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

$I_R = -1.5A \\P_R = 18W \\ P_I = 48 W \\P_V=-66W$
First, we use Ohm's law to find the current: $I_R = \frac{V}{R} = \frac{-12V}{8 \Omega} = \fbox{-1.5A}$ Now, we use the equation for Power: $P=VI$. Recall, if the current goes from the negative to the positive, there is a negative sign in from of this equation. Recalling this, we obtain: $P_R = -(12V)(-1.5A) = \fbox{18W}$ $P_I = (12V)(4A) = \fbox{48W}$ $P_V = -(12V)(5.5A) = \fbox{-66W}$ Note, if the power is negative, the source is supplying energy, while if the power is positive, the source is absorbing energy.