Is the source V_s in the network absorbing or supplying power, and how much?
Image from: Irwin, J. David., and R. M. Nelms. Basic Engineering Circuit Analysis, Tenth Edition. N.p.: John Wiley & Sons, 2010. Print.
Solution:
P_{10v}=(10)(3)=30W
P_{6v}=(6)(3)=18W
P_{9A}=(16)(-9)=-144W
P_{V_s}=(V_s)(6)=6V_sW
P_{8v}=(8)(6)=48W
In the circuit, all power absorbed must equal the power supplied. In our case, one source supplies power (the element with the negative power value), and the power absorbed by the other elements must equal each other. In other words, if we add all of the power together, the ones that supply power and absorb power, it must all equal to 0. Thus, we can write the following equation:
-144+30+18+6V_s+48=0
(Isolate for V_s)
6V_s=48
V_s=8v
Now, let us calculate the power absorbed or supplied by the V_s element.
P_{V_s}=(8)(6)=48W
The value we got for the element is positive, meaning it absorbs power.
This question can be found in Basic Engineering Circuit Analysis, 10th edition, chapter 1, question 1.42.
