Help please guys and gals ?

[Mathmasteriw]

Hi Everyone,
Really pulling my hair out on this one,
Been stuck for over a week , I’ve tryed everything and this is the most correct version I can come up with,
Any help greatly appreciated as usual ??
Thanks ?

 

[Romsek]

Your integration is incorrect. Check your numbers.

 

[Dr.Peterson]

Didn't you ever finish this?

Any pointers on this?

Hi Everyone, Stuck again! Can anybody tell me where i’m going wrong on this please? Dose any one have an example of a similar question? Am I on the right track? Thanks again ?

www.freemathhelp.com

As I said there, your answer was mostly right, you just had a couple details of the work wrong. You've corrected the unit, but you still have the C (which I think is because your book is wrongly using an indefinite integral), and you dropped the 1/a, which I don't think I explicitly pointed out before.

Here, I'll just show it to you:

[MATH]i_L=\frac{1}{L}\int_0^T \cos(100t)dt = \frac{1}{0.01}\int_0^{0.9} \cos(100t)dt = \frac{1}{0.01}\left[\frac{1}{100}\sin(100t)\right]_0^{0.9}\\ = \frac{1}{0.01}\left[\frac{1}{100}\sin(100(0.9))-\frac{1}{100}\sin(100(0))\right]= \frac{1}{0.01}\frac{1}{100}\sin(90) = \frac{1}{0.01}\frac{1}{100}0.893996 = 0.894\text{ A}[/MATH]​

 

[Mathmasteriw]

Didn't you ever finish this?

Any pointers on this?

Hi Everyone, Stuck again! Can anybody tell me where i’m going wrong on this please? Dose any one have an example of a similar question? Am I on the right track? Thanks again ?

www.freemathhelp.com

As I said there, your answer was mostly right, you just had a couple details of the work wrong. You've corrected the unit, but you still have the C (which I think is because your book is wrongly using an indefinite integral), and you dropped the 1/a, which I don't think I explicitly pointed out before.

Here, I'll just show it to you:

[MATH]i_L=\frac{1}{L}\int_0^T \cos(100t)dt = \frac{1}{0.01}\int_0^{0.9} \cos(100t)dt = \frac{1}{0.01}\left[\frac{1}{100}\sin(100t)\right]_0^{0.9}\\ = \frac{1}{0.01}\left[\frac{1}{100}\sin(100(0.9))-\frac{1}{100}\sin(100(0))\right]= \frac{1}{0.01}\frac{1}{100}\sin(90) = \frac{1}{0.01}\frac{1}{100}0.893996 = 0.894\text{ A}[/MATH]​

Oh my Doc !! Thank you so much!??
Nope , been on this for soooo long! One of those problems that no matter how many times I've tried it or studied that nothing seemed to work! I guess that's what happens sometimes and when you stress yourself out.
?Lovely work Doc I really appreciate that, I will write it up now and include your magic.

 

[Steven G]

Please do not write that cos(ax) = (1/a)sin(ax) + c. First of all, for a given value for a, cos(ax) will not have an arbitrary constant at the end. 2nd of all, YOU know that it is [math]\int cos(ax)dx\ that\ equals\ \dfrac{1}{a}\sin(ax) + c[/math]