- JEE Main 2019: Two identical capacitors, each of capacitance C, are charged to a potential difference V and then connected in parallel. The total energy stored in the system is:
(A) CV2/2 (B) 2CV2/3 (C) 3CV2/2 (D) 4CV2/3
Answer: (C)
Explanation: When the capacitors are connected in parallel, the equivalent capacitance is C + C = 2C. The total energy stored in the system is given by U = (1/2)CV2 + (1/2)CV2 = 3CV2/2.
- JEE Main 2018: A capacitor of capacitance C is charged by connecting it to a battery of emf V. After charging it is disconnected from the battery. A dielectric slab of dielectric constant K is now introduced between the plates of the capacitor. The work done in the process of inserting the slab is:
(A) (1 - 1/K)CV2 (B) (1 - 1/K2)CV2 (C
