Question

The current in a wire varies with time according to the relationship

$\mathbf{I}\mathbf{=}\mathbf{55}\mathbf{}\mathbf{A}\mathbf{-}\left(0.65\frac{A}{s^2}\right){\mathbf{t}}^{\mathbf{2}}$. (a) How many coulombs of charge pass a cross section of the wire in

the time interval between $\mathbf{t}\mathbf{=}\mathbf{0}$and role="math" localid="1655721302619" $\mathbf{t}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{s}$? (b) What constant current would transport the

same charge in the same time interval?

Short answer

a) The charge passing through the wire between 0-8 seconds is 329 C

b) The constant current that would transport this charge is 41.0A

Step-by-step solution

Write the formula of current.

Consider the formula for the current:

$\mathbf{I}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{t}}$

Here, Q is the charge and t is the time.

Calculation of charge passing between 0-8 seconds.

(a)

Consider the relation for the time and the current is:

$I=55A-\left(0.65\frac{\mathrm{A}}{{\mathrm{s}}^2}\right)t^2$

Apply the formula for the current to determine the charge.

$$\begin{gathered} △Q=\left[55-\left(0.65\right)t^2\right]dt \\ Q={\int }_0^8\left[55-\left(0.65\right)t^2\right]dt \\ Q={\left[55t-\frac{1}{3}\left(0.65t^3\right)\right]}_0^8 \\ Q=440-111 \end{gathered}$$

Solve further as,

$Q=329\mathit{C}$

Calculation of the constant current in the wire.

(b)

Solve for the current in the wire.

$$\begin{gathered} I=\frac{\mathrm{Q}}{\mathrm{t}} \\ I=\frac{330\mathrm{C}}{8\mathrm{s}} \\ I=41.0\mathrm{A} \end{gathered}$$

Therefore, the value of the current is 41.0 A .