The charge flowing through a conductor varies with time as
`q= 8 t - 3t^(2)+ 5t^(3)`
Find (i) the initial current (ii) time after which the current reaches a maximum value of current

Answer

(i)` l = 8A`,(ii)` 0.2 s`,(iii)`7.4A`

Explanation

Given `q = 8 t - 3t^(2) + 5t^(3)`
(i) current `I = (dq)/(dt) = 8 - 6t + 15t^(2)`…..(i)
When `t = 0 I = 8A`
(ii) `(dI)/(dt) = - 6+30t`
For `l` to the maximum or minimum
`(dI)/(dt) =0 or - 6 + 30 t = 0 or t = 0.2 s`
(iii) putting this value of `t` in (i) we get
`I = 8 - 6 xx 0.2 + 15(0.2)^(2) = 8 - 1.2 + 0.6 = 7.4 A`
As this value of `l` is less than at `t = 0`, it will be minimum value , so minimum value of current `= 7.4A`