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Unformatted text preview: ragsdale (zdr82) – HW6 – ditmire – (58335) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The currents are flowing in the direction in dicated by the arrows. A negative current denotes flow opposite to the direction of the arrow. Assume the batteries have zero inter nal resistance. 17 . 3 Ω 21 . 6 Ω 5 . 4 V 19 . 9 V I I I Find the current through the 17 . 3 Ω resistor and the 5 . 4 V battery at the top of the circuit. Correct answer: 1 . 46243 A. Explanation: Let : R 1 = 17 . 3 Ω , R 2 = 21 . 6 Ω , E 1 = 5 . 4 V , and E 2 = 19 . 9 V . R 1 R 2 E 1 E 2 I 1 I 2 I 3 At nodes, we have I 1 − I 2 − I 3 = 0 . (1) Pay attention to the sign of the battery and the direction of the current in the figure. Us ing the lower circuit in the figure, we get E 2 + I 2 R 2 = 0 (2) so I 2 = −E 2 R 2 = − 19 . 9 V 21 . 6 Ω = − . 921296 A . Then, for the upper circuit E 1 − I 2 R 2 − I 1 R 1 = 0 . (3) E 1 + E 2 − I 1 R 1 = 0 . I 1 = E 1 + E 2 R 1 = 5 . 4 V + 19 . 9 V 17 . 3 Ω = 1 . 46243 A . Alternate Solution: Using the outside loop −E 1 − E 2 + I 1 R 1 = 0 (4) I 1 = E 1 + E 2 R 1 . 002 (part 2 of 2) 10.0 points Find the current through the 21 . 6 Ω resistor in the center of the circuit. Correct answer: − . 921296 A. Explanation: From Eq. (2) I 2 = − E 2 R 2 = − 19 . 9 V 21 . 6 Ω = − . 921296 A . 003 (part 1 of 3) 10.0 points E C D A B i 1 2 r i 2 9 r i 3 9 r i 4 2 r i 5 I ragsdale (zdr82) – HW6 – ditmire – (58335) 2 Find the ratio i 1 i 2 , where I is the current entering and leaving the battery. Hint: Apply the Kirchhoff’s law to the loop ACDA . 1. i 1 i 2 = 2 7 2. i 1 i 2 = 5 9 3. i 1 i 2 = 4 3 4. i 1 i 2 = 4 9 5. i 1 i 2 = 5 6 6. i 1 i 2 = 3 2 7. i 1 i 2 = 1 2 8. i 1 i 2 = 9 2 correct 9. i 1 i 2 = 4 7 10. i 1 i 2 = 7 6 Explanation: Let : R 1 = 2 r , R 2 = 9 r , R 3 = R 2 = 9 r , and R 4 = R 1 = 2 r . E C D A B i 1 R 1 i 2 R 2 i 3 R 3 i 4 R 4 i 5 I Basic Concept: DC Circuit. Solution: Based on Kirchhoff’s law, the equation for the loop ACDA is given by − i 1 R 1 + i 2 R 2 = 0 ⇒ i 1 i 2 = R 2 R 1 = 9 r 2 r = 9 2 . 004 (part 2 of 3) 10.0 points Find the magnitude of the current i 5 which flows from C to D . 1. i 5 = 1 4 I 2. i 5 = 1 5 I 3. i 5 = 7 11 I correct 4. i 5 = 1 11 I 5. i 5 = 1 8 I 6. i 5 = 3 7 I 7. i 5 = 3 11 I 8. i 5 = 3 13 I 9. i 5 = 1 13 I 10. i 5 = 1 3 I Explanation: i 1 = 9 2 i 2 I = i 1 + i 2 = 9 2 i 2 + i 2 = 11 2 i 2 , therefore i 2 = 2 11 I , and i 1 = 9 11 I . Following a similar analysis, one finds that i 4 i 3 = 9 2 , so that i 3 = 2 11 I and i 4 = 9 11 I . ragsdale (zdr82) – HW6 – ditmire – (58335) 3 Note: The junction equation at D is i 2 + i 5 = i 4 ⇒ i 5 = i 4 − i 2 , or = i 3 − i 1 = 2 11 I − 9 11 I = − 7 11 I  i 5  = 7 11 I ....
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This note was uploaded on 10/28/2010 for the course PHY 12343 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Turner

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