Answer
The power of battery 1 is $~~1.26~W$
Work Step by Step
Let $i_1$ be the current through $R_1$ and let's assume the direction of the current is to the right.
Let $i_2$ be the current through $R_2$ and let's assume the direction of the current is to the left.
Let $i_3$ be the current through $R_3$ and let's assume the direction of the current is down.
Using the junction rule:
$i_3 = i_1+i_2$
Using the loop rule counterclockwise around the right loop:
$1.00-2.00~i_2-5.00~i_3 = 0$
$1.00-2.00~i_2-5.00~(i_1+i_2) = 0$
$1.00-7.00~i_2-5.00~i_1 = 0$
$-5.00+35.0~i_2+25.0~i_1 = 0$
Using the loop rule clockwise around the left loop:
$3.00-4.00~i_1-5.00~i_3 = 0$
$3.00-4.00~i_1-5.00~(i_1+i_2) = 0$
$3.00-9.00~i_1-5.00~i_2 = 0$
$21.0-63.0~i_1-35.0~i_2 = 0$
We can add these two equations to find $i_1$:
$16.0-38.0~i_1 = 0$
$i_1 = \frac{16.0}{38.0}$
$i_1 = 0.421~A$
The current through resistance 1 is $~~0.421~A$
The current provided by battery 1 is $0.421~A$
We can find the power of battery 1:
$P = i~\mathscr{E}_1$
$P = (0.421~A)(3.00~V)$
$P = 1.26~W$
The power of battery 1 is $~~1.26~W$
