How Do You Find Equivalent Capacitance in This Complex Circuit?

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Homework Help Overview

The discussion revolves around finding the equivalent capacitance in a complex circuit involving multiple capacitors with different configurations, including series and parallel arrangements. The original poster presents a circuit with a 20 V battery and various capacitors, seeking assistance in calculating the total capacitance.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method of combining capacitors in series and parallel, with some questioning the original poster's calculations and assumptions. There are attempts to clarify the steps taken in the calculations, and participants suggest re-evaluating the arrangement of capacitors.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the problem and emphasizing the importance of showing work for clarity. There is recognition of errors in previous calculations, and some participants are exploring different branches of the circuit to determine the best approach.

Contextual Notes

Participants note the complexity of the circuit and the need for careful consideration of the arrangement of capacitors. There is mention of potential confusion regarding the values of certain capacitors and the necessity of redrawing the circuit after each calculation step.

majormaaz
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Homework Statement


In Fig. 26-32, the battery has a potential difference of 20 V.
http://www.webassign.net/hrw/26_32.gif

Homework Equations


If Capacitors are in series, then (1/C1) + (1/C2) = (1/Cnew)
If Capacitors are in parallel, then you just add them, C1 + C2 = Cnew

The Attempt at a Solution


I'm not sure how to get the equivalent capacitance from here.
Would I just add the 12/7 to the 2uF parallel from it to make one capacitor? What happens then to the 2uF originally in series with the 12/7 uF?
This is what I did: Combine the 3 and 4 uF capacs, then the 12/7 and 2 uF(parallel), then that and 3 uF (series) and then that with the remaining 2 uF and 3 uF, in series.
I ended up with 0.696, which is apparently wrong. Is there a correct way how to resolve all these capacitors? PEMDAS for circuits? Any help will be appreciated
 
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Here's a picture showing where I'm stuck at:
 

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Uh ... how did you get 4 in series with 4 = 12/7 ?
 
phinds said:
Uh ... how did you get 4 in series with 4 = 12/7 ?

...Thaaaat appears to be my problem. Thanks! I'll have a go at this again
 
So, I did the work again, fixing that 3 uF into a 4 uF, and I ended up with getting 12/17 uF, but that's not right either.
Could somebody advise me as to which branches/capacitors I should deal with first, and so on?
 
majormaaz said:
So, I did the work again, fixing that 3 uF into a 4 uF, and I ended up with getting 12/17 uF, but that's not right either.
Are you using a calculator?

It is pretty obvious when capacitors are in series, so you might as well start there. (But it doesn't matter where you start.) Re-draw the circuit after each step.

So you have 2x 4µF in series, what equivalent capacitance is that?

I can see a pair of 3µF in parallel.
 
majormaaz said:
So, I did the work again, fixing that 3 uF into a 4 uF, and I ended up with getting 12/17 uF, but that's not right either.
Could somebody advise me as to which branches/capacitors I should deal with first, and so on?

You need to show your work so we can see how it is that you are getting these nonsensical answers

Show step by step how you reduce two 4-uf's in series to a single equivalent cap.
 
I hope this picture helps! It goes from left to right, just in case you didn't know.
EDIT: Sorry, wrong picture
 

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Here we go! My REVISED work.
 

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  • #10
You KEEP not showing your work, just the end results of each step. Let me repeat myself:

You need to show your work so we can see how it is that you are getting these nonsensical answers
 
  • #11
You overlooked my hint? ("I can see a pair of 3µF in parallel")
 

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