# Thread: find charge, power, energy

1. ## find charge, power, energy

The current entering the positive terminal of a device is $\displaystyle i(t) = 3e^{-2t}[A]$
and the voltage across the device is $\displaystyle v(t) = 5\frac{di}{dt} [V]$

(a) Find the charge delivered to the device between $\displaystyle t=0$ and $\displaystyle t=2s$

I would presume we use the intergal

$\displaystyle Q \triangleq\int_{t_0}^t{idt}$

but didn't know how to use the $\displaystyle v(t) = 5\frac{di}{dt} V$ with this

the other related question are. (b) Calculate the power absorbed and (c) Determin the energy absorbed in $\displaystyle 3s$.

2. Your idea with (a) is correct. Just carry that out. You don't need v(t) for part (a), but you do for part (b), and possibly (c). What do you get?

3. ## books ans was 1.297C

$\displaystyle Q = \int_{t=0}^2{3e^{-2t}}dt = 1.47253C$

well somewhere theres something
the books ans was $\displaystyle 1.297C$

4. Hmm. Yeah, I get

$\displaystyle Q=\dfrac{3}{2}\left(1-e^{-4}\right)\approx 1.47253.$

I think the book is wrong on that one. Does the book have an analytical expression, or just the approximation?

5. ## the same as you suggested

the answer for that question in book is given as
$\displaystyle (a) 1.297C, (b) -90e^{-4t}, (c)-22.5J$

i will see if I can do (b) and (c)

the example they had in the book was pretty much the same as you suggested.
so maybe a typo in the answers.

6. I agree with (b)'s answer. How is the book defining "energy absorbed"? Is it just the integral of the power over the desired time interval?

7. ## (b) calculate the power absorbed

their definition of power is...."the rate of expending or absorbing energy, measured in Watts (W)"

anyway this is how I did (b) assuming yours also...

Since $\displaystyle P=vi$ and

$\displaystyle v=5\frac{di}{dt} = -30e^{-2t}[V]$

$\displaystyle i=3e^{-2t}[A]$

Then

$\displaystyle P=(-30e^{-2t})(3e^{-2t})=-90e^{-4t}[W]$

however in (c) "Determine the energy absorbed in 3s" I thot the setting$\displaystyle t=3$ in $\displaystyle -90e^{-4t}$ would give$\displaystyle P$ in Joules$\displaystyle [J]$
but

$\displaystyle -90e^{-12} =-0.0005529 [J]$ isn't $\displaystyle -22.5 [J]$ which is the ans

8. I agree with your (b) procedure. For (c), I think you'll find that my hint in post # 6 will do the trick. Integrate from t = 0 to t = 3.

9. ## =-22.5[j]

$\displaystyle \int_{t=0}^3(-90e^{-4t}) \Rightarrow \left(\frac{45}{2}\left(\frac{1}{e^{12}}-1\right)\right) \Rightarrow -22.49986 \Rightarrow -22.5[J]$

i should know but why was it necessary to take the $\displaystyle \int_{t=0}^3$

10. Your computations look correct to me.

Power is equal to the time derivative of the work. Work has the same units as energy, and since there isn't any other energy floating around that I'm aware of, I postulated that the work must be equal to the integral of the power, by the fundamental theorem of the calculus. So that's why I thought the integral was the right thing to do. In addition, there was a clue: the "energy absorbed in 3 seconds". Whenever anyone uses a phrase like that, my mind automatically goes towards an integral, since the rate at which the energy is absorbing might vary over the 3 seconds. In that case, you must do an integral.

Make sense in a sort of hand-waiving manner? Maybe someone else here on the MHF could produce a more rigorous proof.