Answer
The direction of the current $i_1$ is up.
Work Step by Step
The equivalent resistance of $R_7$ and $R_8$ is $R_{78} = 4.0~\Omega$
The equivalent resistance of $R_1, R_2$ and $R_{78}$ is $R_{1278} = 2.0~\Omega+2.0~\Omega+4.0~\Omega = 8.0~\Omega$
The equivalent resistance of $R_6$ and $R_{1278}$ is $R_{12678} = 4.0~\Omega$
The equivalent resistance of $R_3$ and $R_{12678}$ is $R_{123678} = 4.0~\Omega+4.0~\Omega = 8.0~\Omega$
The equivalent resistance of $R_4$ and $R_5$ is $R_{45} = 3.0~\Omega+1.0~\Omega = 4.0~\Omega$
Let's assume the direction of $i_1$ is up.
Let $i_3$ be the current through battery 2 and let's assume the direction of the current is to the right.
Let $i_4$ be the current through resistance 3 and let's assume the direction of the current is to the left.
Using the junction rule:
$i_4 = i_1+i_3$
Using the loop rule counterclockwise around the right loop:
$8.0-4.0~i_3-16 = 0$
$4.0~i_3 = -8.0$
$i_3 = -2.0~A$
The negative sign shows that the direction of the current through battery 2 is to the left.
Using the loop rule counterclockwise around the left loop using $R_{123678} = 8.0~\Omega$:
$16-8.0~i_4 = 0$
$8.0~i_4 = 16$
$i_4 = 2.0~A$
We can find $i_1$:
$i_4 = i_1+i_3$
$i_1 = i_4 -i_3$
$i_1 = (2.0~A) -(-2.0~A)$
$i_1 = 4.0~A$
We assumed that the direction of the current $i_1$ is up, and we found the current to be $~~i_1 = 4.0~A$
Since the value we found for $i_1$ is positive, the direction of the current $i_1$ is up.
