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In the figure shown , after the switch ...

In the figure shown , after the switch 's' is turned from postion 'A' to postion 'B' the energy dissipated in the circuit in terms of capactance 'C' and total charge 'Q' is :

A

`3/8 Q^(2)/C`

B

`5/8 Q^(2)/C`

C

`1/8 Q^(2)/C`

D

`3/4 Q^(2)/C`

Text Solution

Verified by Experts

The correct Answer is:
A
Energy dissipated when switch is thrown from 1 to 2.

Q = CV
`V. = (CV + 0)/(C + 3C) = V/4`
`H =1/2 CV^(2) -[1/2 C xx (V/4)^(2) + 1/2 3C (V/4)^(2)]`
`H = 1/2 CV^(2)[1 -{1/16 + 3/16}]`
`=1/2CV^(2) xx 3/4 = 3/8CV^(2) = 3/8 xx Q^(2)/C = 3/8 Q^(2)/C`
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