1.

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

Answer»

`(CV^2)/(12)`
`(7CV^2)/(8)`
`(5CV^2)/(4)`
`(3CV^2)/(8)`

Solution :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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