Current Electricity

JEE Main level, single correct, medium

15 of 380

Question 15 of 380

The charge flowing in a conductor changes with time as Q(t)=αt−βt2+γt3Q\left(t\right)=\alpha t-\beta {t}^{2}+\gamma {t}^{3}, where α,β\alpha ,\beta and γ\gamma are constants. Minimum value of current is:

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Answer: (D) α−β23γ\alpha -\dfrac{{\beta }^{2}}{3\gamma }

Given here, Q=(αt−βt2+γt3)Q=\left(\alpha t-\beta {t}^{2}+\gamma {t}^{3}\right)
Current flowing through conductor is given by
i=dQdt=(α−2βt+3γt2)...(1)i=\dfrac{dQ}{dt}=\left(\alpha -2\beta t+3\gamma {t}^{2}\right)...\left(1\right)
For current to be minimum, didt=0\dfrac{di}{dt}=0
So, didt=(6γt−2β)=0\dfrac{di}{dt}=\left(6\gamma t-2\beta \right)=0
⇒t=β3γ\Rightarrow t=\dfrac{\beta }{3\gamma }
Substituting in equation (1),
i=(α−2βt+3γt2)=α−2β(β3γ)+3γ(β3γ)2=α−2β23γ+3β29γ=(α−β23γ)i=\left(\alpha -2\beta t+3\gamma {t}^{2}\right)=\alpha -2\beta \left(\dfrac{\beta }{3\gamma }\right)+3\gamma {\left(\dfrac{\beta }{3\gamma }\right)}^{2} =\alpha -2\dfrac{{\beta }^{2}}{3\gamma }+3\dfrac{{\beta }^{2}}{9\gamma } =\left(\alpha -\dfrac{{\beta }^{2}}{3\gamma }\right)

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