When two capacitors are connected in series and connected across 4kV line, the energy stored in the system is 8 J. the same capacitors, if connected in parallel across the same line, the energy stored in 36 J. find the individual capacitances.

Explanation

V=4kV=4000 V
In series connection ,energy stored =`(1)/(2)C,V^(2)`
i.e ,`8=(1)/(2)C_(s)xx(4000)^(2)`
or `C_(s)=(16)/(16xx10^(6))`

`C_(s)=1muF`
In parallel connection energy stored =`(1)/(2)C_(P)V^(2)`
i.e., 36 =`(1)/(2)C_(P)(4000)^(2)`
`C_(P)=(72)/(16xx10^(6))`
`C_(p)=4.5 muF`
If `C_(1)` and `C_(2)` are the capacitances of the two capacitors in parallel connection `C_(1)+C_(2)=C_(p)`
i.e., `C_(1)+C_(2)=4.5`......(1)
In series connection `(1)/(C_(1))+(1)/(C_(2))=(1)/(C_(s))`
i.e., `(C_(1)_C(2))/(C_(1)C_(2))=(1)/(C_(s))`
i.e., `(4.5)/(C_(1)C_(2))=(1)/(1)`
`C_(1)C_(2)=4.5`.........(2)
Using the relation `(C_(1)-C_(2))^(2)=(C_(1)+C_(2))^(2)-4C_(1)C_(2)`
`(C_(1)-C_(2))^(2)=(4.5)^(2)-4xx4.5`
`(C_(1)-C_(2))^(2)=20.25-18`
`(C_(1)-C_(2))^(2)=2.25`
`C_(1)-C_(2)=1.5` ............(3)
Adding (1)and (3)
`2C_(1)=6`
C_(1)=3muF`
(3) `implies 3-C_(2)=1.5`
`C_(2)=1.5 muF`