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.

The power absorbed by the BOX

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

 

This question can be found in Basic Engineering Circuit Analysis, 10th edition, chapter 1, question 1.13.

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