# How to do this node analysis

1. Aug 8, 2008

### Cricket.sign

1. The problem statement, all variables and given/known data
Using mesh analysis work out the currents flow through each of the three meshes. From these current values calculate the voltage of node A and node B with respect to the ground node. XX V has a value of 13.

http://www.flickr.com/photos/29385059@N07/2744576925/

2. Relevant equations
KCL
KVL

3. The attempt at a solution
Well I figured out the mesh analysis easy enough. Obtaining, I1=.046, I2=.398 and I3=-.5908. Got to the node analysis and haven't got a clue. I came up with

1. V1/15 = 3 + [V1 -(V2+30)/40] + [V1-V2]/10

2. [V1-V2]/10 + [V1 -(V2+300)/40] = V2/20 + [(V2+17)/25]

Which gave me V1 = 187.5
V2 = 105.5

2. Aug 9, 2008

### CEL

In your two equations there is a term where you mix voltages and currents:
1. [V1 -(V2+30)/40] where V1 is a voltage and (V2+30)/40 is a current
2. [V1 -(V2+300)/40] the same

3. Aug 13, 2008

### Cricket.sign

[V1 -(V2+30)/40] is actually incorrectly input. It should be (V1-V2-30)/40. Not sur if that changes anything, but I think it should

4. Aug 14, 2008

### CEL

It changes, but it is not correct yet. According to KCL, the sum of the currents out of a node must be zero. Using your reference it should be:
$$\frac{V_1}{15}+3+\frac{V_1-V_2-30}{40}+\frac{V_1-V_2}{10}=0$$