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Unformatted text preview: 1 APPENDIX C PC.1 Because the capacitor voltage is zero at t = 0, the charge on the capacitor is zero at t = 0. Then using Equation 3.5 in the text, we have t dx dx t i t q t t 3 3 ) ( ) ( = = + = For s, 2 = 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 15V 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 . 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 2...
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This note was uploaded on 04/03/2008 for the course EE 442/448 taught by Professor Kumar during the Spring '08 term at Iowa State.
 Spring '08
 Kumar
 Volt

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