Answer
The current in resistance 3 is $~~1.25~A$
Work Step by Step
We can find an expression for the current in $R_1$:
$i_1+i_2 = i_3$
$i_1 = i_3-i_2$
We can apply the loop rule on the left loop of the circuit:
$10.0-4.00~i_1 - 4.00~i_3 = 0$
$10.0-4.00~(i_3-i_2) - 4.00~i_3 = 0$
$10.0-8.00~i_3 - 4.00~i_2 = 0$
We can apply the loop rule on the right loop of the circuit:
$5.00-4.00~i_2 - 4.00~i_3 = 0$
$10.0-8.00~i_2 - 8.00~i_3 = 0$
We can subtract the second expression from the first expression:
$0+4.00~i_2 = 0$
$i_2 = 0$
We can find $i_3$:
$10.0-8.00~i_3 - 4.00~i_2 = 0$
$10.0-8.00~i_3 - 0 = 0$
$8.00~i_3 = 10.0$
$i_3 = 1.25~A$
The current in resistance 3 is $~~1.25~A$
