collin37

Let I1 be the current flows through the 2-ohm resistor from left to right,
I2 be the current through the 4-ohm resistor downward,
I3 be the current through the 8-ohm resistor from right to left.

Hence, I1 + I3 = I2
i.e. I3 = (I2 - I1)  ------------- (1)

Consider the left loop,
3 + 6 = 2(I1) + 4(I2)
i.e. I1 = [9 - 4(I2)]/2  ------------- (2)

Consider the right loop,
9 + 6 = 8(I3) + 4(I2)
15 = 8(I2 - I1) + 4(I2)
15 = 12(I2) - 8(I1)
i.e. 15 = 12(I2) - 8[9-4(I2)]/2
solve for I2 gives I2 = 1.821 A

Therefore, the current in the 4-ohm resistor is 1.821 A and is flowing downward.