1.

The circuit is completed at t=0, then heat loss in R=4Omega resistance, till the capacitor gets fully charged.

Answer»

`64 MU J`
`384 mu J`
`320 mu J`
`576 mu J`

Solution :Let `R+2=R_(1)`
`q(t)=(3CE)/2(1-e^(1/(2R_(1)C)))`
`I=(dq)/(dt)=(3E)/(4R_(1))e^(-1/(2R_(1)C))`
HEAT loss in `R_(1)=intI^(2)R_(1)dt`
`=576muJ`
so, heat loss in `R=4Omega` is `=576xx2/3=384muJ`


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