Question

A 2.20-µF and a 4.16-µF capacitor are connected to a 54.0-V battery. What is the total charge supplied to the capacitors when they are wired in the following ways?

(a) in parallel with each other

_________ C

(b) in series with each other

____________C

Answer 1

A) If parallel, add all capacitance...
Total C = C1 + C2
C = 2.20+4.16 = 6.36 µF...
given Q = CV
Q = (6.36x 10^-6) x 54
Q = 3.434 x 10^-4               
Q = 3.434 x 10^-4 coulombs

(B) In series, add the reciprocal of all capacitance...

1/C = 1/C1 + 1/C2
1/C = 1/2.2 + 1/(4.16)
1/C = 0.69493
so, C = 1/0.69493= 1.4389 µF...
Q = CV
Q = 1.4389 x 10^-6 x 54     
Q = 7.77 x 10^-5 Coulombs.