# Nodal analysis, setting up the equations

1. Sep 14, 2010

### gomezfx

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

[PLAIN]http://img844.imageshack.us/img844/8508/circuit.png [Broken]

2. Relevant equations
KCL set ups

3. The attempt at a solution
I am having trouble setting up the KCL equations

The 30V voltage source on the left with the resistor is messing me up, how do I incorporate that?

Do I just do 30/15 and assume a 2A current source is going into node A?

Here are my equations:

VD = reference node = 0 V

KCL @ Node B
(VB-VA)/25 + (VB-VC)/50 - 1 + 4 = 0

KCL @ Node C
(VC-VD)/50 + (VC-VB)/50 + 1 = 0

KCL @ Node A?
(VA-VB)/25 + (VA-VD)/31.25 - 4 + ?? = 0

Last edited by a moderator: May 4, 2017
2. Sep 14, 2010

### ehild

You have 30V at the other end of the 15 ohm resistor. So the current through it is (VA-30)/15.

ehild

3. Sep 14, 2010

### gomezfx

Ok, then the entire equation for voltage at node A is:

(VA-30)/15 + (VA-VB)/25 + (VA-VD)/31.25 - 4 = 0?

After I solve for all the voltages at the nodes, how do I find power at the independent voltage a current sources?

Do I just use voltage drops?
IE, Power at 4A = (VA-VB)*4

4. Sep 14, 2010

### ehild

This is correct.

Yes, knowing the potential difference U across the terminals of a source and the current I flowing through it, the power is U*I.

ehild