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PHY 303L (9)
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Department
Physics
Course
PHY 303L
Professor
Storr
Semester
Summer

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rashid (sar3553) – HW6 DC Circuits – storr – (37537) 1 This print-out should have 15 questions. c Multiple-choice questions may continue on the next column or page – find all choices 18 Ω before answering. a 60 Ω 001 10.0 points b A loop circuit has a resistance of1R and a current of 1.9 A. The current is reduced to 41 Ω 83 Ω 1.7 A when an additional 3 Ω resistor is added in series with1R . 96 V What is the value of 1 ? Assume the inter- d S 1 nal resistance of the source of emf is zero. Find the resistance between points a and b. Correct answer: 25.5 Ω. Correct answer: 12.9024 Ω. Explanation: Explanation: c R1 3 Let : I1= 1.9 A, R a b I2= 1.7 A, and R = 3 Ω. 2 R2 R4 Let the current with R be I and the cur- EB 1 1 d S 1 rent with additional resistanc2 R b2 I . Then since the emf is a constant and the internal resistance of the battery is zero, we have Let : R1= 18 Ω, R2= 41 Ω, R3= 60 Ω, R4= 83 Ω, and I1R 1 I 2R +1R ) 2 I R = I R + I R E = 96 V. 1 1 2 1 2 2 Ohm’s law is V = I R. I2 R 1 I − I R2 A good rule of thumb is to eliminate junc- 1 2 tions connected by zero resistance. = 1.7 A (3 Ω) 1.9 A − 1.7 A R 1 = 25.5 Ω . a R2 R 3 b d c R 002 10.0 points 4 EB Four resistors are connected as shown in the figure. rashid (sar3553) – HW6 DC Circuits – storr – (37537) 2 The series connection of 2 and R 3ives the equivalent resistance Let : R 1 65.0 Ω, R 23= R 2 R 3 R 2 13.0 Ω, = 41 Ω + 60 Ω R = 13.0 Ω, 3 = 101 Ω. R 4 65.0 Ω, and Req= 63.0 Ω. The total resistanceab between a and b can be obtained by calculating the resistance in R R the parallel combination of the resistors R , 3 4 1 R 4 and R23 i.e., R R 1 2 1 1 1 1 = + + R ab R1 R 2 R 3 R4 R E R4(R 2 R )3+ R R1+ 4 (R 1 R2) 3 S = R 1 (4 + 2 ) 3 R1and R 2re in series: R = R 1 (4 + 2 ) 3 ab R4(R 2 R )3+ R R1+ 4 (R 1 R2) 3 R 12= R 1 R =265 Ω + 13 Ω = 78 Ω The denominator is as are 3 and R4: R4(R 2 R )3+ R R1+ 4 (R 1 R2) 3 R = R + R = 13 Ω + 65 Ω = (83 Ω)[41 Ω + 60 Ω] + (18 Ω)(83 Ω) 34 3 4 = 78 Ω . + (18 Ω)[41 Ω + 60 Ω] = 11695 Ω , Then R 12and R 34are in parallel: −1 so the equivalent resistance is R = 1 + 1 1234 R R (18 Ω)(83 Ω)[41 Ω + 60 Ω] 12 34 R ab= 2 1 1 −1 (11695 Ω ) = + 78 Ω 78 Ω = 12.9024 Ω . = 39 Ω and R is then in series wit1234 003 10.0 points Req = R + R 1234 The equivalent resistance of the circuit in R = R eq− R1234= 63 Ω − 39 Ω the figure is eq= 63.0 Ω. = 24 Ω . 13 Ω 65 Ω 004 (part 1 of 2) 10.0 points 65 Ω 13 Ω E R S 1 Ω 3 V Find the value of R. . V . 17 Ω . 9 42 17 16.4 Ω 12.1 Ω Correct answer: 24 Ω. Explanation: rashid (sar3553) – HW6 DC Circuits – storr – (37537) 3 What is the current through 12.1 Ω bottom- This now gives 2 loop equations and 1 junction right resistor? equation. The loop equations are Correct answer: 1.12793 A. V − I R − I R = 0 1 34 34 1 1 Explanation: V2+ I 34R 34− I2R 2 0 c d b e and the junction equation is I34+ I2= I 1 E R R E 1 3 4 2 Substituting 34 = I1−I 2n to the loop equa- R 1 R 2 tions, we have a g f h (R1+ R )34−1R 34 2= V 1 Let : R1= 16.4 Ω, −R 34 1+ (R 2 R )34= 2 . 2 R2= 12.1 Ω, R = 42.1 Ω, Rewriting and multiplying by factors, we have 3 2 R4= 17 Ω, (R1+ R 34 34 1− R 34I2= R 34 V1 −(R +1 )R34 34 1+(R +1 )(34+R )2 34 2 E1= 9.9 V, and = (R + R )V . 1 34 2 E2= 17.3 V. Subtracting, we have 2 Basic Concepts: Kirchhoff’s Laws [(R1+ R )34 + 2 ) −34 ]I 34 2 X = R 34V1+ (R 1 R )34, 2 V = 0 around a loop. for X I = 0 at a circuit junction. D = (R 1 R )34 + 2 ) − 34 2 Solution: A key simplification is to realize 34 = (16.4 Ω + 12.11 Ω)(12.1 Ω + 12.11 Ω) that R3and R a4e connected in parallel and 2 can be combined immediately, before applying − (12.11 Ω) 2 Kirchoff’s rules. = 543.575 Ω , c d b e I I we have 1 2 R 34V1 (R1+ R )34 2 34 34 I2= + E1 R I E2 D D (12.11 Ω)(9.9 V) R 1 R 2 = 543.575 Ω2 a f (16.4 Ω + 12.11 Ω)(17.3 V) h g + 543.575 Ω2 The combined resistance is given by = 1.12793 A . 1 1 1 R 3 R 4 R = R + R = R R 34 3 4 3 4 R 3 4 005 (part 2 of 2) 10.0 points R 34= R3+ R 4 What is the power dissipated in 17 Ω right- (42.1 Ω)(17 Ω) centered resistor? = 42.1 Ω + 17 Ω = 12.11 Ω. Correct answer: 0.784579 W. rashid (sar3553) – HW6 DC Circuits – storr – (37537) 4 Explanation: E1 In a similar way we can solve for the current r1 through R giving 34 I1 R 2 −1R V 1 2 a r2 b I34= R R + R (R + R ) 1 2 34 1 2 I2 (12.1 Ω)(9.9 V) − (16.4 Ω)(17.3 V) = E2 r (16.4 Ω)(12.1 Ω) + (12.11 Ω)(16.4 Ω + 12.1 Ω) 3 = −0.301578 A. I3 The voltage across R4is just Applying Kirchhoff’s loop rule clockwise around the right loop, V 4 V 34 = R 34 34 E1− r 1 1 r I2 20 so the power is V 2 P = 4 r1 1− E 1 (7 Ω)(2.4 A) − 29 V R 4 I2= r = (4.8 Ω) 2 2 = (−3.6521 V) = −2.54167 A, 17 Ω directed from b toward a with magnitude = 0.784579 W . 2.54167 A . 006 (part 1 of 2) 10.0 points 007 (part 2 of 2) 10.0 points Find I3. In th
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