# Find voltage between two nodes

by builder_user
Tags: nodes, voltage
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Mentor
P: 11,689
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.
Attached Files
 Schem1.pdf (67.3 KB, 2 views)
 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.
 Mentor P: 11,689 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 off-diagonal 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 z12 and z21 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).
P: 196
 Quote by gneill 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 off-diagonal 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 z12 and z21 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).
I've got it.I compared elements of matrix with my equations a few minutes ago.

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