Question #191698

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.


Expert's answer

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






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