Nodal Analysis

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#1 21677291 14 Apr 2017 04:07

Cheni Hermon Cheni Hermon

Anonymous

Post #1
21677291 14 Apr 2017 04:07

Hello,

I have a question in my HW, and i'm not sure about how to decide about my nodes.

I set nodes 2,3,6,7 as my nodes, while node 7 as the reference node.

I feel like it's aint true at all, and I would like to get some help..

Thanks for help
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#2 21677292 16 Apr 2017 20:28

Dean Franks Dean Franks

Anonymous

Post #2
21677292 16 Apr 2017 20:28

Things that are connected by ideal wires are all parts of the same node, so this circuit has 4 nodes from an analysis standpoint, 1, 2, 7 and the node below r2 (which can probably be ignored if you combine all three resistors into their equivalent first).
5,6,7,8 are all the same node and 2,3,4 are all part of the same node.

Dean
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#3 21677293 17 Apr 2017 01:32

Cheni Hermon Cheni Hermon

Anonymous

Post #3
21677293 17 Apr 2017 01:32

Thank you very much!

If I need to solve this circuit by node analysis - than I can't combine the resistors (because my target is finding the whole currents and voltages in this circuit )- and if i'll do so, I'll stay with only one node, am I?
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#4 21677294 17 Apr 2017 10:00

Dean Franks Dean Franks

Anonymous

Post #4
21677294 17 Apr 2017 10:00

This problem is easiest to solve in two phases. First, you can combine R(L), R2, R3 and R4 into a single resistor to solve the initial currents and voltages. Once you have the voltage between 2 and 7, you can then put all the resistors back in and solve for the currents and voltages across the individual resistors.

Also, you can choose any node as reference, but your choice of 7 is the best choice for this circuit.
#5 21677295 21 Apr 2017 18:32

vergel atoc vergel atoc

Anonymous

Post #5
21677295 21 Apr 2017 18:32

nodes 2,3,4 are the same node. and 5,6,7,8 are on the same node too. this problem is easily solved by combining R3,R4 and R2 to RL. 1 is an independent node so you can leave it as is. you will be left with 3 nodes. node 1, node 2,3,4, and node 5,6,7,8.
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