mjhjmtu

Find the power of each source and provide your answers based on passive convention

so the problem was that the powers didnt add up to zero at the end, so I got some of the current and voltage values wrong...

Also, I dont know which power values are suppose to be negatives.

1) The first problem I can see is when using Cramer's rule. The correct expression should be (using your equations):

$$i_1=dfracmatrix0.006&1\-15&-1000bigg=-3textmA $$

You can then find $i_2=9textmA$.

2) Also you are forgetting to square the current to find the power in the resistors.

With that:

$$(2textkOmega)(-3textmA)^2+(1textkOmega)(9textmA)^2+(15textV)(-3textmA)+(-6textmA)(9textV) =0$$

Notice that I had to place a minus sign for the last term $-(6textmA)(9textV)$ and that is because the current is "comes out" of the terminal labeled as + in your schematic, which means the current source is supplying power. The voltage is actually supplying power too but you have the reference direction for the current going into the + term, that is why we didn't have to place a minus sign in front of that term, but anyway the current came back with a negative sign so everything still holds.

I'm not sure where you went wrong, but the thing can be solved fairly quickly using superposition.

First set the current source to 0 and just solve with the voltage source present.
$15 rm V/3 rm kOmega=5 rm mA$. So we'll say $i_1a=-5 rm mA$ and $i_2a=5 rm mA$.

Now set the voltage source to 0, and just consider the current source. This can be solved by inspection because using the current divider rule, the 2 kohm resistor will get 1/3 of the current and the 1 kohm resistor wil get 2/3 of the current. So $i_1b = 2 rm mA$ and $i_2b = 4 rm mA$.

Combining the two solutions, $i_1 = -3 rm mA$ and $i_2 = 9 rm mA$.

From there you can easily calculate the current through the voltage source ($i_1$) and the voltage across the current source ($1 rm kOmegatimes 7 rm mA$), and then you can calculate the power in or out of each component.

While superposition is, I believe, the best way to approach this problem (as The Photon shows), since there is just one unknown node voltage, node voltage analysis is almost as good.

If you find the node voltage $V_A$ at the junction of the current source and two resistors, you have all you need to find the powers.

The node voltage equation is:

$$frac15,mathrmV - V_A2,mathrmkOmega + 6,mathrmmA = fracV_A1,mathrmkOmega$$

Solving for $V_A$ is just algebra:

$$V_A = left(frac15,mathrmV2,mathrmkOmega + 6,mathrmmAright)/left(frac12,mathrmkOmega + frac11,mathrmkOmega right)= 9,mathrmV$$

Can you take it from here?