Question

The current entering the positive terminal of a device isi(t)=(6e^{-2t} times 10^{-3}) A and the voltage across the device is v(t)=(10\frac{ di }{ dt })V. Determine the energy absorbed in 3s.

Answer 1

\[i(t)=(6e^{-2t} \times 10^{-3})A\]

Answer 2

\[v(t)=(10\frac{ di }{ dt })V\]

Answer 3

I have an answer and if you give me a second I will upload the picture. It does not match the textbook and I am just curious where I went wrong.

Answer 4

\[\text{Power} = \int\limits_{0}^{3}v(t) * i(t) dt\]

Answer 5

I have never seen power presented that way... this is what I did....

Answer 6

Hard to read the link when it is turned by 90 degrees.

Answer 7

sorry give me one second and I will try to edit it

Answer 8

My iPhone switches them around sometimes

Answer 9

The Power that you calculated as a product of v and i is correct.
Since both v and i vary with t, the power that you computed is instantaneous power.
When you multiply instantaneous power by delta_t you get the energy over time delta_t.
You integrate them from t = 0 to t = 3 seconds to sum up all the instantaneous energy.
The units of energy will be in Joules.

Answer 10

I worked out the problem out with the integral you showed and your right. When you explain it the way you did above it makes sense. Furthermore when I do it that way the answer is correct. Thank you. Are you a teacher? Or a student?

Answer 11

Neither. Computer Science major. Just volunteering here.

Answer 12

Wow.... that is what I am really struggling with. I have to do a linux project using an ssh but since I have a Mac my professor says I am at a disadvantage because he is not familiar with them. Thanks for all your help

Answer 13

You are welcome.