# The power absorbed by the BOX

The power absorbed by the BOX is $2e^{-2t}$ W. Calculate the amount of charge that enters the BOX between 0.1 and 0.4 seconds.

Image from: J. D. Irwin and R. M. Nelms, Basic engineering circuit analysis, 10th ed. Hoboken, NJ: John Wiley, 2011.

#### Solution:

Remember that:

$p=vi$

(Where, $p$ is power, $v$ is voltage, and $i$ is current)

$i=\dfrac{p}{v}$

Substitute our equations:

$i=\dfrac{2e^{-2t}}{4e^{-t}}$

$i=0.5e^{-t}$

To find the amount of charge that entered during the time interval, we must remember the following:

$i=\dfrac{d(q(t))}{dt}$

$d(q(t))=i\,dt$

(Take the integral of both sides)

$\,\displaystyle q(t)=\int^{t_2}_{t_1}i\,dt$

(Substitute our current equation)

$\,\displaystyle q(t)=\int^{0.4}_{0.1}(0.5e^{-t})\,dt$

$q(t)=-0.5e^{-t}\Big|^{0.4}_{0.1}$

$q=0.117$ C