# Nodal Analysis on Electrical Circuit

1. Feb 3, 2013

### gl0ck

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

3. The attempt at a solution
According to the similar problem
Node X:
I1=(22-Vx)/66
I2=(66-Vx)/44
Il=Vx/264
I3=I1+I2+Il
Node Y:
I4=Vy/240
I5=6
I4=I3+I5

After the calculations I get 1.08 for Vx, which I think is wrong.

Thanks

2. Feb 3, 2013

### Staff: Mentor

Yeah, Vx is definitely greater than 1.08V.

3. Feb 3, 2013

### gl0ck

If the direction of I2 is that shown on the picture, then shouldn't it be I2=(Vx-66)/44
and for I3, I3=(Vy-Vx)/24. If these assumptions are correct, then the calculations are easy, but I think the problem is somewhere in the expressions of the currents.

4. Feb 3, 2013

### Staff: Mentor

Yes, they look fine.
With the expressions that you now have for the currents you should be able to write and solve the two node equations (based on KCL at each node).

5. Feb 4, 2013

### gl0ck

Are the equations for the X node:
I1+I2+Il = I3?
and for the Y node:
I4=I3+I5?
Because I don't understand how the current direction should be expressed by the Kirchoff's law.
which will made the equations look like these:
((22-Vx)/66)+((Vx-66)/44)+(Vx/264)=((Vy-Vx)/24)
and
Node Y:
Vy/240=((Vy-Vx)/24)+6

6. Feb 4, 2013

### Staff: Mentor

If you choose to assume the current directions as depicted in the diagram, then some are flowing into the X node and some are flowing out. According to KCL the sum of the ones flowing in must balance the sum of those flowing out.

Regarding the math of current directions, suppose that you had a couple of nodes A and B with potentials Va and Vb as follows:

The current flowing from node A into node B would be given by $Iab = \frac{Va - Vb}{R}$ .

The current flowing from node B into node A would be given by $Iba = \frac{Vb - Va}{R}$ .

In other words, you specify the assumed current direction by the order in which you take the difference in the potentials.

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• ###### Fig1.gif
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Last edited: Feb 4, 2013
7. Feb 5, 2013

### gl0ck

I2+Il=I1+I3
I3+I5=I4
Are these the correct equations for the currents?

Last edited: Feb 5, 2013
8. Feb 5, 2013

### Staff: Mentor

Assuming the current directions as depicted in the diagram, the first equation looks okay but the second equation does not look okay. Note that both I3 and I4 are shown flowing OUT of node Y. For I5 (which isn't defined on the circuit diagram) you can just use a constant 6 Amps.

9. Feb 5, 2013

### gl0ck

Is it possible Vx to be 160.3V and Vy to be 276.63V?
I found plausible values for the currents:
I1=(22-160.3)/66=-2.095A
I2=(160.3-66)/44=2.143A
I3=(276.63-160.3)/24=4.847A
Il=160.3/264=0.607A
I4=276.63/240=1.152A

10. Feb 5, 2013

### Staff: Mentor

Yes, it's possible. In fact it's more than likely

11. Feb 5, 2013

### gl0ck

Thank you for the help. :)