- #1
builder_user said:Do I need second Kirhgoff rules' equations for every loop?
gneill said:That's not the value I get for i1. Better check your derivation.
builder_user said:Ok.But what's the next?
gneill said:Very sorry. I see that your figure of 3.7A for i1 and about 5.2V for the voltage between nodes 1 and 2 look okay.
gneill said:Again, sorry about that.
15 pages of A4? That seems like a lot for just three loops. What method are you using to solve the equations?
gneill said: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).
Voltage between two nodes is the difference in electrical potential energy per unit charge between those two nodes.
Voltage between two nodes can be calculated by dividing the difference in potential energy between the two nodes by the amount of charge that flows between them.
The unit of measurement for voltage between two nodes is volts (V).
The voltage between two nodes can be affected by the amount of charge, the distance between the nodes, and the type of material between the nodes (e.g. conductor, insulator).
The voltage between two nodes determines the rate of flow of electricity. A higher voltage will result in a greater flow of electricity, while a lower voltage will result in a slower flow of electricity.