JEE MAIN 2019 SAMPLE QUESTION.
Two capacitor with capacitance c1 and c2 are charged to a potential v1 and v2 respectively. When they are connected in parallel the ratio of their respective charges
Answers
Explanation:
) Suppose C1, C2 are capacities of 2 condensers charged to potentials V1 and V2 respectively. Then total charges before sharing= C1V1 + C2 V2
If V is the common potential on sharing charges, then total charge after sharing= C1V + C2 V = ( C1 + C2 ) V
As no charge is lost in the process of sharing, therefore:-
( C1 + C2 ) V = C1V1 + C2 V2
So, common potential is given by :- V= C1V1+C2V2/C1+ C2 ......(1)
(b) As by formula :- Charge is given by :- Q = C V
So, charge on first capacitor is given by :- Q1 = C1V1
and charge on second capacitor is given by :- Q2 = C2 V2
(c) As energy stored in capacitor is given by E = 1/2CV2
Total energy before sharing charges, E1=1/2C1V12+ 1/2C2V22 .............(2)
Total energy after sharing charges, E2 = 1/2( C1+ C2) V2 .............(3)
Putting value of common potential V from equation (1)in equation (3) :-
we get E2 = (C1V1+C2V2 )'2/2( C1 + C2 )
Calculating E1 - E2 = C1C2/(V1−V2)22(C1+C2)= Total energy of the system of two condensers after connection.
which is positive
i.e., E1 - E2 > 0 or E1 > E2 or E2 < E1.
This proves that total energy after contact is somewhat less than total energy before contact.
Answer:
Two capacitor with capacitance c1 and c2 are charged to a potential v1 and v2 respectively. When they are connected in parallel the ratio of their respective charges