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I can't solve this simple circuit, made of 5 capacitors.I know i only have to use parallel and series rules. The solution in the book is $4/3$ C

Community
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tommycautero
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2 Answers2

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This is a balanced Wheatstone Bridge. Hence no current will flow through the $3C$ capacitor. Thus you can simply remove the central wire and have a system of the $2$ branches on the top and on the bottom. $$$$The $2$ capacitors in the top branch are in series, and hence have a net capacitance of $\dfrac{1}{C_{net, top}}=\dfrac11+\dfrac12=\dfrac32C\Rightarrow C_{net,top}=\dfrac23C$.$$$$ Similarly the bottom branch has a capacitance of $C_{net,bottom}=\dfrac23 C$. $$$$Now the two branches on the top, and on the bottom, are in parallel, and hence have a net capacitance of $C_{net,circuit}=C_{net,top}+C_{net,bottom}=\dfrac23+\dfrac23=\dfrac43 C$

User1234
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  • Thank you very much! But how can one deduce that a capacitor(like the 3C one here) doesn't contribute? Is there a general rule or something? – tommycautero Apr 26 '18 at 09:01
  • Yes, it is a general rule for balanced Wheatstone Bridges. – User1234 Apr 26 '18 at 09:02
  • Sorry, I've got to go to the doctor right now. I'll attach the link for Wheatstone bridges for Capacitors later. However it's a very similar concept used for Wheatstone Bridges for Resistors. You can read up on them here:https://www.electronics-tutorials.ws/blog/wheatstone-bridge.html – User1234 Apr 26 '18 at 09:06
  • thanks, good luck with the doctor, you've been my hero! – tommycautero Apr 26 '18 at 09:08
  • The essential principle is that in a BALANCED Wheatstone Bridge, there is NO CURRENT through the 'central branch'. – User1234 Apr 26 '18 at 09:08
  • Thanks for your wishes too!:) – User1234 Apr 26 '18 at 09:09
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It's similar to working with resistances but the other way round as seen here. Try and tackle it by "collapsing" the circuit bit by bit replacing every two capacitors with one that has the same capacitance as the configuration of the two.

The capacitor in the middle has no potential difference across it (due to the symmetry of the circuit along the horizontal axis) so it does not contribute anything to the circuit.

Jepsilon
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