Hw17_pdf - mcconnell (kam2342) – homework17 – Turner...

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: mcconnell (kam2342) – homework17 – Turner – (60230) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points Three batteries and four resistors are con- nected in a loop as shown below. 1673 Ω 26 V 710 Ω 26 V 1916 Ω 2608 Ω 20 V a b What is the magnitude of the current in the loop? Correct answer: 0 . 00289561 A. Explanation: R 3 E 3 R 4 E 1 R 1 R 2 E 2 a b I Let : R 1 = 1916 Ω , R 2 = 2608 Ω , R 3 = 1673 Ω , R 4 = 710 Ω , E 1 = 26 V , E 2 = 20 V , and E 3 = 26 V . In this problem, three emfs are connected in series, so E total = E 2 + E 3- E 1 = 20 V + 26 V- 26 V = 20 V R T = R 1 + R 2 + R 3 + R 4 = 1916 Ω + 2608 Ω + 1673 Ω + 710 Ω = 6907 Ω . The current in the circuit is counter- clockwise because E 2 + E 3 > E 1 , so I = E total R T = 20 V 6907 Ω = . 00289561 A . 002 (part 2 of 3) 10.0 points What is the potential V ab from point a to point b ? Caution: Include the sign in your answer. Correct answer:- 19 . 0998 V. Explanation: V ab = E 2- I R 2- I R 1- E 1 = 20 V- (0 . 00289561 A) (2608 Ω)- (0 . 00289561 A) (1916 Ω)- 26 V =- 19 . 0998 V . 003 (part 3 of 3) 10.0 points What is the voltage drop across the top, left resistor? Correct answer: 4 . 84436 V. Explanation: V 3 = I R 3 = (0 . 00289561 A) (1673 Ω) = 4 . 84436 V . 004 10.0 points 6 . 9 V 1 . 2 V 4 . 3 V I 1 1 . 2 Ω 3 . 5 Ω I 2 7 . 4 Ω I 3 9 . 4 Ω Find the current I 1 in the 1 . 2 Ω resistor at the bottom of the circuit between the two mcconnell (kam2342) – homework17 – Turner – (60230) 2 power supplies. Correct answer: 0 . 644088 A. Explanation: E 1 E 2 E 3 I 1 R A R B I 2 R C I 3 R D At a junction (Conservation of Charge) I 1 + I 2- I 3 = 0 . (1) Kirchhoff’s law on the large outside loop gives ( R A + R B ) I 1 + R D I 3 = E 1 + E 2 . (2) Kirchhoff’s law on the right-hand small loop gives R C I 2 + R D I 3 = E 3 . (3) Let : R A = 1 . 2 Ω , R B = 3 . 5 Ω , R C = 7 . 4 Ω , R D = 9 . 4 Ω , E 1 = 6 . 9 V , E 2 = 1 . 2 V , and E 3 = 4 . 3 V . Using determinants, I 1 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1- 1 E 1 + E 2 R D E 3 R C R D vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 1- 1 R A + R B R D R C R D vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle Expanding along the first row, the numera- tor is D 1 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1- 1 E 1 + E 2 R D E 3 R C R D vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 0- 1 vextendsingle vextendsingle vextendsingle vextendsingle E 1 + E 2 R D E 3 R D vextendsingle vextendsingle vextendsingle vextendsingle + (- 1) vextendsingle vextendsingle vextendsingle vextendsingle E 1 + E 2 E 3 R C vextendsingle...
View Full Document

This note was uploaded on 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas.

Page1 / 8

Hw17_pdf - mcconnell (kam2342) – homework17 – Turner...

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