Charge on a capacitor in a series

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acdurbin953
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Homework Statement



Circuit diagram with a battery V on the left, capacitors C1 on the top wire and C5 on the bottom wire, and C2 and C3 in series with C4 in parallel between nodes b and a; nodes c, b and a are labeled.

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A circuit is constructed with five capacitors and a battery as shown. The values for the capacitors are: C1 = C5 = 3.9 μF, C2 = 3.9 μF, C3 = 5.7 μF, and C4 = 4.2 μF. The battery voltage is V = 12 V.

What is Q5, the charge on capacitor C5?

Homework Equations


C=Q/V

The Attempt at a Solution


Okay, I thought that voltage across series capacitors was all the same across each one. Treating the parallel system as one capacitor of equivalent capacitance, I thought you could just divide the voltage by 3, and then multiply by C5 to get the charge on C5. I tried this and the answer is wrong. I'm really confused.
 
acdurbin953 said:

Homework Statement



Circuit schematic with a voltage source V on the left and five capacitors labeled C1 through C5, with nodes labeled a, b and c. C1 is in the top wire, C2 and C3 are in series in a vertical branch between nodes b and a, C4 is in a parallel branch on the right, and C5 is in the bottom wire.

[/B]
A circuit is constructed with five capacitors and a battery as shown. The values for the capacitors are: C1 = C5 = 3.9 μF, C2 = 3.9 μF, C3 = 5.7 μF, and C4 = 4.2 μF. The battery voltage is V = 12 V.

What is Q5, the charge on capacitor C5?

Homework Equations


C=Q/V

The Attempt at a Solution


Okay, I thought that voltage across series capacitors was all the same across each one. Treating the parallel system as one capacitor of equivalent capacitance, I thought you could just divide the voltage by 3, and then multiply by C5 to get the charge on C5. I tried this and the answer is wrong. I'm really confused.

The voltage across capacitors in series is not the same, unless the capacitances are equal. What is the resultant of C2, C3 and C4? Is it equal to 3.9 μF?
I think you mix charge with voltage. The charge is same on series capacitors .