Answer
The direction of the current in battery 2 is up.
Work Step by Step
Let $i_1$ be the current through battery 1 and let's assume the direction of the current is down.
Let $i_2$ be the current through battery 2 and let's assume the direction of the current is up.
Let $i_3$ be the current through battery 3 and let's assume the direction of the current is up.
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:
$4.0-1.0~i_3+2.0~i_2-4.0-1.0~i_3 = 0$
$-2.0~i_3+2.0~i_2 = 0$
$i_2 = i_3$
$i_2 = i_1-i_2$
$i_2 = 0.50~i_1$
Using the loop rule counterclockwise around the left loop:
$4.0-2.0~i_2-1.0~i_1-2.0-1.0~i_1 = 0$
$2.0-2.0~i_2-2.0~i_1 = 0$
$2.0-2.0~(0.50~i_1)-2.0~i_1 = 0$
$2.0-3.0~i_1 = 0$
$i_1 = 0.667~A$
The current in battery 1 is $~~0.667~A$
We can find $i_2$:
$i_2 = 0.5~i_1 = (0.5)(0.667~A) = 0.33~A$
The current in battery 2 is $~~0.33~A$
We assumed that the direction of the current in battery 2 is up, and we found the current to be $0.33~A$
Since the value we found for $i_2$ is positive, the direction of the current in battery 2 is up.
