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A current flowing through a wire depends on time as $I = 3{t^2} + 2t + 5$. The charge flowing through the cross section of the wire in time from $t = 0$ to $t = 2\sec $ is
(A) $22C$
(B) $20C$
(C) $18C$
(D) $5C$

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Last updated date: 27th Feb 2024
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Hint: The current is defined as the rate of flow of charge through the cross section of a conductor. The current given in the above question is not a constant, but is a function of time. Therefore, the charge flowing in the given interval of time can be calculated by integrating the expression for the current in the interval.

Formula Used: We will be using the following formula,
$i = \dfrac{{dq}}{{dt}}$
where $i$ is current, $q$ is charge and the time is $t$

Complete step-by-step solution:
We know that the current flowing in a conductor is defined as the rate of flow of charge through a cross section of the conductor. That is,
$i = \dfrac{{dq}}{{dt}}$
Multiplying by $dt$ both the sides, we have
$dq = idt$
Integrating both the sides, we have
$\int_0^q {dq} = \int_{{t_1}}^{{t_2}} {idt} $
According to the question, the current is $i = 3{t^2} + 2t + 5$, ${t_1} = 0$ and . Substituting these above we get
$\int_0^q {dq} = \int_0^2 {\left( {3{t^2} + 2t + 5} \right)dt} $
$ \Rightarrow \left[ q \right]_0^q = \left[ {3\left( {\dfrac{{{t^3}}}{3}} \right) + 2\left( {\dfrac{{{t^2}}}{2}} \right) + 5t} \right]_0^2$
Substituting the limits, we get
$q - 0 = \left[ {3\left( {\dfrac{{{{\left( 2 \right)}^3}}}{3}} \right) + 2\left( {\dfrac{{{{\left( 2 \right)}^2}}}{2}} \right) + 5\left( 2 \right) - 3\left( {\dfrac{{{{\left( 0 \right)}^3}}}{3}} \right) - 2\left( {\dfrac{{{{\left( 0 \right)}^2}}}{2}} \right) - 5\left( 0 \right)} \right]$
$ \Rightarrow q = 8 + 4 + 10$
On solving, we finally get
$q = 22C$
Thus, the charge flowing through the cross section of the wire in the given time interval is equal to $22C$

Hence, the correct answer is option A.

Note: The definition of the current used in the above solution is for the instantaneous current. The current definition is commonly known for the average current in a time interval. But we were given the current as a function of time and not as a constant. This means that we were given the instantaneous expression for the current. So we had to use the definition for the instantaneous current.
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