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The charge flowing through a resistance ...

The charge flowing through a resistance R varies with time t as `Q=at-bt^(2)`, where a and b are positive constants The total heat produced in R is -

(A) `(a^3R)/(3b)`
(B) `(a^3R)/(2b)`
(C) `(a^3R)/(b)`
(D) `(a^3R)/(6b)`

A

`(a^(3)R)/(3b)`

B

`(a^(3)R)/(2b)`

C

`(a^(3)R)/(b)`

D

`(a^(3)R)/(6b)`

Text Solution

Verified by Experts

The correct Answer is:
D
Current passing through resistance R ,
`I=(dQ)/(dt)=a-2bt`
Now , I will be zero when `a-2bt=0" or, " t=(a)/(2b)`
`therefore ` Total heat produced in R
`= int_(0)^(t)=I^(2)Rdt=int_(0)^((2)/(2b))(a-2bt)^(2)Rdt`
`=int_(0)^((a)/(2b))(a^(2)R+4b^(2)Rt^(2)-4abRt)dt`
`=[a^(2)Rt+(4b^(2))/(3)Rt^(3)-(4abR)/(2)t^(2)]_(0)^((a)/(2b))=(a^(3)b)/(6b)`
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