Use the Node Voltage method to solve

AI Thread Summary
The discussion focuses on using the Node Voltage method to find voltages VCB and VDG in a given circuit. Participants outline their attempts to set up equations using Kirchhoff's Current Law (KCL) and Ohm's Law, emphasizing the importance of simplifying the circuit by combining series and parallel resistances. A key point is the realization that certain node voltages can be eliminated to simplify calculations, leading to a clearer path to solving for VC. Ultimately, one participant successfully calculates VC as 2.5 V using an equivalent circuit approach. The conversation highlights the iterative nature of solving circuit problems and the importance of verifying assumptions about voltage potentials.
cavalieregi
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Homework Statement


Gl4flF7.jpg

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.
upload_2014-11-6_1-31-21.png

This is what I have done. Not sure if correct?
upload_2014-11-6_1-28-52.png

How would I find VB? Is VA really = 15V.
NOTE: VD = 2VC so it can be eliminated.
 
cavalieregi said:
Simplified Circuit.
View attachment 75137
This is what I have done. Not sure if correct?
View attachment 75135
How would I find VB? Is VA really = 15V.
NOTE: VD = 2VC so it can be eliminated.

Actually I think I have it just wait.
 
Simplified Circuit.
upload_2014-11-6_1-31-21-png.75137.png

This is what I have done.
upload_2014-11-6_1-51-21.png

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.
nodvoltmet.JPG
 
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!
 
  • #10
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
 
  • #11
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!
 
  • #12
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?
 
  • #13
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
 
  • #14
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.
 
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