Going through previous tests before the exam checking what I did wrong. Can’t figure it out:
With that one, not worried about the part marked wrong at the top, worried about P(Vx). He marks it wrong and writes the “correct” formula, but I used that exact formula to figure out the power. The only thing I could figure is wrong would be the sign, but it seems correct.
Same trouble there, figuring out the power. Seems right, though (in case you can’t see it, the dependent current source = (Vx/ 8) and the DVS = (15kIx).
In the first one, it’s really hard to make out the subscripts, but near as I can tell, you figured the numbers wrong. If the dependent source is Vx, it’s not 15V, it’s the voltage across the 12Ω resistor, right? You did have the formula right, though.
In the second case, it appears you didn’t simplify. The question asked for the answers to be put in terms of the node voltages. For the first one, you need to substitute in for Vx, and for the second one, for the current.
Good luck on the final (unless this was it).
I did use the voltage across the 12k resistor. Note right above the P(Vx) equation, I got the # for Vx (-21V), which was the correct answer, and used that in the formula. When you click on the image, does it magnify? That should help you see things more clearly.
I see what you mean for the second one. Thanks.
Sorry; I wrote the wrong number, but the voltage was still figured wrong.
There’s an easier way to get Vx, since it’s just the voltage across a resistor. You know the resistance ( 12Ω ) and the current (-5A)[sup][/sup], just multiply the two.
I think the formula may have been written there to indicate how you got partial credit, or the grader just didn’t catch your derivation of Vx. But you’d have to ask them to know for sure.
[sup][/sup] I’m guessing this was a typo on the exam, right? I don’t see a ‘k’ next to the resistance values, and since the mA is crossed out, I’m using that for numbers. Even magnified, it’s still hard to read.
He screwed up with the mA, it was supposed to be in Amps. He’s nice when it comes to giving us easier math, and mixing ohms and milliamps or kOhms and amps gets messy.
I didn’t have an issue finding Vx. I got that part right. Finding the power absorbed/supplied by the dependend source Vx, “P(Vx)” is what was supposedly wrong. Still have an hour or so before I have to go to the exam.
If they told you that you figured the power formula wrong, that wasn’t correct.
Since you have limited time, I’ll just tell you : Vx = ( 12Ω )*(-Ix) = -60V
In the formula you figured, you were using the wrong answer from the first part. There wasn’t anything wrong with your algebra, just the numbers you used.
One more comment on this problem. I don’t know if you noticed, but the current didn’t need to be specified once the voltage was given (or the other way around). A single equation, going all around the single loop, would let you figure the loop current from the independent voltage source. Giving both, and the way the problem is phrased, seems a bit misleading.
Hell, I DID screw up the value of Vx. If we screw one value up, there’s no double jeopardy, so we get credit if we do they rest of a problem correctly with the wrong values. I guess they just marked the wrong thing as being incorrect. Thanks.