Question

In fig. `C_(1) = 20 muF, C_(2) = 30 muF and C_(3) = 15 muF` and the insulated plate of `C_(1)` is at a potential of 90 V, one plate of `C_(3)` being earthed. What is the potential difference between th plates of `C_(2)` three capacitors being connected in series ?

Answer 1

we given
`C_(1)=20muF, C_(2)=30 muF, C_(3)=15 muF,V=90V`
`1/C=1/C_(1)+1/C_(2)+1/C_(3)Rightarrow1/C = 1/20 +1/30 +1/15 = ( 3+2+4)/60 = 9/60 = 3/20 Rightarrow 1/C = 3/20 C=20/3 = 6.6 muF`
When the capaciter connected in series, charge on each capacitor is same. Let q be the charge on each capacitor, if ` V_(1),V_(2) and V_(3)` are potential difference across these, capacitors.
` V_(1)+V_(2) +V_(3) = 90`
`or q/(20xx10^(-6)) +q/(30xx10^(-6))+q/(15xx10^(-6)) =90, (3q +2q +4q)/(60xx 10^(-6)) =90, (9q)/(60 xx 10^(-6)) = 90`
`q=(overset(10)cancel(90))/cancel(9)xx60 xx10^(-6)= 600xx10^(-6)= 6.0 xx 10^(-4)C`
Energy stored, `U=1/2CV^(2)=1/(cancel(2))xxoverset(3.0)cancel6.0xx10^(-4)xx30xx30=27000 xx 10^(-4) = 2.7` J