Find the charges on the three capacitors connected to a battery as shown in figure.
Take `C_1 = 2.0 uF , C_2 = 4.0 uF , C_3 = 6.0 uF and V` = 12 volts.
Solution
The capacitances of three capacitors are C1= 2 μF, C2= 4 μF and C3= 6 μF and the voltage of the battery (V) is 12 V.
As the capacitors are connected in parallel, the equivalent capacitance is given by
Ceq = C1 + C2 + C3
= `(2+4+6) uF = 12 uF = 12 xx 10^-6 "F"`
Due to parallel connection, the potential difference across each capacitor is the same and is equal to 12 V.
Therefore, the charge on each capacitor can be calculated as follows :
The charge on the capacitor of capacitance C1= 2 μF is given by
`Q_1 = C_1V = (2 xx 10^-6) xx 12 "C" = 24 xx 10^-6 "C" = 24 "uC"`
Similarly, the charges on the other two capacitors are given by
`Q_2 = C_2V = (4 xx 10^-6) xx 12 "C" = 48 xx 10^-6 "C" = 48 "uC"`
and
`Q_3 = C_3V = (6 xx 10^-6) xx 12 "C" = 72 xx 10^-6 "C" = 72 "uC"`
