Answer to Question #191698 in Civil and Environmental Engineering for John Prats

Question

Find the equivalent capacitance of a 4.20-µF capacitor and an 8.50-µF capacitor when they are connected (a) in series and (b) in parallel.

Answer 1

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors'

Thus equivalent capacitance in parallel connection=4.20+8.50=12.70µF

When capacitors are connected in series, the total capacitance is given by

1/Ct=(1/C1)+(1/C2)

1/Ct=(1/4.20)+(1/12.70)

1/Ct=0.23+0.078

1/Ct=0.23+0.078=0.30

Ct=3.33µF