86. (a) We note that we obtain the maximum value in Eq. 31-28 when we set tfd====π2ω1414600 00417().sor 4.17 ms. The result is εmmsin(sin ().π/2) =9036 0°=V. (b) At t= 4.17 ms, the current is sin ()sin (90( 24.3 ))(0.164A) cos(24.3 )0.1495A0.150 A.diItIφ=−=°−−°=°=≈using Eq. 31-29 and the results of the Sample Problem. Ohm’s law directly gives (0.1495A)(200 )29.9V.RviRΩ=(c) The capacitor voltage phasor is 90° less than that of the current. Thus, at t= 4.17 ms, we obtain sin(90( 24.3 ) 90 )
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.