818_Physics ProblemsTechnical Physics

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

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158 Direct Current Circuits P28.73 Ve tRC = ε so ln VR C t F H G I K J = F H G I K J 1 . A plot of ln 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 24 186 10 ., xy ii = 244, y i = 403 N = 8 Slope = = ∑∑ Nx y x y x i i c h ch ej 2 2 00118 . Intercept = = x x y x i i i 2 2 2 00882 c h . ts VV V a f a f bg ∆∆ ln 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 . . V t F H G I K J =+ . (b) Thus, the time constant is τ == = = RC 11 84 7 slope s . . and the capacitance is C R × = µ 84 7 10 0 10 847 6 . . . s F . P28.74 (a) For the first measurement, the equivalent circuit is as shown in Figure 1. R RRR R ab y y y ==+= 1
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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