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 -b t^(2)`
`i=a-2b t" " { "for "i=0" " t=(a)/(2b)}`
From Joules law of heating
`dH= i^(2)Rdt=H= int_(0)^(a//2b)(a-2b t)^(2)Rdt`
`H=[((a-2b t)^(3)R)/(-3(2b))]_(0)^(a//2b)=(a^(3)R)/(6b)`