# Node voltages, need help with equations

1. Mar 4, 2015

### TheRedDevil18

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

Find the node voltages (diagram below)

2. Relevant equations

3. The attempt at a solution

Somehow when compared to the textbook equations they seem to be wrong, I don't know why, anyways here they are

Node A: i1-i2-ix = 0
(Va-Vc)G4 - (Va-Vb)G2 - Va*G1 = 0

Node B: i2+is-iy = 0
is = iy-i2
= Vb*G3 - (Vb-Va)G2

Node C: Vc = Vs

Are these equations correct ?, if not then why ?

2. Mar 4, 2015

### phinds

How many equations do you have in how many unknowns? Would you be able to solve the set?

3. Mar 4, 2015

### TheRedDevil18

The voltage source can be taken as a constant or given, Vc = Vs, so two equations and two unknowns. I'm not too worried about solving it, I just want to know if they are set up correctly

4. Mar 4, 2015

### phinds

Yes, they look OK. I'm so used to doing loop equations that it just seemed off at first.

5. Mar 4, 2015

### Staff: Mentor

For the current's directions handwritten on the schematic, the second of the two lines quoted is not correct.
The equation for i1 should be that i1 = (Vc-Va)G4
determined by saying that current flows from the higher potential to the lower.

There is a sign error in your Node B equations, also.

6. Mar 4, 2015

### TheRedDevil18

Oh, thank you, I see it now. I had no idea that I had to follow the currents path when working out the potential difference but I see what you mean now. My equations are correct now. Thanks for the replies guys

7. Mar 4, 2015

### donpacino

I know they teach you to draw and label each current in the circuit, but in my opinion that only opens you up to making more mistakes and it takes longer. If you don't need to directly solve for a current, don't solve for it.

An easy way to do solve these types of equations is to do nodal analysis by doing a KCL at each UNKNOWN node. When you do the KCL allways do
0=sum of current into the node
or
0=sum of current out of the node

For me it makes it much easier, as it can prevent silly mistakes.

If you do need to solve for a current, I have found it is easiest to find all the node voltages, then use ohms law to find the current.

8. Mar 4, 2015

Ok, thanks