The charge flowing through a conductor varies with time as $Q = at - b{t^2}$ . Then, the current
A) Decreases linearly with time
B) Reaches a maximum and then decreases
C) Falls to zero at time $t = \dfrac{a}{{2b}}$
D) Changes at a rate $ - 2b$
Answer
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Hint:We are given with current in terms of time. We need to find the manner in which the current changes. Current is the amount of charge flowing through a conductor per unit time. Differentiate the equation of charge with respect to time to get the relation between current and time.
Complete step by step solution:We are given with charge flowing through a conductor as a function of time and we need to find the expression of current. As we know that current is the rate of charge flowing through a conductor, let us differentiate the given relation of charge with respect to time.
The given equation is:
$Q = at - b{t^2}$
Here, $Q$ is the charge flowing through a conductor.
$a,b$ are constants
$t$ denotes time.
Differentiating the given expression with respect to time, we get:
$\dfrac{{dQ}}{{dt}} = \dfrac{{at}}{{dt}} - \dfrac{{b{t^2}}}{{dt}}$
$ \Rightarrow \dfrac{{dQ}}{{dt}} = I = a - 2bt$
As $\dfrac{{dQ}}{{dt}} = I$
It is clear from the expression $I = a - 2bt$ that the current is linearly decreasing. Hence, option B is incorrect and thus, option A is the correct option.
Put $I = 0$ , we get
$a - 2bt = 0$
$ \Rightarrow t = \dfrac{a}{{2b}}$
This implies that the current will be zero at time $t = \dfrac{a}{{2b}}$
Thus, option C is also correct.
Rate of current will be given as:
$\dfrac{{dI}}{{dt}} = \dfrac{{d\left( {a - 2bt} \right)}}{{dt}}$
$ \Rightarrow \dfrac{{dI}}{{dt}} = - 2b$
This implies that the current changes at the rate of $\dfrac{{dI}}{{dt}} = - 2b$ .
Therefore, option D is also correct.
Option A, C and D are the correct options.
Note:We were given with charge as a function of time. Current is the rate of change of charge flowing through a conductor. The derivative of the charge as a function of time with respect to time will give the current as a function of time. The slope of this linear function of time is negative and so we conclude that the current is decreasing with time.
Complete step by step solution:We are given with charge flowing through a conductor as a function of time and we need to find the expression of current. As we know that current is the rate of charge flowing through a conductor, let us differentiate the given relation of charge with respect to time.
The given equation is:
$Q = at - b{t^2}$
Here, $Q$ is the charge flowing through a conductor.
$a,b$ are constants
$t$ denotes time.
Differentiating the given expression with respect to time, we get:
$\dfrac{{dQ}}{{dt}} = \dfrac{{at}}{{dt}} - \dfrac{{b{t^2}}}{{dt}}$
$ \Rightarrow \dfrac{{dQ}}{{dt}} = I = a - 2bt$
As $\dfrac{{dQ}}{{dt}} = I$
It is clear from the expression $I = a - 2bt$ that the current is linearly decreasing. Hence, option B is incorrect and thus, option A is the correct option.
Put $I = 0$ , we get
$a - 2bt = 0$
$ \Rightarrow t = \dfrac{a}{{2b}}$
This implies that the current will be zero at time $t = \dfrac{a}{{2b}}$
Thus, option C is also correct.
Rate of current will be given as:
$\dfrac{{dI}}{{dt}} = \dfrac{{d\left( {a - 2bt} \right)}}{{dt}}$
$ \Rightarrow \dfrac{{dI}}{{dt}} = - 2b$
This implies that the current changes at the rate of $\dfrac{{dI}}{{dt}} = - 2b$ .
Therefore, option D is also correct.
Option A, C and D are the correct options.
Note:We were given with charge as a function of time. Current is the rate of change of charge flowing through a conductor. The derivative of the charge as a function of time with respect to time will give the current as a function of time. The slope of this linear function of time is negative and so we conclude that the current is decreasing with time.
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