- #1

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**A certain load is characterized by an instantaneous power that increases linearly from zero to 1 kW over a 3-hour period. Determine the total energy in (a) kWh and (b) joules.**

**p(t) = [dw(t)] / dt**

W = F p(t)dt T2 is upper limit, T1 is lower limit.

W = F p(t)dt T2 is upper limit, T1 is lower limit.

**One of my problems is my calculus is super weak at this point, but I am curious if there is a way to solve this problem without calculus for one, and two...if it truly does take calculus, how do I go about doing it. I did take calculus but I lost the majority of my calculus skills over the course of two years.**

I first tried using simple geometry and made a triangle, 0-10,800seconds for the x dimension and 0-1kW for the Y dimension. Of course this lead my answer to 10,800Joules. I know this is completely wrong.

I understand the equation of Energy = Watts x Seconds. This is the instant power equation though, so I am unaware of how to do the function of time equation.

The answer for (a) kWh = 1.5kWh. The answer for (b) is 5.4 x 10^6J.

I first tried using simple geometry and made a triangle, 0-10,800seconds for the x dimension and 0-1kW for the Y dimension. Of course this lead my answer to 10,800Joules. I know this is completely wrong.

I understand the equation of Energy = Watts x Seconds. This is the instant power equation though, so I am unaware of how to do the function of time equation.

The answer for (a) kWh = 1.5kWh. The answer for (b) is 5.4 x 10^6J.