Answer
$I_{sc} = -3A$
Work Step by Step
We use Kirchhoff's Voltage Law, which states that the voltage around any closed loop in a circuit is 0, to find:
$-15 + v_x = 0 \\ v_x = 15V$
We now apply Ohm's law, which states that $I=\frac{V}{R}$, to find:
$I_x = \frac{15V}{10\Omega} = 1.5A$
We now find the other current, which is:
$I = aV_x = .3 \times 15 = 4.5A$
Finally, we apply Kirchhoff's Current Law, which states that the sum of the currents entering and leaving the junction is 0, to find:
$-I_{sc} + 1.5 - 4.5 = 0 \\ I_{sc} = 1.5 - 4.5 \\ I_{sc} = \fbox{-3A}$
