I am stuck in a problem where we are supposed to calculate the potential difference across each capacitor after the switch S has been closed, but I am uncertain how to calculate the equivalent capacitance.

I understand that the two pairs of capacitors are connected in parallel, and the pair themselves are connected in series when S is open. What does the circuit look like when S is closed? Thanks in advance!

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    \$\begingroup\$ Capacitance between A and B when S is closed is 4.5uF. There are like two 9uF cap in series, one between A and DC and one between DC and B. \$\endgroup\$
    – user208862
    Commented Apr 25, 2021 at 12:51

1 Answer 1


Yes, @michal-podmanicky is right, from a static point of view the result is 4.5uF. When you short circuit the switch (status = closed, switch ideal) you have two parallel cap = 6+3 uF, and then each group of 2 is in series : 9//9 = 9/2 = 4.5uF.

Maybe you wanted to know what happens during the transient of closing the switch? I ask because you speak of potential difference and not equivalent capacitance. I this case you need to provide the external circuit (e.g. voltage source and equivalent resistance) and the switch resistance. It can be solved as a Wheatstone bridge, where capacitors are in place of the resistors you usually see.


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