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)`
By the definition of current `I = (dQ)/(dt)`
`I =a- 2bt` (for `I = 0 rArr t=a/(2b)`)
From Joule.s law of heating `dH = i^(2)Rdt`
`H = int_(0)^(a//2b) Rdt, H =((a-2bt)^(3)R)/(-3 xx 2b) = (a^(3)R)/(ab)`