Unconventional capacitors connection

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Homework Help Overview

The problem involves analyzing two configurations of capacitors, specifically a π (Pi) configuration and a "T" configuration, to demonstrate their equivalence through specific capacitance equations. The original poster expresses difficulty in understanding the relationships between potential differences and electric charges in these arrangements.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to express capacitors as impedances and suggest starting with resistors for simplicity. There are questions about isolating potential differences and the implications of applying the same voltages to both circuits. Some participants propose investigating ABCD parameters related to 2-port networks.

Discussion Status

There are multiple lines of reasoning being explored, including the potential use of 2-port network analysis. Some participants have provided guidance on possible approaches, while others are awaiting confirmation from a specific user before proceeding further. The discussion remains open with various interpretations being considered.

Contextual Notes

The problem is noted to be challenging and originates from a textbook that does not explicitly cover 2-port networks, raising questions about the appropriateness of the problem's context.

Drako Amorim
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Homework Statement



Hello everyone. The problem is related to these two capacitor connections:
capc.jpg

In the enunciate it is said that these two are equivalent and it is requested to prove that:

  1. C1=(CxCy+CyCz+CzCx)/Cx
  2. C2=(CxCy+CyCz+CzCx)/Cy
  3. C1=(CxCy+CyCz+CzCx)/Cz

Homework Equations



C=Q/Vab (Capacitance definition)

The Attempt at a Solution



My problem with the solution is that I do not know how to analyze them, because of the two potentials differences.

My progress:

It is seen that the two potential differences and the two electric charges are the same in the two arrangements, so I tried to isolate Vac and Vbc in function of the electric charges: to no avail.

I noticed that in the first image, the capacitor at the top will have an electric charge merged from the two originals, that I call q3, so the left capacitor will have a charge of q1 - q3 (or q1+q3?) and the another will have a charge of q2+q3(or q2-q3?), I believe of that order do not import.

Can you give me a light on how to do this? Sorry for my poor English, I'm not proficient.
 

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Drako Amorim said:

Homework Statement



Hello everyone. The problem is related to these two capacitor connections:
View attachment 111748
In the enunciate it is said that these two are equivalent and it is requested to prove that:

  1. C1=(CxCy+CyCz+CzCx)/Cx
  2. C2=(CxCy+CyCz+CzCx)/Cy
  3. C1=(CxCy+CyCz+CzCx)/Cz

Homework Equations



C=Q/Vab (Capacitance definition)

The Attempt at a Solution



My problem with the solution is that I do not know how to analyze them, because of the two potentials differences.

My progress:

It is seen that the two potential differences and the two electric charges are the same in the two arrangements, so I tried to isolate Vac and Vbc in function of the electric charges: to no avail.

I noticed that in the first image, the capacitor at the top will have an electric charge merged from the two originals, that I call q3, so the left capacitor will have a charge of q1 - q3 (or q1+q3?) and the another will have a charge of q2+q3(or q2-q3?), I believe of that order do not import.

Can you give me a light on how to do this? Sorry for my poor English, I'm not proficient.
The first configuration is a π (Pi) configuration, and the second is a "T" configuration. I think the best way to work with this is to express the capacitors as impedances. Figure out the math to make the input and output impedances the same, and the transfer functions the same.

You could also do it first for resistors, since the math is a bit simpler for resistors. Maybe try it with resistors and post your work here so we can comment? :smile:
 
Drako Amorim said:
I tried to isolate Vac and Vbc in function of the electric charges
Not sure what you mean by that.
You understand that the question is saying that whatever the voltages applied, if you apply the same voltages to both circuits you will see the same external charge flows, right?
What equations can you write for the two circuits?
 
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Judging by the diagrams it occurs to me that this problem may originate in a course section studying 2-port networks. If that is the case there may be some benefit to investigating the so-called ABCD parameters.

I won't offer any further hints until @Drako Amorim returns to the thread to confirm.
 
gneill said:
Judging by the diagrams it occurs to me that this problem may originate in a course section studying 2-port networks. If that is the case there may be some benefit to investigating the so-called ABCD parameters.

I won't offer any further hints until @Drako Amorim returns to the thread to confirm.

This problem came from the twelfth edition Sears & Zemansky Electromagnetism, is one of the challenging problems. I think that I solved it and I will transcribe it, but I think that problem is unfair, because there's no 2-port network in the main text
 
Drako Amorim said:
This problem came from the twelfth edition Sears & Zemansky Electromagnetism, is one of the challenging problems. I think that I solved it and I will transcribe it, but I think that problem is unfair, because there's no 2-port network in the main text
Well there are several possible approaches to solving the problem as given. The 2-port network approach occurred to me as possibly being intended and likely comparatively efficient.
 
Redraw the Delta circuit for a Pi

1.jpg


And I conclude that:

Vac = (q1 - q3)/Cy
Vbc = (q2 + q3)/Cx

Vab = q3/Cz = Vac + Vcb = Vac - Vbc

q3/Cz = (q1 - q3)/Cy - (q2 + q3)/Cx

q3 (Cy Cz + Cx Cz + Cx Cy)/(Cx Cy Cz) = q1/Cy - q2/Cx

∴q3 = Cx Cy Cz / (CyCz + CxCz + Cx Cy) [q1/Cy - q2/Cx] ≡ K (q1/Cy - q2/Cx)

∴ ( I ): Vac = q1/Cy - K q1/Cy2 + K q2/(Cx Cy)
( II ): Vbc = q2/Cx + K q1/(CyCx) - K q2/(Cx2)

In the second circuit:

2.jpg


( III ): Vac = V1 + V3 = q1(1/C1 + 1/C3) + q2/C3

( IV ): Vbc = V2 + V3 = q1 (1/C3) + q2 (1/C2 + 1/C3)

Equalizing the coefficients:

( III ) = ( I ):

  • 1/C3 = K/(Cx Cy) → C3 = (Cx Cy + Cy Cz + Cz Cx)/Cz
  • 1/C1 + 1/C3 = 1/Cy - K/Cy2 → C1 = (Cx Cy + Cy Cz + Cz Cx)/Cx
( VI ) = ( II ):

  • 1/C2 + 1/C3 = 1/Cx - K/Cx2 → C2 = (Cx Cy + Cy Cz + Cz Cx)/Cy
As we wanted to prove.
 
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Looks good.
 

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