0132130084_c - APPENDIX C PC.1 Because the capacitor...

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689 APPENDIX C PC.1 Because the capacitor voltage is zero at t = 0, the charge on the capacitor is zero at = 0. Then using Equation 3.5 in the text, we have t dx dx t i t q t t 3 3 0 ) ( ) ( 0 0 = = + = For s, 2 µ = t we have C 6 10 2 3 ) 3 ( 6 = × × = q PC.2 Refer to Figure PC.2 in the book. Combining the 10- resistance and the 20- resistance we obtain a resistance of 6.667 , which is in series with the 5- resistance. Thus, the total resistance seen by the 15-V source is 5 + 6.667 = 11.667 . The source current is 15/11.667 = 1.286 A. The current divides between the 10- resistance and the 20- resistance. Using Equation 2.27, the current through the 10- resistance is A 8572 . 0 286 . 1 10 20 20 10 = × + = i Finally, the power dissipated in the 10- resistance is W 346 . 7 10 2 10 10 = = i P PC.3 The equivalent capacitance of the two capacitors in series is given by F 4 / 1 / 1 1 2 1 = + = C C C eq The charge supplied by the source is C 800 10 4 200 6 = × × = = V C q eq
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This note was uploaded on 11/20/2011 for the course ELE 220 taught by Professor Anas during the Spring '11 term at American Dubai.

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0132130084_c - APPENDIX C PC.1 Because the capacitor...

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