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

We use the equation for power, which is given by $P=-VI$, where I is the current and V is the voltage. Thus, we find: $P_a = -10 \times 2 = \fbox{-20W}$ We must find the voltage through B using Kirchhoff's Voltage Law: $-V_b - 10 + 4 = 0 \\ V_b = -6V$ This allows us to find power: $P_b = -(-6) \times 2 = \fbox{12W}$ Since 1 Amp goes through element C, it follows: $P_c = 4 \times 1 =\fbox{ 4W}$ (Note, there is no negative, for the current starts at the positive, meaning that the battery is being charged.) We now consider element D: $P_d= 4 \times 1 = \fbox{4W }$ (Note, there is no negative, for the current starts at the positive, meaning that the battery is being charged.) We add all of these to find: $\fbox{P = -20W + 12W + 4W + 4W = 0W}$ Thus, power is conserved.