## Electrical Engineering: Principles & Applications (6th Edition)

$I_2 = 8A \\ I_8 = 2A$
First, we use Ohm's law, which states that $I=\frac{V}{R}$, to obtain: $I_2 = \frac{V}{2} \\ I_8 = \frac{V}{8}$ We know that these two currents add to 10 Amps, so we find: $\frac{V}{8} + \frac{V}{2} = 10 \\ \frac{5V}{8} = 10 \\ V = 16 \ V$ Plugging this value back into the equation, we find: $I_2 = \frac{16}{2} = \fbox{8A}\\ I_8 = \frac{16}{8} = \fbox{2A}$