Nodal Analysis on Electrical Circuit

In summary, the conversation involves a discussion about a problem with node equations and current directions in a circuit. The person asking for help is struggling with understanding how to express current directions using Kirchoff's law and is unsure if their calculations are correct. The expert provides guidance on setting up the equations and suggests using assumed current directions to solve the problem. Eventually, plausible values for the currents are found.
  • #1
gl0ck
85
0

Homework Statement


Nodal.png


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
 
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  • #2
gl0ck said:

Homework Statement


View attachment 55350

The Attempt at a Solution


According to the similar problem
Node X:
I1=(22-Vx)/66
I2=(66-Vx)/44 <--- Check the indicated direction of I2.
Il=Vx/264
I3=I1+I2+Il <--- Why not I3 in terms of node voltages Vx and Vy?
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
Yeah, Vx is definitely greater than 1.08V.

Can you show more of your work? How about writing out the node equations?
 
  • #3
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.

Thanks for the reply
 
  • #4
gl0ck said:
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.
Yes, they look fine.
If these assumptions are correct, then the calculations are easy, but I think the problem is somewhere in the expressions of the currents.
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
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
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:

attachment.php?attachmentid=55378&stc=1&d=1360016286.gif


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.
 

Attachments

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Last edited:
  • #7
I2+Il=I1+I3
I3+I5=I4
Are these the correct equations for the currents?
 
Last edited:
  • #8
gl0ck said:
I2+Il=I1+I3
I3+I5=I4
Are these the correct equations for the currents?

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
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
gl0ck said:
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

Yes, it's possible. In fact it's more than likely :smile:
 
  • #11
Thank you for the help. :)
 

1. What is nodal analysis and why is it used in electrical circuits?

Nodal analysis is a method used to analyze and solve complex electrical circuits. It is based on Kirchhoff's current law, which states that the sum of currents entering a node must be equal to the sum of currents leaving that node. Nodal analysis is used to determine the voltages at different nodes in a circuit, making it a valuable tool for circuit analysis and design.

2. How do you set up a nodal analysis equation?

To set up a nodal analysis equation, you must first identify all the nodes in the circuit and label them. Then, using Kirchhoff's current law, you can write an equation for each node by summing the currents entering and leaving that node. This will result in a system of equations that can be solved to determine the voltages at each node.

3. What are the advantages of using nodal analysis?

Nodal analysis is a powerful tool for solving complex circuits because it allows for easy analysis of circuits with multiple voltage sources and current sources. It also provides a systematic approach for solving circuits and can be used for both DC and AC circuits. Additionally, nodal analysis is more efficient and accurate than other methods, such as mesh analysis.

4. What are the limitations of nodal analysis?

Nodal analysis can become complex and time-consuming for circuits with a large number of nodes. It also requires a good understanding of electrical principles and circuit analysis techniques. Nodal analysis is also not suitable for circuits with nonlinear components, such as diodes or transistors.

5. How do you handle dependent sources in nodal analysis?

Dependent sources, such as voltage-controlled voltage sources or current-controlled current sources, can be handled in nodal analysis by using a dummy variable. This variable represents the dependent source and can be eliminated from the equations once the solution is obtained. The value of the dummy variable is then determined using the solution obtained from the nodal analysis equations.

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