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^(2)R)/(27b)`
C. `(a^(3)R)/(3b)`
D. None of these

Answer

A

Explanation

`I=(dQ)/(dt)=a-2bt`
`I=0at t=(a)/(2b)`
`therefore"Heat"=underset(0)overset(a/(2b))intI^(2)Rdt=underset(0)overset(a/(2b))int(a-2bt)^(2)Rdt`
`=R[a^(2)t+(4b^(2)t^(3))/(3)-(4abt^(2))/(2)]^(a/(2b))`
`=R[(a^(2)xxa)/(2b)+(4b^(2)xxa^(3))/(24b^(3))-(4aba^(2))/(8b^(2))]`
`=R[(a^(3))/(2b)+(a^(3))/(6b)-(a^(3))/(2b)]=(Ra^(3))/(6b)`