1.21 - 2 4 t(W p = 5 W t = p 25 s = 5cos 2 4 × 25 = 5 W t...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 1.21 The voltage across and current through a certain device are given by v ( t ) = 5cos ( 4 π t ) V , i ( t ) = 0 . 1cos ( 4 π t ) A . Determine: (a) The instantaneous power p ( t ) at t = 0 and t = 0 . 25 s. (b) The average power p av , defined as the average value of p ( t ) over a full time period of the cosine function (0 to 0.5 s). Solution: (a) p ( t ) = vi = ( 5cos4 π t )( 0 . 1cos 4 π t ) = 0 . 5cos 2 4 π t (W)
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 4 t (W) . p ( ) = . 5 W @ t = p ( . 25 s ) = . 5cos 2 ( 4 × . 25 ) = . 5 W @ t = . 25 s . (b) p av = 1 T i T p ( t ) dt = 1 . 5 i . 5 . 5cos 2 4 t dt = 1 8 [ sin4 t cos4 t + 4 t ] | . 5 = 1 4 (W) ....
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern