{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

60303-Appendix_C

60303-Appendix_C - APPENDIX C PC.1 Because the capacitor...

This preview shows pages 1–3. Sign up to view the full content.

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 614

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
PC.4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 3

60303-Appendix_C - APPENDIX C PC.1 Because the capacitor...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online