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The current in a wire varies with time according to the relation $I=55A-\left(0.65\dfrac{A}{s^2}\right)t^2$.

  • How many coulombs of charge pass a cross-section of the wire in the time interval between $t=0\,s$ and $t=8.5\,s$? Express your answer using two significant numbers.
  • What constant current would transport the same charge in the same time interval?Express your answer using two significant numbers.

The primary objective of this problem is to calculate the amount of charge that could pass through a cross-section in the given interval of time, as well as the constant current that will transfer the charge.

Electric charge is a vital property of matter carried by certain fundamental particles which govern how the particles react to a magnetic or electric field. Electric charge can be either negative or positive and appears in precisely defined natural units and cannot be created or destroyed. It is therefore conserved.

Expert Answer

To begin with this problem, use integration to determine the charge that passes through the cross-section during the given time interval. Then, using the relationship between current, time interval, and charge, calculate the current.

The given equation of current can be plotted against time as:

Geogebra export

1- Given

Electric current  $I=55A-\left(0.65\dfrac{A}{s^2}\right)t^2$

Initial time  $t_1=0\,s$

Final time  $t_2=8.5\,s$

The charge that passes through a cross-section in a given time interval is
$Q=\int\limits_{t_1}^{t_2}\,I dt$

$Q=\int\limits_{0\,s}^{8.5\,s}\,\left(55A-\left(0.65\dfrac{A}{s^2}\right)t^2\right) dt$

$Q=[55t\,A]_{0\,s}^{8.5\,s}-\left[\dfrac{0.65}{3}\dfrac{A}{s^2}\cdot t^3\right]_{0\,s}^{8.5\,s}$

$Q=467.5\,C-133.06\,C$

$Q=334.44\,C$

( where $C=As$ )

Consequently, the amount of charge that passes through a cross-section in the given time interval is $334.44\,C$.

2- The following equation gives the constant current.

$I=\dfrac{\Delta Q}{\Delta t}$

Because the amount of charge is the same in the given interval, therefore, $\Delta Q=Q$ and

$I=\dfrac{Q}{t_2-t_1}$

In the above equation, substitute the given values for $Q$, $t_1$, and $t_2$.

$I=\dfrac{334.44\,C}{8.5\,s-0\,s}$

$=39.35\,A$

( where $A=\dfrac{C}{s}$ )

Hence, the constant current required to transport the charge is $39.35\, A$.

Consider an example to obtain a charge amount using the separation of variables method.

Example 1

What will be the amount of charge (in Coulombs) through the cross-section of a wire in the interval $t_1=2\,s$ to $t_2=6\,s$ when the current is expressed by the equation $I=3t^2-2t+1$?

Given

$I=3t^2−2t+1$

Since

$I=\dfrac{dQ}{dt}$

(Because $\Delta$ represents the finite variability of a quantity, therefore, we have replaced $\Delta $ by $d$.)

$dQ=I\,dt$

$\int dQ=\int\limits_{2}^{6}(3t^2−2t+1)\,dt$

$Q=\left[\dfrac{3t^3}{3}-\dfrac{2t^2}{2}+t\right]_2^6$

$Q=\left[ (216-8)- (36-4)+(6-2)\right] $

$Q=180\,C$

Example 2

A car battery generates $530\, C$ of charge in $6\, s$ when its engine is started, what will be the current $(I)$?

Since,

$I = \dfrac{\Delta Q}{\Delta t}$

Substituting the values for time and charge in the above formula of current yields

$ I = \dfrac{\Delta Q}{\Delta t}=\dfrac{530\,C}{6\,s}=88.33\,\dfrac{C}{s} $

$I=88.33\,A$

  Images/mathematical drawings are created with GeoGebra.

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