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.

#### 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}

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

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