15 - Kapoor (mk9499) oldhomework 15 Turner (60230) 1 This...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Kapoor (mk9499) oldhomework 15 Turner (60230) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points 4 . 7 3 . 7 16 . 8 26 . 8 V 13 . 4 V Find the current through the 16 . 8 lower- right resistor. Correct answer: 1 . 0229 A. Explanation: r 1 r 2 R E 1 E 2 A D E B C F i 1 i 2 I Let : E 1 = 26 . 8 V , E 2 = 13 . 4 V , r 1 = 4 . 7 , r 2 = 3 . 7 , and R = 16 . 8 . From the junction rule, I = i 1 + i 2 . Applying Kirchhoffs loop rule, we obtain two equations: E 1 = i 1 r 1 + I R (1) E 2 = i 2 r 2 + I R = ( I- i 1 ) r 2 + I R =- i 1 r 2 + I ( R + r 2 ) , (2) Multiplying Eq. (1) by r 2 , Eq. (2) by r 1 , E 1 r 2 = i 1 r 1 r 2 + r 2 I R E 2 r 1 =- i 1 r 1 r 2 + I r 1 ( R + r 2 ) Adding, E 1 r 2 + E 2 r 1 = I [ r 2 R + r 1 ( R + r 2 )] I = E 1 r 2 + E 2 r 1 r 2 R + r 1 ( R + r 2 ) = (26 . 8 V) (3 . 7 ) + (13 . 4 V) (4 . 7 ) (3 . 7 ) (16 . 8 ) + (4 . 7 ) (16 . 8 + 3 . 7 ) = 1 . 0229 A . 002 (part 1 of 3) 10.0 points Consider the circuit in the figure. 15 . 0 V 3 . 3 3 . 3 7 . 3 7 . 3 4 . 5 15 . 0 1 . 9 Find the current in the 1.9 resistor. Correct answer: 1 . 33841 A. Explanation: E R 1 R 2 R 3 R 4 R 5 R 6 R 7 Kapoor (mk9499) oldhomework 15 Turner (60230) 2 Let : R 1 = 3 . 3 , R 2 = 3 . 3 , R 3 = 7 . 3 , R 4 = 7 . 3 , R 5 = 4 . 5 , R 6 = 15 . 0 , R 7 = 1 . 9 and V = 15 . 0 V . R 3 and R 4 are in parallel: 1 R 34 = 1 R 3 + 1 R 4 R 34 = parenleftbigg 1 R 3 + 1 R 4 parenrightbigg- 1 = parenleftbigg 1 7 . 3 + 1 7 . 3 parenrightbigg- 1 = 3 . 65 R 2 and R 34 are in series: R 234 = R 2 + R 34 = 3 . 3 + 3 . 65 = 6 . 95 R 5 and R 6 are in parallel: 1 R 56 = 1 R 5 + 1 R 6 R 56 = parenleftbigg 1 R 5 + 1 R 6 parenrightbigg- 1 = parenleftbigg 1 4 . 5 + 1 15 parenrightbigg- 1 = 3 . 46154 R 56 and R 7 are in series: R 567 = R 56 + R 7 = 3 . 46154 + 1 . 9 = 5 . 36154 R 234 and R 567 are in parallel: 1 R 234567 = 1 R 234 + 1 R 567 R 234567 = parenleftbigg 1 R 234 + 1 R 567 parenrightbigg- 1 = parenleftbigg 1 6 . 95 + 1 5 . 36154 parenrightbigg- 1 = 3 . 02665 R 1 and R 234567 are in series: R eq = R 1 + R 234567 = 3 . 3 + 3 . 02665 = 6 . 32665 The current in the circuit is I = V R eq = 15 V 6 . 32665 = 2 . 37092 A V 234567 = I R 234567 = (2 . 37092 A) (3 . 02665 ) = 7 . 17595 V Here the current is I 7 = I 567 = V 234567 R 567 = (7 . 17595 V) (5 . 36154 ) = 1 . 33841 A ....
View Full Document

This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

Page1 / 7

15 - Kapoor (mk9499) oldhomework 15 Turner (60230) 1 This...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online