Substitute the above value into equation (4).

$\begin{array}{l}300I_2+100(0.0273)=6.0\\ 300I_2=6.0-2.73\\ I_2=0.0109\text{A}\end{array}$

Put$0.0109\text{A}$ for $I_2$and $0.0273\text{A}$for $I_3$ into equation (1).

$\begin{array}{c}{I}_1=I_2+I_3\\ =(0.0109+0.0273)\text{A}\\ =0.0382\text{A}\end{array}$

Since the magnitude of the current flowing through the resistance$R_1$ is $0.0382\text{A}$, by using the diagram and the junction rule,you can say that the direction of the current flowing through the resistance$R_1$ is downward.

After solving the equations$\left(1\right)$ and $\left(2\right)$simultaneously,you got the valueof thecurrent flowing through the resistance$R_2$ as $0.0109\text{A}$.

Using the diagram and the junction rule,you can say that the direction of the current flowing through the resistance $R_2$ is rightward.

After solving the equations$\left(1\right)$ and $\left(2\right)$simultaneously, you got the value of current flowing through the resistance$R_3$ as $0.0273\text{A}$.

Using the diagram and the junction rule,you can say thatthe direction of the current flowing through the resistance$R_3$ is leftward.

Electric potential across point Acan be given as,

$\begin{array}{c}{V}_A=I_1{R}_1\\ =0.0382\text{V}\times 100\Omega \\ =+3.82\text{V}\end{array}$

Hence,the electric potential at point A is.$+3.82\text{V}$