Question

The figure below shows three capacitors of capacitance 3µF, 2µF, and 6µF. Connected to a 12v supply circuit.

Calculate:

1. The total capacitance of the circuit. 
2. The total charge stored in the circuit. 
3. The potential differences across 2µF capacitor.

Answer 1

1. CP=2+6=8µf
   CT=3x8 = 24 = 2.18 µf.
         3+8    11
2.  Q =CV 
   = 2.18 X 10^-6 X 12
   = 2.616 X 10^-5 C
3. p.d across 2 µf = 2.616 x 10^-6
                                  8 x 10^-6
   = 3.27V