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Help with Circuits problem (KVL, KCL)by EE is cool
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#1
Aug3013, 09:08 PM

P: 6

1. The problem statement, all variables and given/known data
Find v. A value for R is not needed. Circuit is attached. 2. Relevant equations KVL, KCL 3. The attempt at a solution I have no idea where to begin with this one. I found an answer using source transformations by transforming the 4A current source into a 12V voltage source in series and the 2A current source into an 8V voltage source in series. At this point there's two loops, and I used KCL to find the current in the right loop to be 4A counterclockwise, and from there I used KVL in the supermesh loop to determine that v = 56V. First of all, is this correct? Second, we haven't officially learned source transformations yet so I shouldn't be using them. I am clueless when it comes to current source behavior, so can someone explain to me how this would be done without source transformations? Thanks! 


#2
Aug3013, 09:30 PM

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Hi EE is cool.
Draw a large diagram, because you can do all of the work on the diagram, no complicated simultaneous equations needed. First question: how much current is flowing into the left hand side of resistor "R"? Second question: how much current is flowing out of the right hand side of resistor "R"? Those two questions might give you sufficient clue to be able to finish this. 


#3
Aug3013, 09:33 PM

P: 4

Well by KCL it would be 4A in and also 4A out right? If V = IR, I still can't find V because I don't know R.



#4
Aug3013, 09:35 PM

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Help with Circuits problem (KVL, KCL)
How much current is flowing through the 4Ω resistor? 


#5
Aug3013, 09:40 PM

P: 4

2A, which I guess means 4A flows into the parallel circuit at the top. Here I run into the fact that I don't know how current sources work again. Am I supposed to determine how much of that 4A splits into each branch? I am surely missing something here.
Oops, wrong account haha. 


#6
Aug3013, 09:52 PM

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Current sources are nothing to be frightened of. Think of an element carrying a ammeter that is indicating "there is 4A flowing here" and pointing to its direction. You don't need to know anything more about it.



#7
Aug3013, 10:02 PM

P: 6

Huh, that's a much easier way to think about it. So does that mean there's no current through the 3 ohm resistor since an ammeter has negligible resistance? But then, the diagram would only make sense if the current source went in the other direction wouldn't it?



#8
Aug3013, 10:16 PM

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It makes sense which ever direction, actually. The question you have shows 4A flowing to the right in that top source. So you need some logical way to arrive at a magnitude & direction for the current in the 3Ω. 


#9
Aug3013, 10:22 PM

P: 6

Well there's 4A going into the topleftmost junction, and there's 4A going into that parallel circuit, which makes sense. If there's 4A going against that current due to the current source, there must be 8A going to the left across the 3 ohm resistor?



#10
Aug3013, 10:48 PM

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#11
Aug3013, 10:56 PM

P: 6

Oh nice, that gives me 56V too. I love it when the light bulb finally comes on. Thanks a bunch! :)



#12
Aug3113, 12:13 AM

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Current comes into the left of R, yet its right side is more positive? That must make it a negative resistance. This might point to a shortcoming on the part of whoever set this question.... 


#13
Aug3113, 12:40 AM

P: 6

No I actually don't, it's just the answer I got from doing KVL once I got all the currents and from when I tried the source transformations. I assumed it was correct since I received the same answer from both methods, but maybe it's not. You probably know better than I do.



#14
Aug3113, 01:54 AM

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56V is the answer. But this makes the voltage across element "R" to have the opposite sign to what it would be for an ordinary resistance. (So the only conclusion is that it is a negative resistance, R Ω.)



#15
Aug3113, 10:31 AM

P: 6

Well, it's the weekend so it'll be a few days before I can bring it up with him. I definitely will though.



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