Difficult Introductory Electricity Circuit Problem - Capacitors

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Homework Help Overview

The discussion revolves around a challenging introductory electricity circuit problem involving capacitors. Participants are exploring the configuration of capacitors and their relationships in a circuit that resembles a Wheatstone bridge.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss rearranging the circuit and applying relevant equations like Q = C V. There is a focus on the difficulty of using standard methods due to the unique arrangement of capacitors, which are not in simple series or parallel configurations. Questions about the nature of the capacitors' values and their impact on calculations are also raised.

Discussion Status

Some participants have offered equations and suggested approaches, while others express uncertainty about their methods and seek clarification. There is an ongoing exploration of different interpretations of the problem, particularly regarding the implications of having different capacitance values.

Contextual Notes

Participants are navigating constraints related to the complexity of the circuit and the lack of straightforward series or parallel relationships among the capacitors. The discussion includes considerations of whether the capacitors have the same or different capacitances.

Emc2fma
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Difficult Introductory Electricity Circuit Problem -- Capacitors

Homework Statement



MAvH81u.png


The Attempt at a Solution



This problem was very difficult, and the most I could do was re-arrange the circuit into a make-shift Wheatstone bridge with all capacitors instead of resistors/voltmeter.

Any ideas?
 
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What you need is a relevant equation ! Hence the template...

I propose Q = C V There are five of them, plus one for the equivalent over-all C, so you should get enough material to solve for the latter.

That's how we do parallel, series, etc. of two capacitors. This one is a little more involved.
 
I've already attempted doing that - I think the big problem is that since no two of them are in series/parallel with each other, it's very difficult to use the standard method.
 
Show what you did, it might be just right, only a bit confusing
 
I scrapped my approach because it wasn't working.

Do you mean I should calculate the charge on the capacitors using Q = VC?
 
Emc2fma said:

Homework Statement



[ IMG]http://i.imgur.com/MAvH81u.png[/PLAIN]

The Attempt at a Solution



This problem was very difficult, and the most I could do was re-arrange the circuit into a make-shift Wheatstone bridge with all capacitors instead of resistors/voltmeter.

Any ideas?
Are the capacitances all different, or do all capacitors have the same capacitance?
 

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They're all different...does that make a difference? I figure the calculation would be the same.
 
Emc2fma said:
They're all different...does that make a difference? I figure the calculation would be the same.
The special case of all being equal allows you to make a drastic simplification right from the start, basing this on symmetry of the circuit. For this reason, the special case cannot qualify as a "challenge problem". :smile:
 
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Emc2fma said:

Homework Statement



MAvH81u.png


The Attempt at a Solution



This problem was very difficult, and the most I could do was re-arrange the circuit into a make-shift Wheatstone bridge with all capacitors instead of resistors/voltmeter.

Any ideas?

I just write the KCL equations for complicated questions like this.
 

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