17 - Kapoor (mk9499) homework17 Turner (60230) 1 This...

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Unformatted text preview: Kapoor (mk9499) 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. 1961 28 V 689 26 V 1524 2321 23 V a b What is the magnitude of the current in the loop? Correct answer: 0 . 00384911 A. Explanation: R 3 E 3 R 4 E 1 R 1 R 2 E 2 a b I Let : R 1 = 1524 , R 2 = 2321 , R 3 = 1961 , R 4 = 689 , E 1 = 26 V , E 2 = 23 V , and E 3 = 28 V . In this problem, three emfs are connected in series, so E total = E 2 + E 3- E 1 = 23 V + 28 V- 26 V = 25 V R T = R 1 + R 2 + R 3 + R 4 = 1524 + 2321 + 1961 + 689 = 6495 . The current in the circuit is counter- clockwise because E 2 + E 3 > E 1 , so I = E total R T = 25 V 6495 = . 00384911 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:- 17 . 7998 V. Explanation: V ab = E 2- I R 2- I R 1- E 1 = 23 V- (0 . 00384911 A) (2321 )- (0 . 00384911 A) (1524 )- 26 V =- 17 . 7998 V . 003 (part 3 of 3) 10.0 points What is the voltage drop across the top, left resistor? Correct answer: 7 . 54811 V. Explanation: V 3 = I R 3 = (0 . 00384911 A) (1961 ) = 7 . 54811 V . 004 10.0 points 5 . 2 V 1 . 9 V 3 . 2 V I 1 . 3 2 . 1 I 2 6 . 1 I 3 9 . 6 Find the current I 1 in the 0 . 3 resistor at the bottom of the circuit between the two Kapoor (mk9499) homework17 Turner (60230) 2 power supplies. Correct answer: 0 . 839048 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) Kirchhoffs law on the large outside loop gives ( R A + R B ) I 1 + R D I 3 = E 1 + E 2 . (2) Kirchhoffs law on the right-hand small loop gives R C I 2 + R D I 3 = E 3 . (3) Let : R A = 0 . 3 , R B = 2 . 1 , R C = 6 . 1 , R D = 9 . 6 , E 1 = 5 . 2 V , E 2 = 1 . 9 V , and E 3 = 3 . 2 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 vextendsingle...
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17 - Kapoor (mk9499) homework17 Turner (60230) 1 This...

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