<^< Capacitors in Parallel | Course Index | Variable Capacitors >^>
Example 1:
Capacitors of 1µF and 2.2µF are connected in series.
Determine the effective capacitance of the circuit.
Using "product over sum" gives:
C = (C1 x C2) / (C1 + C2)
thus C = (1 x 2.2) / (1 + 2.2) µF
therefore C = 2.2 / 3.2 = 0.69µF
Example 2:
Capacitors of 2.2µF, 3.3µF and 4.7µF are connected in parallel.
Determine the effective capacitance of the circuit.
The effective capacitance will be given by:
C = C1 + C2 + C3 = 2.2 + 3.3 + 4.7 = 10.2 µF
<^< Capacitors in Parallel | Course index | Variable Capacitors >^>
