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(D) The charge is given by $Q(t) = \alpha t - \beta t^2 + \gamma t^3$.The current $i$ is the rate of change of charge with respect to time: $i = \frac

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$\alpha - \frac{\beta^2}{3\gamma}$

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(D) The charge is given by $Q(t) = \alpha t - \beta t^2 + \gamma t^3$.The current $i$ is the rate of change of charge with respect to time: $i = \frac{dQ}{dt} = \alpha - 2\beta t + 3\gamma t^2$.To find the minimum current,we find the derivative of current with respect to time and set it to zero: $\frac{di}{dt} = -2\beta + 6\gamma t = 0$.Solving for $t$,we get $t = \frac{2\beta}{6\gamma} = \frac{\beta}{3\gamma}$.Now,substitute $t = \frac{\beta}{3\gamma}$ back into the expression for current:$i_{min} = \alpha - 2\beta \left(\frac{\beta}{3\gamma}\right) + 3\gamma \left(\frac{\beta}{3\gamma}\right)^2$$i_{min} = \alpha - \frac{2\beta^2}{3\gamma} + 3\gamma \left(\frac{\beta^2}{9\gamma^2}\right)$$i_{min} = \alpha - \frac{2\beta^2}{3\gamma} + \frac{\beta^2}{3\gamma}$$i_{min} = \alpha - \frac{\beta^2}{3\gamma}$.

The charge flowing in a conductor changes with time as $Q(t) = \alpha t - \beta t^2 + \gamma t^3$. Where $\alpha, \beta$,and $\gamma$ are constants. The minimum value of current is:

Similar Questions

The charge flowing in a conductor varies with time as $Q = at - bt^2$. Then for current,which statement is incorrect?
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$(B)$ reaches a maximum and then decreases
$(C)$ falls to zero after time $t = a/2b$
$(D)$ changes at a rate $-2b$

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