Find the equivalent capacitance of the circuit

AI Thread Summary
The discussion focuses on finding the equivalent capacitance between points A and B in a given circuit. Participants emphasize the importance of redrawing the circuit to simplify the analysis, particularly identifying capacitors C1 and C2 as being in parallel. After several iterations and clarifications, it is confirmed that C1 and C2 combine to form C12, which is in series with C4, leading to the final equivalent capacitance of 5C/3. The conversation highlights the necessity of clear circuit representation for accurate problem-solving in circuit analysis. The correct approach was validated through collaborative discussion and verification against textbook answers.
Krushnaraj Pandya
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Homework Statement


Find equivalent capacitance between A and B of the given figure (pardon camera quality)

Homework Equations


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The Attempt at a Solution


marking junctions, From left to right, starting from original point A-and ending at B I wrote potential at them as follows. A-(A)-(x)-(A)-(B)-B. I labelled the capacitor out of line as C4, and moving from left to right, the ones in line as C1, C2 and C3.
I noticed C1 and C2 are connected across same potential difference and therefore are in parallel connection, I don't know where to place their equivalent capacitor though, I'd appreciate some help, thank you
 

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This is a standard circuit (whether with resistors or capacitors) that is used to see if you really understand how to redraw circuits (which is a VERY necessary skill if you are going to learn circuit analysis).

So far, in saying "I noticed C1 and C2 are connected across same potential difference and therefore are in parallel connection", you only have it part right. Keep going.

EDIT: OOPS. I looked at this too quickly. It has one more component than I am used to. Have to relook.

OK, you're right so far, you just have to redraw the circuit then it will become apparent how to simplify.

ALWAYS redraw such circuits otherwise you're likely to just confuse yourself.
 
As you said, C1 and C2 are in parallel. So call that C12 and redraw the circuit.
You will have to do one one capacitor series equivalence and one more capacitor parallel equivalence.
 
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That second lower wire has no use. I decided to replace capacitors with resistor symbols and saw that C4 and ((C2,C3) are in parallel which in turn is in series with C1. I got the answer, verified by my textbook as 5C/3
 
.Scott said:
As you said, C1 and C2 are in parallel. So call that C12 and redraw the circuit.
You will have to do one one capacitor series equivalence and one more capacitor parallel equivalence.
I was confused where to place C12, but this won't be necessary, I can write C234 and place it in series with C1
 
any insights? I did it the correct way-right?
 
Krushnaraj Pandya said:
I was confused where to place C12, but this won't be necessary, I can write C234 and place it in series with C1
##C_{234}## ?
No. I think you're heading down the wrong path.
Do ##C_{12}## first. Then you will have a 3-capacitor circuit.
Next will be ##C_{124}##. Then ##C_{1234}## and you're done.
 
Krushnaraj Pandya said:
I was confused where to place C12, but this won't be necessary, I can write C234 and place it in series with C1
EXACTLY the kind of confusion I was talking about. Stop writing words and REDRAW THE CIRCUIT. That should always be the first thing you do on this kind of problem.
 
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.Scott said:
C234 ?
No. I think you're heading down the wrong path.
Do C12 first. Then you will have a 3-capacitor circuit.
Next will be C124. Then C1234 and your done.
1)C4 is connected across 'D' and 'B'.
2)C2 and C3 are in series
3)C23 is in parallel with C4
4)C234 is then in series with C1.

I have the correct answer, but am I wrong in the steps?
 
  • #10
##C_2## and ##C_3## are not in series.
After you did ##C_{12}## you needed to redraw.
##C_3## goes directly from A to B, so there is nothing in series with it.

Here is you net list:
##C_1##: A to D
##C_2##: A to D
##C_3##: A to B
##C_4##: D to B
 
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  • #11
.Scott said:
##C_2## and ##C_3## are not in series.
After you did ##C_{12}## you needed to redraw.
##C_3## goes directly from A to B, so there is nothing in series with it.

Here is you net list:
##C_1##: A to D
##C_2##: A to D
##C_3##: A to B
##C_4##: D to B
are C12 and c4 in series? I think I got it. but it was barely apparent after redrawing, just the net list provoked the thought
 
  • #12
Got it perfectly! 1) C1 and C2 in parallel from A to D, therefore C12 from A to D.
2)C4 from D to B and C12 from A to D, so in series.
3)C124 from A to B, so in parallel with C3
Finally the correct answer is 5C/3.
Thank you very much for your help.
 
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  • #13
Krushnaraj Pandya said:
Got it perfectly! 1) C1 and C2 in parallel from A to D, therefore C12 from A to D.
2)C4 from D to B and C12 from A to D, so in series.
3)C124 from A to B, so in parallel with C3
Finally the correct answer is 5C/3.
Thank you very much for your help.
And here are the drawings : :smile:

upload_2018-8-8_7-26-59.png
 

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