The charge flowing through a resistance `R` varies with time `t as Q = at - bt^(2)`. The total heat produced in `R` is

Options

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

Answer

A

Explanation

`Q=at-bt^(2)`
`i=(dQ)/(dt)=a-2bt`
`i=0 implies a-2bt=0 implies t=a/(2b)`
`H=int_(0)^(a/(2b))i^(2)Rdt=Rint_(0)^(a//2b)(a-2bt)^(2) dt`
`=Rint_(0)^(a//2b)(a^(2)-4abt+4b^(2)t^(2)) dt`
`=R|a^(2)t-2abt^(2)+(4b^(2)t^(3))/3|_(0)^(a//2b)`
`=R|a^(2).a/(2b)-2ab. (a^(2))/(4b^(2))+4/3b^(2).(a^(3))/(8b^(3))|`
`=(a^(3)R)/(6b)`