Two capacitors with capacity C1 and C2 are charged to potential V1 and V2 respectively and then connected in parallel. Calculate the common potential across the combination, the charge on each capacitor, the electrostatic energy stored in the system and the change in the electrostatic energy from its initial value.

_{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)

2. Charge on each capacitor after sharing of charges is given by :-

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

_{1}= C

_{1}V

and charge on second capacitor is given by :- Q

_{2}= C

_{2}V

where V is the common potential given by eq. 1

3.As energy stored in capacitor is given by E = 1/2Cv

^{2}

Total final electrostatic energy stored in the system of two capacitors, when both are connected in parallel combination is

E

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

putting value of V from eq 1 in above equation, we get :

E

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

4. Initial electrostatic energy before sharing of charges in both the conductors is :

E

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

Change in electrostatic energy is given by :

$\Delta E={E}_{1}-{E}_{2}\phantom{\rule{0ex}{0ex}}onsolving,weget:\phantom{\rule{0ex}{0ex}}\Delta E=\frac{{C}_{1}{C}_{2}({V}_{1}-{V}_{2}{)}^{2}}{2({C}_{1}+{C}_{2})}$

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