# Finding Power (Simple Question)

Discussion in 'Homework Help' started by Student01, Mar 4, 2010.

1. ### Student01 Thread Starter Active Member

Apr 15, 2009
35
0
Hi all,

I'm stuck on the following two questions:

1)An eletrical element draws a current of i(t) = 10cos4t A and has a voltage of v(t) = 120cos4t V across it. Find energy absorbed by the element in 2s.

2)The voltage across a device and the current through it are
v(t) = 5cos2t V i(t) = 10 (1 - e^0.5t) A

Calculate the power consumed by this device at t = 1 s.

I know P = intregral( i(t) x v(t) dt) but I don't know how to integrate the product of two trigonometric functions, or the product of a trigonometric and exponential function. Am I going about this the wrong way?

2. ### Heavydoody Active Member

Jul 31, 2009
140
11
I don't know if integration is the correct method since I haven't started AC yet. It does sound logical though. I can tell you how to integrate the first part. If you get it in the form of cos^2(θ), you can use the half angle formula to create a relatively simple integral. You will end up using two substitutions, don't forget to update your limits of integration accordingly. Try it and you will see.

3. ### Heavydoody Active Member

Jul 31, 2009
140
11
The way the second part is worded, it sounds like they are looking for an instantaneous value. However, if they want from 0 to 1, it can be integrated, but it gets hairy. You end up using integration by parts twice and some creative algebra (as well as just slightly less than a tree's worth of paper).

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
In the first case you would need to integrate the instantaneous power function p(t)=v(t)*i(t) over the 2 second interval.

I agree with Heavydoody that in the second case you only need to compute the instantaneous power at time t=1sec. No integration is required for that case.

5. ### Eng_Bandar Active Member

Feb 27, 2010
50
1
Q1

I think the problem in integration
I will give the key and easy to complete it
you can see Attachments

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Last edited: Apr 15, 2010
6. ### Student01 Thread Starter Active Member

Apr 15, 2009
35
0
Ok, thanks for the help guys. So it turns out I just needed the right trigonometric identity to simplify the first problem, and was mis-interpreting the second question.