Question

Two capacitors are connected parallel to each other and connected to the battery with voltage 25 V . Let C1 = 9.5 μF ,C2 = 5.0 μF be their capacitances. Suppose the charged capacitors are disconnected from the source and from each other, and then reconnected to each other with plates of opposite sign together.By how much does the energy of the system decrease?

Answer 1

Value of the two capacitors -

C1 = 9.5 μ F

C2 = 5.0 μ F

Initially the two capacitors are in parallel.

So, its equivalent capacity = C1 + C2 = (9.5 + 5) μ F = 14.5 μ F

Therefore, the initial stored energy = 0.5 C V^2 = 0.5 *14.5 x 10^-6 *25^2 = 4.53 x 10^-3 J

And -

q1 = c1 V = 9.5 x 10^-6 x 25 = 2.37 x 10^-4 C

q2 = c2 V = 5 x 10^-6 x 25 = 1.25 x 10^-4 C

So,

q1- q2 = 1.12 x 10^-4 C is the total charge in both the capacitors after disconnected and joined oppositely.

And, C = 14.5 μ F

Energy stored in both = 0.5 q² / C = 0.5 (1.12 x 10^-4) ² / (14.5 x 10^-6) = 4.32 x 10^-4 J

Therefore, decrease in the energy of the system = 4.53 x 10^-3 J - 4.32 x 10^-4 J

= 4.098 x 10^-3 J (Answer).