Two capacitors C1 and C2 are charged to potential V1 and V2 resp. and then connected in parallel. Calculate common potential, charge on each capacitor and energy in the system after connection??

_{1}

_{, }C

_{2}

_{ }are capacities of 2 condensers charged to potentials V

_{1}and V

_{2}

_{ }respectively. Then total charges before sharing= C

_{1}V

_{1 + }C

_{2}

_{ }V

_{2}

_{ }

If V is the common potential on sharing charges, then total charge after sharing= C

_{1}V + C

_{2}

_{ }V = ( C

_{1}

_{ + }C

_{2}

_{ }) V

As no charge is lost in the process of sharing, therefore:-

( C

_{1}

_{ + }C

_{2}

_{ }) V = C

_{1}V

_{1}

_{ + }C

_{2}

_{ }V

_{2}

_{ }

So, common potential is given by :- V= $\frac{{C}_{1}{V}_{1}+{C}_{2}{V}_{2}}{{C}_{1}+{C}_{2}}$ ......(1)

(b) As by formula :- Charge is given by :- Q = C V

So, charge on first capacitor is given by :- Q

_{1}

_{ = }C

_{1}V

_{1}

and charge on second capacitor is given by :- Q

_{2 = }C

_{2}

_{ }V

_{2}

(c) As energy stored in capacitor is given by E = $\frac{1}{2}C{V}^{2}$

Total energy before sharing charges, E

_{1}=$\frac{1}{2}{C}_{1}{{V}_{1}}^{2}+\frac{1}{2}{C}_{2}{{V}_{2}}^{2}$ .............(2)

Total energy after sharing charges, E

_{2}= $\frac{1}{2}({C}_{1}+{C}_{2}){V}^{2}$ .............(3)

Putting value of common potential V from equation (1)in equation (3) :-

we get E

_{2}

_{ }= $\frac{({C}_{1}{V}_{1}+{C}_{2}{V}_{2}{)}^{2}}{2({C}_{1}+{C}_{2})}$

Calculating E

_{1}

_{ }- E

_{2}

_{ }= $\frac{{C}_{1}{C}_{2}({V}_{1}-{V}_{2}{)}^{2}}{2({C}_{1}+{C}_{2})}$= Total energy of the system of two condensers after connection.

which is positive

i.e., E

_{1}- E

_{2}> 0 or E

_{1}> E

_{2}or E

_{2}< E

_{1}.

This proves that total energy after contact is somewhat less than total energy before contact.

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