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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» Correct Answer - (i)` l = 8A`,(ii)` 0.2 s`,(iii)`7.4A` 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` |
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