InterviewSolution
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InterviewSolution
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The charge flowing in a conductor varies time as, `q=at - 1/2 bt^(2) +1/6 ct^(3)` Where a,b,c are positive constants. Then, find (i) the initial current (ii) the time after which the value of current reaches a maximum value (iii) the maximum or minimum value of current. |
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Answer» (i) Current, `i=(dq)/(dt)` `=d/dt(at - 1/2 bt^(2) +1/6 ct^(3))` `=a-bt +1/2 ct^(2)` When t =0, initial current , i=a (ii) For i to be maximum or minimum, `(di)/(dt)=0= - b +ct` or `t =b/c` Putting this value of t in (i), we have `i=a - b xxb/c + 1/2 c xx b^(2)/c^(2) = a - b^(2)/c + b^(2)/(2c) = a - b^(2)/(2c)` As this value of i is less than that at t=0, it must be minimum. So minimum value of current `a= - (b^(2))/(2c)` |
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