Answer
$I = 1.5A $
$V = 52.5V$
Work Step by Step
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} $
