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Problem 6.34 - Enhanced - with Hints and Feedback At t=0, a series-connected capacitor and inductor are placed across the ter

Part C -5t V, what is the current, i(t), through the inductor? The L = 100 mH inductor in the following figure has an initialPart D + ע The current through the L = 100 mH inductor in the following figure is i = 2t² mA. Find the power absorbed by thePart E The current through the L = 100 mH inductor in the following figure is i = 2(2-et/100) mA. Use the integral of the powPLEASE BOX ALL ANSWERS FOR A THUMBS UP

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Part A: Given 6 -16ooot 1.5e -Hooot 0.5e vo L. dio . at - Jio.de Vo = - [25X1o cu vo - 240oo.e - 16000 t ) + 2000 e chooot? VParto Given L=100mH , i at² mA Power absorbed = V. I Lodi dt Li.di at L. P = (100MH) (2+ma) · (ht) x103 P = (0-1)(8+3) uw P=For part a and c apply capacitor and inductor formulas and for part d apply power formula for given data and part e integrate power to get energy and substitute t=0 in calculated energy value.

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