Equivalent Capacitance: Find C Between A & B | Tips

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The discussion focuses on finding the equivalent capacitance between points A and B in a circuit with identical capacitors. Participants emphasize the importance of using Kirchhoff's laws and suggest redrawing the circuit to simplify the analysis. They discuss identifying parallel and series connections among capacitors, with some confusion arising from the arrangement of nodes and connections. A few participants share their attempts at solving the problem using different methods, including nodal analysis and collapsing nodes, while seeking clarification on discrepancies in their results. Ultimately, the conversation highlights the necessity of understanding circuit behavior rather than just applying formulas.
  • #31
jishnu said:
Sorry my mistake, can you please help me know about the difference in nodal analysis and star to delta or vise-versa conversions. Isn't the thing that I have done in #19 also called mesh / nodal analysis(I am not sure about that).

The second thing you did in #19 is called star-delta conversion.

You did not do this: https://www.electronics-tutorials.ws/dccircuits/dcp_6.html, which is called nodal analysis.
 
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  • #32
OK. Forget what I said in #27. After you did the Y-Δ transformation: https://en.wikipedia.org/wiki/Y-Δ_transform
each capacitor has a value of C/3. But, you didn't get the schematic after the transformation correct. Two of the capacitors are shorted by a wire that was in the pre-transformation topology as shown here. Look carefully and you'll see that there are 4 capacitors of value C/3 in parallel for a final value of (4C)/3

1528788005266.jpg
 

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  • #33
Since you have got the correct answer by two methods, I'll show you what I would consider the most general method, which others suggested early on in the thread.

You can do a nodal analysis like this. First select a reference node and number the rest of the nodes like this:

CapNetwork.jpg

Take the admittance of each capacitor to be just C. Now you can write the admittance matrix Y by inspection using this method:

After you get the admittance matrix, invert it, giving the impedance matrix Z:

CapNetworkNodal.png


The reactance, and hence the inverse of the equivalent capacitance, of each node is equal the value on the main diagonal of the Z matrix.

The value in red, 3/(4 C), is the reactance from node 1 to ground so the equivalent capacitance is (4 C)/3. Similarly, the equivalent capacitance from node 2 to ground is (3 C)/2, and from node 3 to ground it's (12 C)/5. You get all the equivalent capacitances with one matrix calculation. :smile:
 

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  • #34
In my opinion, this could be this way:
upload_2018-6-21_20-17-6.png

Does not it?
 

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  • #35
It could be this way:
upload_2018-6-21_20-19-7.png
 

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  • #36
Babadag said:
In my opinion, this could be this way:
View attachment 227162
Does not it?

This is what jishnu did in post #16 and in the left side of the image of post #19
 
  • #37
I'm sorry, I just looked at the matrix post!
 
  • #38
The Electrician said:
OK. Forget what I said in #27. After you did the Y-Δ transformation: https://en.wikipedia.org/wiki/Y-Δ_transform
each capacitor has a value of C/3. But, you didn't get the schematic after the transformation correct. Two of the capacitors are shorted by a wire that was in the pre-transformation topology as shown here. Look carefully and you'll see that there are 4 capacitors of value C/3 in parallel for a final value of (4C)/3

View attachment 227136
Thanks allot for the clarification. [emoji4]
 
  • #39
jishnu said:
Thanks allot for the clarification. [emoji4]

For extra credit, consider the addition of one more capacitor to the network like this:

CapNetworkX.jpg


Is the equivalent capacitance changed, and if so, what is it?
 

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  • #40
The Electrician said:
For extra credit, consider the addition of one more capacitor to the network like this:

View attachment 227191

Is the equivalent capacitance changed, and if so, what is it?
The capacitance would simply become equivalent to "C", is that correct...!
 
  • #41
jishnu said:
The capacitance would simply become equivalent to "C", is that correct...!

How did you get that? The extra cap makes it difficult to do parallel/series reduction. But nodal analysis is truly general. Any topology can be solved.

The matrix from post #33 only needs a small tweak to represent the extra capacitor.

Here's the result:

CapNetworkNodalX.png


The equivalent capacitance between A and B is (7 C)/5

This shows the power of nodal analysis.
 

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  • #42
The Electrician said:
How did you get that? The extra cap makes it difficult to do parallel/series reduction. But nodal analysis is truly general. Any topology can be solved.

The matrix from post #33 only needs a small tweak to represent the extra capacitor.

Here's the result:

View attachment 227194

The equivalent capacitance between A and B is (7 C)/5

This shows the power of nodal analysis.
This is how I got the answer(attached a rough work)
Can you please provide me resources to know more about formation of the admittance matrix, I couldn't understand much relevant things about its matrix formation from the electronics tutorial link that you have provided.
1529684476876.jpg
 

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  • #43
jishnu said:
This is how I got the answer(attached a rough work)
Can you please provide me resources to know more about formation of the admittance matrix, I couldn't understand much relevant things about its matrix formation from the electronics tutorial link that you have provided. View attachment 227195
'

When using the Y-Δ transformation: https://en.wikipedia.org/wiki/Y-Δ_transform

the elements of the circuit must be treated as resistances (impedances in our case), not conductances (admittances in our case). We have taken the capacitances to be admittances because of the way capacitors combine when they are in parallel, and when in series. When capacitors are in parallel, their values are simply added. This is not the way resistors (or inductors) in parallel combine.

So for the purpose of this exercise, we treat the admittance of a capacitor whose value is 2C as if it were an admittance of value 2C, or an impedance of value 1/(2C). If we want the equivalent capacitance of two capacitors 2C and 3C in parallel, it's just 5C. But, if they're in series, we must calculate 1/Ceqv = 1/(2C) + 1/(3C), which gives Ceqv = (6C)/5

For capacitors in series, add their impedances; for capacitors in parallel, add their admittances.

When using the Y-Δ transformation be sure to have the values used in the calculation in impedance form, not admittance form.

Here are the steps for the solution of the new circuit I gave you using the Y-Δ transformation. In these images the values shown are in admittance form. Even though it's not shown, to use the transformation I took the reciprocals to convert them to impedances before doing the calculations. The results were then converted back to admittance form by taking reciprocals.

CnetPlusA.jpg
CnetPlusB.jpg


Notice how nicely the nodal analysis method with the nodal equations in matrix form took care of all the reductions and algebra automatically.

Regarding more information about the admittance matrix, search on Google for the phrase "admittance matrix by inspection". Also look for "nodal analysis".
 

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