# Use the Node Voltage method to solve

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1. Nov 5, 2014

### cavalieregi

1. The problem statement, all variables and given/known data

From the above circuit find VCB and VDG

Know Data:
VE = 0 (ground)

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

3. 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.

2. Nov 5, 2014

### ehild

The problem highly simplifies if you replace the series and parallel resistances with their resultants.

3. Nov 5, 2014

### cavalieregi

So if I replaced the two between F and G, and the one between C and F right?

4. Nov 5, 2014

### ehild

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.

5. Nov 5, 2014

### cavalieregi

Simplified Circuit.

This is what I have done. Not sure if correct?

How would I find VB? Is VA really = 15V.
NOTE: VD = 2VC so it can be eliminated.

6. Nov 5, 2014

### cavalieregi

Actually I think I have it just wait.

7. Nov 5, 2014

### cavalieregi

Simplified Circuit.

This is what I have done.

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

8. Nov 5, 2014

### ehild

The circuit is equivalent with the one in the picture, except B and D nodes disappearing. You can solve it for Vc.

9. Nov 5, 2014

### cavalieregi

Thanks I have now managed to solve this!

10. Nov 5, 2014

### ehild

No, VA is not 15 V, as we count the potential with respect to E. VA-VB=15 V

11. Nov 5, 2014

### cavalieregi

I figured Va = Vb + 15 and it worked!

12. Nov 5, 2014

### ehild

Splendid! What did you get for VC?
Have you tried to solve the problem also with the very simple equivalent circuit?

13. Nov 5, 2014

### cavalieregi

I got VC = 2.5 V

14. Nov 5, 2014

### ehild

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.