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

$I = 1.5A$ $V = 52.5V$
First, we apply Kirchhoff's junction rule to obtain: $I = .5I_x + I_x \\ I = 1.5I_x$ We now use Ohm's law to find $I_x$: $I_x = \frac{V}{R} = \frac{20V}{20\Omega} = 1A$ Thus, we find $I$: $I = 1.5 I_x = 1.5(1) = \fbox{1.5A}$ We now use Kirchhoff's Voltage Law, which states that the voltage around any closed loop in a circuit is 0, to find: $-v + (15\Omega)(1.5A)+20V + (10\Omega)(1A) = 0 \\ v = (15\Omega)(1.5A)+20V + (10\Omega)(1A)\\ v = \fbox{52.5V}$