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Will rate!Chapter 31, Problem 017 GO In the figure, R = 11.0 12, C = 6.59 pF, and L = 52.0 mh, and the ideal battery has emf = 33.0 V.

Chapter 31, Problem 019 Using the loop rule and deriving the differential equation for an LC circuit find the current (sign i

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R RAHOM, L=52,0 mm = 11.ou, L = 52.0 m H = 52410-3 H c = 6.59+10 of Emf, E = 33.0vtm when capacitor becomes fully charged. charge on capacitor 1% - CE n- when switch is thrown to poshon b. At led sine t charwhere az a te w e general solution is (H) - 0 cos (wtto) At t=0 &q=% % = 8 coslo) At too & Iso I a- de - owsin (ast +O) o - gD I = -de - qw sin(at) Cumery amplitude - 2W - CE. ( 2f) - ce. I are e - 16.59410 x 33 5241013 -0.371A)

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