Question

two capacitors c1=6if and c2=12uf are in series across a 180 volts d.c supply. calculate the charges on c1 and c2

Answer 1

A capacitance of a capacitor means having a charge on one coulomb establishing a potential difference of 1 volt between its plates.

C=Q/V

Therefore, Q = C*V

Where Q is the charges in coulombs, C is the capacitance in Farad and V is the voltage.

Since the capacitors are connected in series, their equivalent will be:

$C_{eq}=(\frac{1}{6uF}+\frac{1}{12uF}) ^{-1}=4uF.$

Note: You can see that the series connected capacitors formula is the same as the parallel connected resistors formula

Therefore, $Q=4*10^{-6}*180=7.2*10^{-4}C.$

Since $C_{2}=2C_{1}$, $Q_{2}=2Q_{1}$ as C and Q are directly proportional.

Therefore, $Q_{1}=2.4*10^{-4}C$ and $Q_{2}=4.8*10^{-4}C$