The first thing is to find the currents in all the branches.
By applying I = U / R in the 25 Ω resistor we have I = 10 / 25 = 0.4 A
By Kirchoff’s law in the center loop, we have 5 + 10*I –10 = 0
So I = 5 / 10 = 0.5 A
Now we have:
We can say that we have:
a. Power delivered by source S1 is P = U*I = 10*0.9 = 9 W
Power delivered by source S2 is P = -U*I = - 5*1 = - 5 W (or considered as received)
Power delivered by the current source P =U*I = 5*0.5 = 2.5W
b. Power delivered to the 25 Ω resistor is P = 25*(0.4)2 = 4 W
Power delivered to the 10 Ω resistor is P = 10*(0.5)2 = 2.5 W
c. Power delivered = 9 – 5 + 2.5 = 4 + 2.5 = 6.5 W
Power received by the resistors = 4 +2.5 =6.5 W
The energy is conserved here.