14. Complex circuits-solutions

14. Complex circuits-solutions - gilvin (jg47854) 14....

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Unformatted text preview: gilvin (jg47854) 14. Complex circuits meyers (21235) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. This homework is due Tuesday, March 29, at midnight Tucson time. 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. 9 . 9 19 . 4 4 V 15 . 4 V I I I Find the current through the 9 . 9 resistor and the 4 V battery at the top of the circuit. Correct answer: 1 . 9596 A. Explanation: Let : R 1 = 9 . 9 , R 2 = 19 . 4 , E 1 = 4 V , and E 2 = 15 . 4 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 =- 15 . 4 V 19 . 4 =- . 793814 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 = 4 V + 15 . 4 V 9 . 9 = 1 . 9596 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 19 . 4 resistor in the center of the circuit. Correct answer:- . 793814 A. Explanation: From Eq. (2) I 2 =- E 2 R 2 =- 15 . 4 V 19 . 4 =- . 793814 A . 003 10.0 points gilvin (jg47854) 14. Complex circuits meyers (21235) 2 4 . 4 3 . 4 15 . 6 25 V 12 . 5 V Find the current through the 15 . 6 lower- right resistor. Correct answer: 1 . 02459 A. Explanation: r 1 r 2 R E 1 E 2 A D E B C F i 1 i 2 I Let : E 1 = 25 V , E 2 = 12 . 5 V , r 1 = 4 . 4 , r 2 = 3 . 4 , and R = 15 . 6 . From the junction rule, I = i 1 + i 2 . Applying Kirchhoffs loop rule, we obtain two equations: E 1 = i 1 r 1 + I R (1) E 2 = i 2 r 2 + I R = ( I- i 1 ) r 2 + I R =- i 1 r 2 + I ( R + r 2 ) , (2) Multiplying Eq. (1) by r 2 , Eq. (2) by r 1 , E 1 r 2 = i 1 r 1 r 2 + r 2 I R E 2 r 1 =- i 1 r 1 r 2 + I r 1 ( R + r 2 ) Adding, E 1 r 2 + E 2 r 1 = I [ r 2 R + r 1 ( R + r 2 )] I = E 1 r 2 + E 2 r 1 r 2 R + r 1 ( R + r 2 ) = (25 V) (3 . 4 ) + (12 . 5 V) (4 . 4 ) (3 . 4 ) (15 . 6 ) + (4 . 4 ) (15 . 6 + 3 . 4 ) = 1 . 02459 A . 004 (part 1 of 2) 10.0 points An automobile battery has an emf of 15 . 5 V and an internal resistance of 0 . 116 . The headlights have total constant resistance 4 . 95 . What is the potential difference across the headlight bulbs when they are the only load on the battery? Correct answer: 15 . 1451 V. Explanation: Let : E = 15 . 5 V , R i = 0 . 116 , and R h = 4 . 95 ....
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14. Complex circuits-solutions - gilvin (jg47854) 14....

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