{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

818_Physics ProblemsTechnical Physics

818_Physics ProblemsTechnical Physics - 158 P28.73 Direct...

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

View Full Document Right Arrow Icon
158 Direct Current Circuits P28.73 V e t RC = ε so ln ε V RC t F H G I K J = F H G I K J 1 . A plot of ln ε V F H G I K J versus t should be a straight line with slope equal to 1 RC . Using the given data values: FIG. P28.73 (a) A least-square fit to this data yields the graph above. x i = 282 , x i 2 4 1 86 10 = × . , x y i i = 244, y i = 4 03 . , N = 8 Slope = = N x y x y N x x i i i i i i c h c hc h e j c h 2 2 0 011 8 . Intercept = = x y x x y N x x i i i i i i i 2 2 2 0 088 2 e jc h c hc h e j c h . t s V V V a f a f b g ln ε 0 4.87 11.1 19.4 30.8 46.6 67.3 102.2 6.19 5.55 4.93 4.34 3.72 3.09 2.47 1.83 0 0.109 0.228 0.355 0.509 0.695 0.919 1.219 The equation of the best fit line is: ln . . ε V t F H G I K J = + 0 011 8 0 088 2 b g . (b) Thus, the time constant is τ = = = = RC 1 1 0 011 8 84 7 slope s . . and the capacitance is C R = = × = τ µ 84 7 10 0 10 8 47 6 . . . s F . P28.74 (a) For the first measurement, the equivalent circuit is as
Image of page 1
This is the end of the preview. Sign up to access the rest of the 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