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?
The current entering the positive terminal of a device is
and the voltage across the device is
(a) Find the charge delivered to the device between and
I would presume we use the intergal
but didn't know how to use the with this
the other related question are. (b) Calculate the power absorbed and (c) Determin the energy absorbed in .
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...
however in (c) "Determine the energy absorbed in 3s" I thot the setting in would give in Joules
isn't which is the ans
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.