
#19
Mar2411, 06:21 PM

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P: 11,419

Okay, even better.
Attached is an example, using MathCad, of solving the loops by writing the loop equations in matrix form (which can be done by inspection, and is (almost) foolproof). It's very quick. 



#20
Mar2411, 06:27 PM

P: 196

Intersting.I didn't know about it in Mathcad
all results are the same as the results in my method of countour currents thank you for your help. 



#21
Mar2411, 06:45 PM

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P: 11,419

The most interesting part, I think, is that you can write the equation quickly by inspection.
For the impedance matrix, the terms on the diagonal are just the sum of the impedances (resistances in this case) in the given loop. The offdiagonal terms are just the negative of the impedances that are shared by the loops indexed by the matrix entry. So in this case, for example, R2 is shared by loops 1 and 2, so z_{12} and z_{21} are both R2. Note that the impedance matrix is symmetrical about the diagonal, so it's very quick to fill in. The voltage vector is just the sum of the voltage source rises in the given loop (where a rise is taken to mean that a given voltage source goes from  to + in the direction of the loop current). 



#22
Mar2411, 06:50 PM

P: 196




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