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

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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...
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## 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.

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

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