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# oldhw15_pdf - mcconnell(kam2342 – oldhomework 15 –...

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mcconnell (kam2342) – oldhomework 15 – Turner – (60230) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points 3 . 4 Ω 2 . 4 Ω 11 . 6 Ω 27 . 6 V 13 . 8 V Find the current through the 11 . 6 Ω lower- right resistor. Correct answer: 1 . 5 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 = 27 . 6 V , E 2 = 13 . 8 V , r 1 = 3 . 4 Ω , r 2 = 2 . 4 Ω , and R = 11 . 6 Ω . From the junction rule, I = i 1 + i 2 . Applying Kirchhoff’s 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 ) = (27 . 6 V) (2 . 4 Ω) + (13 . 8 V) (3 . 4 Ω) (2 . 4 Ω) (11 . 6 Ω) + (3 . 4 Ω) (11 . 6 Ω + 2 . 4 Ω) = 1 . 5 A . 002 (part 1 of 3) 10.0 points Consider the circuit in the figure. 15 . 0 V 2 . 6 Ω 2 . 6 Ω 6 . 5 Ω 6 . 5 Ω 4 . 5 Ω 13 . 0 Ω 1 . 9 Ω Find the current in the 1.9 Ω resistor. Correct answer: 1 . 47449 A. Explanation: E R 1 R 2 R 3 R 4 R 5 R 6 R 7

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mcconnell (kam2342) – oldhomework 15 – Turner – (60230) 2 Let : R 1 = 2 . 6 Ω , R 2 = 2 . 6 Ω , R 3 = 6 . 5 Ω , R 4 = 6 . 5 Ω , R 5 = 4 . 5 Ω , R 6 = 13 . 0 Ω , R 7 = 1 . 9 Ω and Δ V = 15 . 0 V . R 3 and R 4 are in parallel: 1 R 34 = 1 R 3 + 1 R 4 R 34 = parenleftbigg 1 R 3 + 1 R 4 parenrightbigg - 1 = parenleftbigg 1 6 . 5 Ω + 1 6 . 5 Ω parenrightbigg - 1 = 3 . 25 Ω R 2 and R 34 are in series: R 234 = R 2 + R 34 = 2 . 6 Ω + 3 . 25 Ω = 5 . 85 Ω R 5 and R 6 are in parallel: 1 R 56 = 1 R 5 + 1 R 6 R 56 = parenleftbigg 1 R 5 + 1 R 6 parenrightbigg - 1 = parenleftbigg 1 4 . 5 Ω + 1 13 Ω parenrightbigg - 1 = 3 . 34286 Ω R 56 and R 7 are in series: R 567 = R 56 + R 7 = 3 . 34286 Ω + 1 . 9 Ω = 5 . 24286 Ω R 234 and R 567 are in parallel: 1 R 234567 = 1 R 234 + 1 R 567 R 234567 = parenleftbigg 1 R 234 + 1 R 567 parenrightbigg - 1 = parenleftbigg 1 5 . 85 Ω + 1 5 . 24286 Ω parenrightbigg - 1 = 2 . 76491 Ω R 1 and R 234567 are in series: R eq = R 1 + R 234567 = 2 . 6 Ω + 2 . 76491 Ω = 5 . 36491 Ω The current in the circuit is I = Δ V R eq = 15 V 5 . 36491 Ω = 2 . 79595 A Δ V 234567 = I R 234567 = (2 . 79595 A) (2 . 76491 Ω) = 7 . 73054 V Here the current is I 7 = I 567 = Δ V 234567 R 567 = (7 . 73054 V) (5 . 24286 Ω) = 1 . 47449 A . 003 (part 2 of 3) 10.0 points Find the potential difference across the 1.9 Ω resistor. Correct answer: 2 . 80153 V. Explanation: Solution: Δ V 7 = I 7 R 7 = (1 . 47449 A) (1 . 9 Ω) = 2 . 80153 V . 004 (part 3 of 3) 10.0 points Find the current in the 13.0 Ω resistor. Correct answer: 0 . 379154 A. Explanation: Solution: I 6 = Δ V 6 R 6 = (4 . 92901 V) (13 Ω) = 0 . 379154 A . 005 10.0 points What is the emf E of the battery at the lower left in the figure?
mcconnell (kam2342) – oldhomework 15 – Turner – (60230) 3 9 . 43 Ω 17 .

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