Because I made a mistake in transcribing from your expression and then correcting it in Post #12. I'm having to scroll the screen to see the diagram and then to see my work and my memory is getting bad enough that I shouldn't do that.At node e:
node e: (6A) + (Vc-Vb)/(2 Ω) + (Ve-Vb)/(2 Ω) = 0
You counted the two terms that overlap though, so why wouldn't you do it for Node f?
Sorry I don't see it.
No, I just made a mistake that you caught because you were trying to make sense of what I wrote. Good catch. That is why I went ahead and worked the rest of the problem for you. Look that over and get comfortable with it, then I will show you how to get the necessary equations MUCH quicker.I think I see what you did now, you used super position
Sorry for not doing very much....I should be able to do this stuff easier than I do.No, I just made a mistake that you caught because you were trying to make sense of what I wrote. Good catch. That is why I went ahead and worked the rest of the problem for you. Look that over and get comfortable with it, then I will show you how to get the necessary equations MUCH quicker.
Nothing to be sorry for. I made the mistake and that resulted in you chasing a rabbit down a hole for a bit. This stuff isn't easy at first and it takes a lot of practice working problems to get it down.Sorry for not doing very much....I should be able to do this stuff easier than I do.
I agree that the current coming from Node e into Node d is 6A and not -6A and it does matter because if you settle for getting the magnitude right but ignore a sign mismatch, then that is a VERY wrong answer. So when you see something like that it is important to resolve it.Current coming from node e into node d is 6A not -6A but it does not matter, the value is the same.
The current flowing into Node d from Node b and Node f have to add up to -6A. This is because, when added to the +6A flowing from Node e to Node b the result has to be zero. The key is that I didn't say that they had to add to -6A to be EQUAL to the current coming from Node e (which is +6A), but that they had to add to -6A in order to be CONSISTENT with the current coming from Node e.WBahn said:Close enough and a very good sign. But not done yet. Next let's get the currents flowing into Node d from Node b and f and see if those add to -6A to be consistent with the current coming from Node e.
(Vb-Vd)/(3 Ω) = -1.2267 A
(Vh - Vd)/(1 Ω) = -4.770 A
Sum = -5.997 A
Node b: + Vb[(1/5Ω)+(1/3Ω)+(1/2Ω)] - (0V+40V)[(1/5Ω)] - Vd[(1/3Ω)] - (Ve+40V)[(1/2Ω)] = 0
Node d: - Vb[(1/3Ω)] + Vd[(1/3Ω)+(1/1Ω)] - (0V+30V)[(1/1Ω)] = 6A
Node e: - (Vb-40V)[(1/2Ω)] + Ve[(1/2Ω)+(1/5Ω)] - 0V[(1/5Ω)] = -6A
Node b: + 31Vb - 10Vd - 15Ve = 240V + 600V = 840V
Node d: - Vb + 4Vd = 18V + 90V = 108V
Node e: - 5Vb + 7Ve = -60V - 200V = -260V
I'm impressed how you got Bill to do so much work for you! It certainly shows he is more generous / helpful than you originally thought.Sorry for not doing very much....I should be able to do this stuff easier than I do.
Yup and I certainly feel like an *** for thinking that initially and believe me. Though you only can think about what you see. If nothing else from studying his problem here what hits home the most is when he says "always always, always". I'll never forget that. Never ever...I'm impressed how you got Bill to do so much work for you! It certainly shows he is more generous / helpful than you originally thought.
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