The charge flowing in a conductor changes with time as Q(t) = t - t^2 + t^3, where,, are constants. The minimum value of current is:

Given: Q(t) = t - t^2 + t^3 Find: The minimum value of current. Current is the rate of flow of charge, so I(t) = dQdt = - 2 t + 3 t^2 For the minimum value of current, differentiate I(t) with respect to t and set it equal to zero: dIdt = -2 + 6 t = 0 This gives t = 3 Substitute this value of t into I(t): I = - 2 ( 3) + 3 ( 3)^2 I = - 2 ^23 + ^23 I = - ^23 Therefore, the minimum current is - ^23. The solution states the correct option is B, but this expression matches option D in the listed options, so the option labels on the solution's are inconsistent. Since I(t) = - 2 t + 3 t^2 is a quadratic expression in t with coefficient of t^2 equal to 3, it opens upward when 0. Hence the stationary value at t = 3 is the minimum value of current.

The charge flowing in a conductor changes with time as Q(t) = t -… — Solution | JEE Physics

MCQEasyJEE 2023Electric Current & Drift Velocity

The charge flowing in a conductor changes with time as

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

where $\alpha, \beta, \gamma$ are constants. The minimum value of current is:

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

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

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

Find: The minimum value of current.

Current is the rate of flow of charge, so

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

For the minimum value of current, differentiate $I(t)$ with respect to $t$ and set it equal to zero:

\frac{dI}{dt} = -2\beta + 6\gamma t = 0

This gives

t = \frac{\beta}{3\gamma}

Substitute this value of $t$ into $I(t)$ :

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} + \frac{\beta^2}{3\gamma}

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

Therefore, the minimum current is $\alpha - \frac{\beta^2}{3\gamma}$ . The solution states the correct option is B, but this expression matches option D in the listed options, so the option labels on the solution's are inconsistent.

Why the Extremum is Minimum

Since

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

is a quadratic expression in $t$ with coefficient of $t^2$ equal to $3\gamma$ , it opens upward when $\gamma > 0$ . Hence the stationary value at

t = \frac{\beta}{3\gamma}

is the minimum value of current.

Common mistakes

Differentiating $Q(t)$ incorrectly to get current. Current is $I = \frac{dQ}{dt}$ , not $Q$ itself. Always differentiate the charge function first.
Setting $\frac{dQ}{dt} = 0$ directly to find the minimum current. That gives where current may be zero, not where current is minimum. To minimize current, differentiate $I(t)$ and solve $\frac{dI}{dt} = 0$ .
Choosing the option label from the solution without checking the expression. Here the worked result is $\alpha - \frac{\beta^2}{3\gamma}$ , which matches option D in the given list even though the solution says B.

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