Difficult Introductory Electricity Circuit Problem - Capacitors

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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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  • #2
BvU
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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.
 
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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.
 
  • #4
BvU
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Show what you did, it might be just right, only a bit confusing
 
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I scrapped my approach because it wasn't working.

Do you mean I should calculate the charge on the capacitors using Q = VC?
 
  • #6
SammyS
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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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  • #7
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They're all different...does that make a difference? I figure the calculation would be the same.
 
  • #8
NascentOxygen
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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:
 
  • #9
berkeman
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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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