91Ó°ÊÓ

Consider the value of the voltage sources:

$$\begin{gathered} {\mathrm{Î\mu }}_1=6.00V \\ {\mathrm{Î\mu }}_2=12.0V \end{gathered}$$

Consider the values of the resistance as:

$$\begin{gathered} R_1=200\mathrm{Î\copyright } \\ {R}_2=100\mathrm{Î\copyright } \end{gathered}$$

Consider the concept of Kirchhoff’s voltage law and current law to find the unknown currents and their direction. By applying KVL to both loops, determine two different equations.

Consider the formulas:

According to KVL,$∑V=0$

According to KCL,$∑I_{in}=∑I_{out}$

Applying KVL to first loop:

${\mathrm{Î\mu }}_1+i_1{R}_1−i_2{R}_2=0$…… (1)

Applying KVL to second loop:

${\mathrm{Î\mu }}_2−i_1{R}_1=0$

Hence, to find current$i_1:$

$i_1=\frac{{\mathrm{Î\mu }}_2}{R_1}$

Substitute the values and solve as:

$\begin{array}{c}{i}_1=\frac{12}{200}\\ =0.06A\end{array}$

Size of the current through resistance 1 is$i_1=0.06A$

Current$\left(i_1\right)$is directed in downward direction to complete the one cycle.

Therefore, the direction of the current is in the downward direction.

Substitute the valuesof$i_1$ in equation.$\left(1\right)$

$6.00+(0.06×200)−(i_2×100)=0$

Hence, the value of the current is:

$\begin{array}{c}{i}_2=\frac{18.0}{100}\\ =0.18A\end{array}$

Size of the current through resistance 2 is$i_2=0.18A$

Current $i_2$is directed in leftward direction as suggested by the positive value of the current through the resistance $R_2$.

Current through the battery2

$i=i_1+i_2$

Substitute the values and solve as:

$\begin{array}{c}i=0.06+0.18\\ =0.24A\end{array}$

Size of the current through battery 2 is $i=0.24A$.

The current through battery 2 is directed in upward direction.