#### Answer

The power in the circuit is conserved.

#### Work Step by Step

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.