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Technique to analyze complex circuits? 
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#1
Mar814, 08:29 AM

P: 51

I would like to what is the best way to analyze complex circuit with resistors in series and parallel. I have attached a sample circuit. You can also show it to me with other circuits. Thanks



#2
Mar814, 09:01 AM

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P: 5,465

Hi Tekneek. The best approach is to redraw the circuit, preserving its electrical connections but moving or reorienting the drawing of some elements, perhaps just one at a time. You may have to make many attempts before it becomes clear how the arrangement simplifies. It's a case where your skill at this will improve with practice. (Or so they tell me.)



#3
Mar814, 01:56 PM

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P: 1,991

Nodes connected by simple wire, i.e. nodes with 0 resistance between them are at the same potential and can be condensed into a single node. As you know. This one seems to condense into a fairly trivial circuit.



#4
Mar814, 03:20 PM

P: 51

Technique to analyze complex circuits?



#5
Mar814, 05:08 PM

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P: 1,991

Aaargh why am I telling you?  nobody told me. Next time you'll get something I don't. Nice thing about this is that this step involves no calculations. 


#6
Mar814, 05:35 PM

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P: 7,279

There are only 4 different nodes, because of the connecting links. Two are already labeled a and b on your diagram, so label the other two c and d.
To draw a new diagram, start by drawing 4 dots for the nodes and label them. Then draw each resistor between the correct two nodes. 


#7
Mar814, 05:47 PM

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#8
Mar914, 06:55 PM

P: 51

Soo is it like this...
R1(R2+R2+R4 in parallel)R5 


#10
Mar914, 11:15 PM

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P: 3,735

The dirt simple general rule is this:
It's an iterative process. You redraw the circuit multiple times, in each iteration you combine two elements and replace them with their equivalent. Repeat until it's a trivial circuit. For example, in your circuit: You MIGHT go like this: 1. Observing that R3 is in parallel with R2, draw the circuit with R3 omitted and R2 replaced by (R2//R3) 2. Observing that R4 is in parallel with (R2//R3), redraw the circuit with R4 omitted and (R2//R3//R4) up there where R2 used to be. 3. Observing that you now have a series circuit of three resistors, redraw it as a single resistor from point a to point b of value (R1 + R5 + (R2//R3//R4)). Which is what you got. Now it's a one step solution for current. So you can figure voltage everywhere by reversing the simplifying steps above. That's the how behind the what. You got the idea already, this is just a simple statement of the process. I hope it's easy to remember. Practice, practice, practice. 


#11
Mar1114, 08:40 AM

P: 51




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