# AC Nodal analysis -- help please

1. Apr 7, 2015

1. The problem statement, all variables and given/known data
Determine using nodal analysis the current in branch Z4

2. Relevant equations

3. The attempt at a solution

Not sure how to handle the source v3 my solution so far has came up with answers that are not the same as the answers that I got for mesh analysis.

Node A VA-V1/Z1+VA/Z4+VA-VB/Z2= 0
VA(1/Z1+Z2+Z4)-V1(1/Z1)-VB(1/Z2)=0
VA(0.5-j0.4)-VB(0-j0.2)=60+j0
Node B VB-VA/Z2+VB/Z5+VB-V2/Z3=0
-VA(1/Z2)+VB(1/Z2+Z3+Z5)-V2(1/Z3)=0
-VA(0-j0.2)+VB(0.25-j0.45)=0+j30

After plugging this in to Wolframalpha I get VA = 79.93+j68.3
VB = -10.9+j36.42

Therefore VA/Z4 = -13.66+j15.99 A my answer from mesh = 8.78+j16.59 A I have put this back into the equations to confirm.

Like I say im not sure about the V3 source.Is the potential at node VA the same as V3?

Many thanks for any help.

2. Apr 7, 2015

### Staff: Mentor

No, but VA = VB + V3.

Have you covered supernodes?

3. Apr 7, 2015

### donpacino

first, your equations for nodal analysis are not correct. you are not taking into account the voltage source V3

The goal of nodal analysis is to express all node voltage in the form of all known inputs/independent variables. Due to the voltage source connecting nodes A and B, once we know one we instantly know the other (Va=VB+V3)
Nodal analysis works by doing a KCL at each unknown node. to do a KCL you must know all current that leave the node. We do not know the current through that voltage source.

Given those two points, you cannot use regular nodal analysis to solve this problem. One way is to use something called a 'supernode.'
first define the supernode, which is nodes A and B, and the components that DIRECTLY connect them. Then write a KCL for all currents leaving the supernode and you will have your governing equation.

4. Apr 8, 2015

Thank you both... I have applied a supernode to the circuit which gave me the equation:

(120+j0-VA/2+j0)+(0-VA/0-J5)+(0-VB/0+J4)+(0+J120-VB/4+J0)=0
VA-VB=V3

Therefore VA/Z4 = -9.15+j17.28A :)

I now have to double check my loop analysis. Thanks again for the nudge much appreciated.