Use the Node Voltage method to solve

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cavalieregi
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Homework Statement


Circuit schematic with nodes A to G, a 15V source between A and B, a 5V source between E and G, ground at E, and 1kohm resistors on each branch.

From the above circuit find VCB and VDG

Know Data:
VE = 0 (ground)

Homework Equations


KCL and Ohm's Law are used in Node Voltage Method.

The Attempt at a Solution


I decided I would determine the node voltages then work out VCB and VDG afterwards.
1. KCL and Ohm's law at Node C
## 15 + V_F + V_D = 3 V_C## -(1)​

2. Super Node at EG
2.1 KCL and Ohm's law for all currents entering
## 2(V_F - V_G) = 2 V_E - V_D - V_B##
## 2(V_F - V_G) + V_D + V_B = 2 V_E## -(2)​
2.2 Potentials at Nodes E and G
## V_G = -5V## -(3)​

3. KCL and Ohm's law at Node D
## V_C = 2 V_D## -(4)
​
This is as far as I got I am unsure what to do next so I can get 5 equations so they can all be solved simultaneously.
 
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The problem highly simplifies if you replace the series and parallel resistances with their resultants.
 
ehild said:
The problem highly simplifies if you replace the series and parallel resistances with their resultants.
So if I replaced the two between F and G, and the one between C and F right?
 
cavalieregi said:
So if I replaced the two between F and G, and the one between C and F right?
Yes, and the two resistors between C and E, and also the two resistors in series with the 15 V battery.

In the simplifed circuit, you can find the voltage VC with the node voltage method . Knowing Vc, you can calculate the currents, and knowing the currents, you can determine all node voltages.
 
Simplified Circuit.
Circuit schematic with nodes A to F: 15V source between A and B, R1 1k from A to C, R4 1.5k from C to F, R2 1k from C to D, R3 1k from D to E, R5 1k from B to E, and a 5V source between E and F.

This is what I have done. Not sure if correct?
Handwritten nodal analysis working: KCL at node C giving (VA - VC)/1 = (VC - VF)/(3/2) + (VC - VD)/1, VA = 15V, then super node KCL and KVL steps, with VF = -5V and remaining steps to find VB.

How would I find VB? Is VA really = 15V.
NOTE: VD = 2VC so it can be eliminated.
 
Simplified Circuit.
Circuit schematic with nodes A to F: 15V source between A and B, R1 1k from A to C, R4 1.5k from C to F, R2 1k from C to D, R3 1k from D to E, R5 1k from B to E, and a 5V source between E and F, with ground at E.

This is what I have done.
Handwritten-style typed working for a node voltage analysis: KCL at node C, supernode KCL and KVL equations, substitutions giving Vc = 27.5 V and a final Vb = -230/3 approx 76.67 V.

I think I have made a mistake somewhere.
NOTE: VB is meant to be negative at end where = 76.67V
 
Your solution is very complicated...

The circuit is equivalent with the one in the picture, except B and D nodes disappearing. You can solve it for Vc.
Circuit schematic with nodes A, C, E and F: a 15V source between A and ground, resistors R5 1k and R1 1k in series on the top wire, R4 1.5k from C to F, R23 2k from C down to E, and a 5V source between E and F.
 
ehild said:
Your solution is very complicated...

The circuit is equivalent with the one in the picture, except B and D nodes disappearing. You can solve it for Vc. View attachment 75156
Thanks I have now managed to solve this!
 
cavalieregi said:
Simplified Circuit.
View attachment 75137
This is what I have done. Not sure if correct?
How would I find VB? Is VA really = 15V.
No, VA is not 15 V, as we count the potential with respect to E. VA-VB=15 V
 
ehild said:
No, VA is not 15 V, as we count the potential with respect to E. VA-VB=15 V
I figured Va = Vb + 15 and it worked!
 
cavalieregi said:
Thanks I have now managed to solve this!

Splendid! What did you get for VC?
Have you tried to solve the problem also with the very simple equivalent circuit?
 
ehild said:
Splendid! What did you get for VC?
Have you tried to solve the problem also with the very simple equivalent circuit?
I got VC = 2.5 V
 
Using the equivalent circuit, the node equation for the currents at C is: ##\frac{15-Vc}{2}=\frac{Vc}{2}+\frac{Vc-(-5)}{1.5}## You need only this equation to solve for Vc, which is the same you got.