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Initially capacitor is kept connected to...

Initially capacitor is kept connected to battery of EMF V for substantial period of time. Then suddenly the battery is removed and terminals X Y are connected to ends of a battery of EMF `(V)/(2)` as shown. What is the heat loss in the circuit after new battery is connected?

A

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

B

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

C

`(5CV^(2))/(8)`

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
`W_("batt")=Delta u +" Heat loss"`
`(V)/(2)((3CV)/(2))=[(1)/(2) C((V)/(2))^(2)-(1)/(2)CV^(2)]+" Heat loss"`
Heat loss `=(3CV^(2))/(4) +(3)/(8) CV^(2) =(9)/(8) CV^(2)`
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