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

Answer

c

Explanation

Given charge `= at- bt^(2)`
Current `I = (dQ)/(dt) = a- 2bt`
If the current exits the time `t` then
`0 = a - 2bt or t = a//2b`
Total head produced can be given by
`H = underset(0)overset(t)(int) I^(2)R dt = underset(0)overset(a//2b)(int) (a - 2bt)^(2)R dt`
`=underset(0)overset(a//2b)(int) a^(2) - 4abt + 4b^(2)t^(2)R dt`
`= [a^(20t) - (4bt^(2))/(2)+ (4b^(2)t^(3))/(3)]_(0)^(a//2b) R`
`= a^(2)R//6b`