Question

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

• A. $\alpha-\frac{\gamma^2}{3 \beta}$
• B. $\beta-\frac{\alpha^2}{3 \gamma}$
• C. $\alpha-\frac{\beta^2}{3 \gamma}$
• D. $\alpha-\frac{3 \beta^2}{\gamma}$

Answer 1

The Correct Option is C

The problem involves finding the minimum value of the current derived from the given charge equation. The charge flowing in a conductor changes with time according to the equation:

\(Q(t) = \alpha t - \beta t^2 + \gamma^3\)

To find the current, which is the rate of change of charge with respect to time, we need to differentiate the charge equation \(Q(t)\) with respect to time \(t\):

\(I(t) = \frac{dQ}{dt} = \frac{d}{dt} (\alpha t - \beta t^2 + \gamma^3)\)

Upon differentiating, we get:

\(I(t) = \alpha - 2\beta t\)

To find the minimum value of the current, we observe that the current is linear with time and decreases linearly as \(t\) increases. Thus, finding the minimum current involves setting the derivative of the current \(I(t)\) with respect to \(t\) to zero and solving for \(t\). However, since the term \(\gamma^3\) is not dependent on \(t\), the critical point of this equation where the minimum can occur will be when:

\(I(t) = \alpha - 2\beta t \to t = \frac{\beta}{3 \beta}\)

Substituting \(t = \frac{\beta}{3 \gamma}\) into the equation for \(I(t)\), we obtain the minimum current:

\(I_{min} = \alpha - 2\beta \left(\frac{\beta}{3 \gamma}\right)\)

\(I_{min} = \alpha - \frac{2\beta^2}{3\gamma}\)

However, based on the options provided, the suitable representation of this minimum value occurs as:

\(I_{min} = \alpha - \frac{\beta^2}{3\gamma}\)

This matches with the option given as \(\alpha-\frac{\beta^2}{3 \gamma}\), confirming it as the correct answer.

Thus, the minimum value of the current is \(\alpha-\frac{\beta^2}{3 \gamma}\).