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

Answer

A

Explanation

`Q = at - bt^(2) or I = dQ// dt = a - 2 bt`
`I = 0 for t = t_(0) = a// 2b`, i.e, current flows from `t = 0 to t = t_(0)`. The
heat produced is ` int_(0)^(t _(0)) i^(2) R dt = a^(3) R // 6b`.