A parallel combination of a 1.51 μ F 1.51 μF capacitor and a 2.99 μ F 2.99 μF capacitor is connected in series to a 4.41 μ F 4.41 μF capacitor. This three‑capacitor combination is connected to a 16.5 V 16.5 V battery. Find the charge of each capacitor.

Explanation:

combined capacitor of parallel capacitors

1.51 + 2.99

= 4.5 μF

Resultant capacitance of all capacitor

= 4.5 x 4.41 / ( 4.5 +4.41) μF

= 2.22 μF

Charge on series capacitor

= C X V

= 16.5 X 2.22  μC

= 36.63   μC

Potential on series capacitor (4.41 μF)

= Q / C

= 36.63 / 4.41

= 8.3 V

Potential on each of parallel capacitor

= 16.5 - 8.3

= 8.2 V

Charge on 2.99 μF capacitor

= 2.99 x 8.2 = 24.52 C

Charge on 1.51 μF capacitor

= 1.51 x 8.2 =12.38 C

=