How Do You Calculate the Equivalent Capacitance in This Complex Circuit?

[fccniy]

OOO|---C5----|
OOO|OOOOOOOOO|
A----C1---C2---C3----B
OOOOOOOO|OOOOOOOOOO|
OOOOOOOO|----C4----|

The "C"s stand for capacitor.
The capacitance of C2 is 10 μF and others are 4 μF.

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The book contains neither the answer nor solution.

I tried to apply Kirchhoff's Loop Rule:
q_4 / 4 + q_2 /10 + q_3 / 4 = 0
q_1 / 4 + q_2 /10 + q_5 / 4 = 0
q_1 + q_4 + q_3 + q_5 = 0


Then I got the equivalent capacitance=0. What is the problem with my argument?

 

[OlderDan]

My guess is you have a sign problem. It's not obvious to me that you have assumed one sign for the voltage drop across each capacitor and used it consistently. In fact, it looks very much like you have not. There must be some additional equations based on charge conservation.

 

[kyle8921]

There are a few issues with your argument. First, you have not taken into account the fact that capacitors in series have an inverse relationship, while capacitors in parallel have an additive relationship. In this circuit, C1, C2, and C3 are in series, while C4 and C5 are in parallel. This means that the equivalent capacitance will be different for these two sets of capacitors.

Secondly, your equations seem to be missing some key information. The Kirchhoff's Loop Rule only applies to circuits with resistors, not capacitors. In order to find the equivalent capacitance, you will need to use the formula 1/Ceq = 1/C1 + 1/C2 + 1/C3 + 1/C4 + 1/C5.

Lastly, it is important to note that the value of the equivalent capacitance will not be 0. It will be a non-zero value, as there are capacitors in the circuit with non-zero values. It is possible that your calculations may have resulted in an error, which is why you are getting a value of 0.

To correctly find the equivalent capacitance, you will need to apply the formula mentioned above and take into account the relationships between capacitors in series and parallel. It may also be helpful to redraw the circuit to better visualize the relationships between the capacitors.