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Circuit Analysis Help

depends what level your'e at now before we cna tell you what advanced means.. it doesn't look advanced to me though.. justl ooks like you did algebra on it, which is what you're supposed to do
 
I'm currently up to nodal analysis, so not advanced by any means. Apparently I was supposed to do nodal analysis on this, but I didn't...
 
Maybe transformation? I am too lazy to analyze your work right now(busy) but you can change all those Voltage sources into current sources and such...
 
His email to me was:

I am not quite sure how to grade problem #2 in your mid-term exam. The solution you provided is:
a). Not via nodal analysis;
b). Seemingly based on a few physically incorrect assumptions - but -
it resulted in the correct answer, and in some parts resembled one of the advanced methods of analysis. I wasn't sure whether it was intentional, or if you got the solution by accident.
Therefore, I have tentatively given you a 50% credit for it, but if you come to me before Monday and explain the reasoning behind your method, I will give you full credit for it and, possibly, extra points as well.

So I'm just wondering how I should explain this. Is it incorrect to do Req when the resistors have voltage sources connected to the same line?
 
oooohs i got itS! u decatived the voltage sources by settin them to 0 and hten you got the thevenin resistance of the circuit.

seriously thats it im sobering up a litlte now you jus trgot the thevenin resistance by shorting out the voltage sources
 
Originally posted by: iamtrout
I'm currently up to nodal analysis, so not advanced by any means. Apparently I was supposed to do nodal analysis on this, but I didn't...
Click to expand...

Nodal was like week 2 for us.... You wanna do some nodal analysis there, but it shouldn't be bad... Pick your reference pt as the bottom, and use the nodes up top. You should have enough equations easily...
 
It looks like you were making an incorrect assumption when you said the current in each branch was (V_n/n)/(R_n/n)... that would be true if the node on top was grounded, but as the circuit was drawn, its an unknown. So it would be ((V_n/n)-V_unknown)/(R_n/n)

How you still managed to get a right answer is beyond me. If I were you, I'd take my 50% and run with it.

Edit: Oh, and yes, equivalent resistances are only for when the resistors and nothing else are in parallel. I was thinking you did a Thevenin equivalent of some kind, but I guess you weren't - so I guess that part wasn't right either. He was probably referring to Thevenin equivalents when he said "advanced method".
 
Why must the node on top be grounded? If the node on top was grounded ( - ) and the bottom was ( - ) there would be a short and thus nothing would happen?
 
Originally posted by: iamtrout
Why must the node on top be grounded? If the node on top was grounded ( - ) and the bottom was ( - ) there would be a short and thus nothing would happen?
Click to expand...

Well, if it were grounded, the voltage sources would each push a current through their own branch resistor and back into the ground. A short circuit is when you put a wire (with 0 resistance) directly between two nodes and bypass all the resistances that would normally be in between them - but the current still has to go through the resistor in this case, so its not a short.

Edit: Hopefully that made sense.
 
You were suppose to sacrifice 30 virgins(every male in your class) to appease the might god Norton.
 
Use your bottom node as ground reference, and top as a node for analysis... then use superposition and the summation of those currents / voltages.

 
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