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What is the maximum number of 5.50 uF capacitors that can be connected in parallel with a 6.00 V battery while keeping the total charge stored within the capacitor array below 942 UC? Number If the same number of 5.50 HF capacitors are connected in series with the same 6.00 V battery, how much charge will end up on each capacitor plate? Number

Answer 1

(a) 28

c = q / v =  942uc / 6 = 157 uf

in parallel .. thus capacitance add up .. for max number of capacitors in parallel = 157 /5.5 =28.54

so for keeping .. charge less than 942 ..maximum number of capacitors = 28

(b) 1.1785 uc

q = c*v = 5.5uf /28 * 6 = 1.1785 uc.

Answer 2

let n capacitors are connected in parallel, therefore c=n*5.50uF

charge=c*v;

=n*5.5uF*6V =942 uC

n=171

capacitors in series c=5.5/n uF

charge=6V*5.5/171 uF =0.19298 uC