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

(a) `Q = at - bt^(2)`
`I = a - 2 bt { f o r I = implies t = (a)/(2b)}`
From Joule's law of heating
`dH = i^(2) H dt`
`H = int_(0)^(a//2b) (a - 2 bt)^(2) R dt`
`= ((a - 2 bt)^(3) R)/(- 3 xx 2b) |_(0)^((a)/(2b)) = (a^(3) R)/( 6b)`