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

\(Q=(αt−βt^2+γt^3)\) 

\(i=\frac{dQ}{dt}​=(α−2βt+3γt^2)\) 

\(\frac{di}{dt}​=(3γt−2β)=0\) 

\(⇒t=\frac{\beta}{3γ}\)​ 

\(i=(α−2βt+3γt^2)=\bigg(α−\frac{3γ}{β^2}\bigg)\)

\(\text{The Correct Option is (C):}\) \(\alpha-\frac{\beta^2}{3 \gamma}\)

Answer 2

The Correct Option is (C): \(\alpha - \frac{\beta^2}{3\gamma}\)

For constant current \(I = \frac{q}{t}\).
For a variable current \(I = \frac{dq}{dt}\).

As \(i = \frac{dq}{dt}\)
\(= \frac{d}{dt} \times ( \alpha t - \beta(t^2) + \gamma(t^3))\)

Therefore, \(i = \alpha - 2\beta + 3\gamma(t^2)\)

This i (current) is the function of time.

The minimum value of the current will be when: \(\frac{di}{dt} = 0\)
\(\rightarrow - 2β + 6γt = 0\)
\(\rightarrow 6\gamma t = 2\beta\)
\(\rightarrow t = \frac{2\beta}{6\gamma}\)
\(\rightarrow t = \frac{\beta}{3\gamma}\)

(Current) i will be minimum when
\(= \alpha - 2[\frac{\beta^2}{3\gamma}] + 3\gamma[\frac{\beta^2}{9\gamma^2}]\)
\(= \alpha - \frac{2β^2}{3\gamma} + \frac{β^2}{3\gamma}\)
\(= \alpha - \frac{(\beta^2)}{3\gamma}\)

Therefore, the minimum value of Current \(= \alpha - \frac{(\beta^2)}{3\gamma}\)