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Physics Question on Electric Current

The charge flowing in a conductor changes with time as Q(t)=αt−βt2+γ3Q(t)=\alpha t-\beta t ^2+\gamma^3 Where α,β\alpha, \beta and γ\gamma are constants Minimum value of current is :

A

α−γ23β\alpha-\frac{\gamma^2}{3 \beta}

B

β−α23γ\beta-\frac{\alpha^2}{3 \gamma}

C

α−β23γ\alpha-\frac{\beta^2}{3 \gamma}

D

α−3β2γ\alpha-\frac{3 \beta^2}{\gamma}

Answer

α−β23γ\alpha-\frac{\beta^2}{3 \gamma}

Explanation

Solution

Q=(αt−βt2+γt3)Q=(αt−βt^2+γt^3)

i=dQdt​=(α−2βt+3γt2)i=\frac{dQ}{dt}​=(α−2βt+3γt^2)

didt​=(3γt−2β)=0\frac{di}{dt}​=(3γt−2β)=0

⇒t=β3γ⇒t=\frac{\beta}{3γ}​

i=(α−2βt+3γt2)=(α−3γβ2)i=(α−2βt+3γt^2)=\bigg(α−\frac{3γ}{β^2}\bigg)

The Correct Option is (C):\text{The Correct Option is (C):} α−β23γ\alpha-\frac{\beta^2}{3 \gamma}