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PHY303LHW6S

PHY303LHW6S - rosen(arr956 Homework 06 Chiu(58295 This...

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rosen (arr956) – Homework 06 – Chiu – (58295) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A battery with an emf of 13 V and internal resistance of 0 . 43 Ω is connected across a load resistor R . If the current in the circuit is 2 . 16 A, what is the value of R ? Correct answer: 5 . 58852 Ω. Explanation: Let : E = 13 V , I = 2 . 16 A , and R i = 0 . 43 Ω . The electromotive force E is given by E = I ( R + R i ) R = E I R i = 13 V 2 . 16 A 0 . 43 Ω = 5 . 58852 Ω . 002 (part 2 of 2) 10.0 points What power is dissipated in the internal re- sistance of the battery? Correct answer: 2 . 00621 W. Explanation: The power dissipation due to the internal resistance is P = I 2 R i = (2 . 16 A) 2 (0 . 43 Ω) = 2 . 00621 W . 003 (part 1 of 2) 10.0 points Four resistors are connected as shown in the figure. 92 V S 1 c d a b 17 Ω 42 Ω 52 Ω 82 Ω Find the resistance between points a and b . Correct answer: 12 . 2464 Ω. Explanation: E B S 1 c d a b R 1 R 2 R 3 R 4 Let : R 1 = 17 Ω , R 2 = 42 Ω , R 3 = 52 Ω , R 4 = 82 Ω , and E = 92 V . Ohm’s law is V = I R . A good rule of thumb is to eliminate junc- tions connected by zero resistance. E B a d b c R 1 R 2 R 3 R 4

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rosen (arr956) – Homework 06 – Chiu – (58295) 2 The series connection of R 2 and R 3 gives the equivalent resistance R 23 = R 2 + R 3 = 42 Ω + 52 Ω = 94 Ω . The total resistance R ab between a and b can be obtained by calculating the resistance in the parallel combination of the resistors R 1 , R 4 , and R 23 ; i.e. , 1 R ab = 1 R 1 + 1 R 2 + R 3 + 1 R 4 = R 4 ( R 2 + R 3 ) + R 1 R 4 + R 1 ( R 2 + R 3 ) R 1 R 4 ( R 2 + R 3 ) R ab = R 1 R 4 ( R 2 + R 3 ) R 4 ( R 2 + R 3 ) + R 1 R 4 + R 1 ( R 2 + R 3 ) The denominator is R 4 ( R 2 + R 3 ) + R 1 R 4 + R 1 ( R 2 + R 3 ) = (82 Ω)[42 Ω + 52 Ω] + (17 Ω) (82 Ω) + (17 Ω) [42 Ω + 52 Ω] = 10700 Ω 2 , so the equivalent resistance is R ab = (17 Ω) (82 Ω) [42 Ω + 52 Ω] (10700 Ω 2 ) = 12 . 2464 Ω . 004 (part 2 of 2) 10.0 points What is the current in the 42 Ω resistor? Correct answer: 0 . 978723 A. Explanation: The voltages across R 2 and R 3 , respec- tively, (the voltage between a and b ) are V ab = V 23 = 92 V , and we have I 23 = I 3 = I 2 = V ab R 23 = 92 V 94 Ω = 0 . 978723 A . 005 10.0 points The following diagram shows a closed elec- trical circuit. The ammeter in the center of the resistive network reads zero amperes. E S 1 A 4 Ω 6 Ω 6 Ω R x Find the electric resistance R X . 1. R X = 9 Ω . correct 2. R X = 12 Ω . 3. R X = 21 Ω . 4. R X = 18 Ω . 5. R X = 4 Ω . 6. R X = 7 Ω . 7. R X = 24 Ω . 8. R X = 14 Ω . 9. R X = 5 Ω . 10. R X = 17 Ω . Explanation: E S 1 u x y I u R 1 I u R 2 I R 3 I R x I A 0 A Let : R 1 = 4 Ω , R 2 = 6 Ω , and R 3 = 6 Ω . If the ammeter reads zero I A = 0 A, the two ends of the ammeter should be an equipoten- tial: V u A = V A .
rosen (arr956) – Homework 06 – Chiu – (58295) 3 This means that the potential drop from x or y to each side of the ammeter through either the upper u or lower part of the circuit must be the same. V x V u = I u R 1 (1) V x V = I R 3 (2) V u V y = I u R 2 (3) V V y = I R x (4) Setting the potential across the ammeter equal V u = V ( i.e., Eq. 1 = 2 and Eq. 3 = 4), we have I u R 1 = I R 3 and I u R 2 = I R x , so I u I = R 3 R 1 = R x R 2 R x = R 2 R 3 R 1 = (6 Ω) (6 Ω) 4 Ω = 9 Ω .

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PHY303LHW6S - rosen(arr956 Homework 06 Chiu(58295 This...

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