Question

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. 22 C
B. 20 C
C. 18 C
D. 5 C

Answer 1

The current flowing through the circuit is the rate of flow of charges per unit time. Therefore, rearrange the equation for the charge by integrating it. Substitute the limits of integration as $t = 0$ to $t = 2$sec and determine the charge in coulomb.

Formula used:
Current, $I = \dfrac{{dq}}{{dt}}$
Here, q is the charge and t is the time.

Complete step by step answer:
We have given the equation for current flow, $I = 3{t^2} + 2t + 5$and we have to calculate the charge flow in time ${t_1} = 0\,{\text{s}}$ to ${t_2} = 2\,{\text{s}}$.
We know that the current flowing through the circuit is the rate of flow of charges per unit time. Therefore,
$I = \dfrac{{dq}}{{dt}}$
Here, q is the charge.

Since we want to determine the charge, we have to integrate the above equation. Therefore, integrating the above equation from ${t_1}$ to ${t_2}$, we get,
$\int\limits_{{t_1}}^{{t_2}} {I\,dt} = q$
$\Rightarrow q = \int\limits_{{t_1}}^{{t_2}} {I\,dt}$

Substituting ${t_1} = 0\,{\text{s}}$, ${t_2} = 2\,{\text{s}}$and $I = 3{t^2} + 2t + 5$ in the above equation, we get,
$q = \int\limits_0^2 {\left( {3{t^2} + 2t + 5} \right)\,dt}$
$\Rightarrow q = 3\int\limits_0^2 {{t^2}\,dt + } 2\int\limits_0^2 {t\,dt + } 5\int\limits_0^2 {dt}$
$\Rightarrow q = 3\left( {\dfrac{{{t^3}}}{3}} \right)_0^2 + 2\left( {\dfrac{{{t^2}}}{2}} \right)_0^2 + 5\left( t \right)_0^2$
$\Rightarrow q = 3\left( {\dfrac{{{2^3}}}{3}} \right) + 2\left( {\dfrac{{{2^2}}}{2}} \right) + 5\left( 2 \right)$
$\Rightarrow q = 8 + 4 + 10$
$\therefore q = 22\,{\text{C}}$

Therefore, the charge flowing through the cross section of the wire is 22 C.Hence,option A is the correct answer.

Note: If the current flowing through the wire does not change with the time, the current can be expressed as charge per unit time. Since the current is changing with time, you have to take the rate of flow of charge that is the derivative of the charge with respect to the time. Remember, the finite integrations do not have a constant of integration C.