A 9.0 μF and 4.0 μF capacitors are connected in parallel

[mikebc]

A 9.0 μF and 4.0 μF capacitors are connected in parallel...

Hi, I understand how to do most of this question, I completed "A", but I am confused with question "B". I think that the series capacitor would have full voltage going across it and that the parallel capacitors would have less voltage. But since resistors work oppositely then capacitors and this is the same way to work out resistor voltage I am wondering if I am working it out wrong. Any help would be appreciated.

Question
A 9.0 μF and 4.0 μF capacitors are connected in parallel, and this combination is
connected in series with a 12.0 μF capacitor.
a. What is the net capacitance?
b. If 32 V is applied across the whole network, calculate the voltage
across each capacitor.

I have attached a scan of how I think that it works.

Thanks!

 

[mikebc]

Oops, I forgot to add attachment...

Here is the scan.

 

[Vijay Bhatnagar]

Equivalent capacitance = 6.24 micro F. From Q = CV we get charge Q drawn from the battery = 6.24 x 10^-6 x 32 = 199.68 X 10^-6 C.

When capacitors are in series, the charge on each capacitor is same. 13 micro F capacitor and 12 micro F cap are in series. Hence, Q the charge on each of these is same (199.68 X 10^-6 C).

V = Q/C. Hence voltage across 12 micro F cap = 199.68 X 10^-6/12 x 10^-6 = 16.64 V

Voltage across 13 micro F cap = 199.68 X 10^-6/13 x 10^-6 = 15.36 V

As the voltage across capacitors in parallel is same, voltage across each of 9 & 4 micro F caps will be 15.36 V.

 

[mikebc]

Thank you for explaining that so clearly for me. Some concepts my textbook does not clearly explain, or it is just expected to be understood. Cheers!