Question

A 0.50-?F and a 1.4-?F capacitor (C1 and C2, respectively) are connected in series to a 18-V battery.

Calculate the potential difference across each capacitor.

Calculate the charge on each capasitor.

Calculate the potential difference across each capacitor assuming the two capacitors are in parallel

Calculate the charge on each capasitor assuming the two capacitors are in parallel.

Answer 1

The capacitor c1 is 0.5 micro farad and the capacitor c2 is 1.4 micro farad

the voltage applied across the capacitors is V=18 V

a)

The capacitors are connected in series means 1/c=1/c1+1/c2

1/c=1/0.5+1/1.4=1/2.714 micro farad

c=0.368 micro farad

the charge induced by across the capacitors are Q=CV

Q=(0.368 micro farad)(18 V)=6.632 micro coulomb

potential at first capacitor is V1=Q/c1=(6.632 micro coulomb )/(0.5 micro farad)=13.26 V

potential at second capacitor is V2= Q/c2=(6.632 micro coulomb)/(1.4 micro farad)=4.737 V

potential difference across each capacitor is V=V2~V1=13.26-4.737=8.523 V

b)

each charge on each capacitor is,

Q1=c1V=(0.5 micro farad)(8.523 V)=4.261 micro coulomb

Q2=c2V=(1.4 micro farad)(8.523 V)=11.932 micro coulomb

c)

The capacitors are connected in parallel means c=c1+c2

c=0.5+1.4=1.9 micro farad

the charge induced by across the capacitors are Q=CV

Q=(1.9 micro farad)(18 V)=34.2 micro coulomb

potential at first capacitor is V1=Q/c1=(34.2 micro coulomb )/(0.5 micro farad)=68.4 V

potential at second capacitor is V2= Q/c2=(34.2 micro coulomb)/(1.4 micro farad)=24.428 V

potential difference across each capacitor is V=V2~V1=68.4-24.428=43.972 V

d)

each charge on each capacitor is,

Q1=c1V=(0.5 micro farad)(43.972 V)=21.986 micro coulomb

Q2=c2V=(1.4 micro farad)(43.972 V)=61.56 micro coulomb