Answer
The direction of the current through resistor 2 is to the right.
Work Step by Step
Let $i_1$ be the current through $R_1$ and let's assume the direction of the current is down.
Let $i_2$ be the current through $R_2$ and let's assume the direction of the current is to the right.
Let $i_3$ be the current through $R_3$ and let's assume the direction of the current is to the left.
Using the junction rule:
$i_1 = i_2+i_3$
$i_3 = i_1-i_2$
Using the loop rule counterclockwise around the right loop:
$12.0-300~i_3-100~i_1 = 0$
$12.0-300~(i_1-i_2)-100~i_1 = 0$
$12.0-400~i_1+300~i_2 = 0$
$24.0-800~i_1+600~i_2 = 0$
Using the loop rule clockwise around the left loop:
$6.00-200~i_2-100~i_1 = 0$
$18.0-600~i_2-300~i_1 = 0$
We can add these two equations to find $i_1$:
$42.0-1100~i_1 = 0$
$i_1 = \frac{42.0}{1100}$
$i_1 = 38.2~mA$
The current through resistance 1 is $~~38.2~mA$
We can find $i_2$:
$6.00-200~i_2-100~i_1 = 0$
$6.00-200~i_2-100~(38.2~mA) = 0$
$200~i_2 = 2.18$
$i_2 = \frac{2.18}{200}$
$i_2 = 10.9~mA$
The current through resistance 2 is $~~10.9~mA$
We assumed that the direction of the current through resistor 2 is to the right, and we found the current to be $10.9~mA$
Since the value we found for $i_2$ is positive, the direction of the current through resistor 2 is to the right.
