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

Answer

A

Explanation

`Q = a t - bt^(2)`
`I = (dQ)/(dt) = a-2bt, i=0 at t = a/(2b)`
`H = int_(0)^(t=(a)/(ab)) i^(2)Rdt = R int_(0)^(a/2b) (a-2bt)^(2) dt`
`= R int_(0)^(a/2b) (a^(2)-4abt+4b^(2)t^(2))dt`
`R|a^(2)t - 4ab(t^2)/(2)+(4b^2)t^3))/(3)|_(0)^(a/2b)`
`=R[a^(2) . a/(2b) -(4ab)/(2) * (a^2)/(4b^2) +(4b^2)/(3) * (a^3)/(8b^3)]`
`=(a^3R)/(b) [1/2 - 1/2 + 1/6] = (a^3R)/(6b)`.