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In the circuit what is the change of to...

In the circuit what is the change of total electrical energy stored in the capacitors when the key is pressed ?

A

`(CV^2)/(12)`

B

`(7CV^2)/(8)`

C

`(5CV^2)/(4)`

D

`(3CV^2)/(8)`

Text Solution

Verified by Experts

The correct Answer is:
C
c. Initially, all three capacitors are in parallel
`E_i = 1/2 xx 3CV^2 = 3/2 CV^2`
When key is closed, two capacitors are in series.
`E_f = 1/2 C (V/2)^2 xx 2 = (CV^2)/4`
`Delta E = E_i - E_f = 5/4 CV^2` .
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