Chapter_26_Solutions - 1 1 26.1: a) Req = + = 12.3 . 32 20...

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26.1: a) R eq = 1 32 + 1 20 1 = 12.3 . b) I = V R eq = 240 V 12.3 = 19.5 A. c) I 32 = V R = 240 V 32 = 7.5 A; I 20 = V R = 240 V 20 = 12 A.
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26.2: R eq = 1 R 1 + 1 R 2 1 = R 1 + R 2 R 1 R 2 1 R eq = R 1 R 2 R 1 + R 21 . R eq = R 1 R 2 R 1 + R 2 < R 1 and R eq = R 2 R 1 R 1 + R 2 < R 2
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26.3: For resistors in series, the currents are the same and the voltages add. a) true. b) false. c) I same, R different so P different; false. d) true. e) V = IR. I same, R different; false. f) Potential drops as move through each resistor in the direction of the current; false. g) Potential drops as move through each resistor in the direction of the current, so V P = I 2 R. b > V c ; false. h) true.
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26.4: a) False, current divides at junction a . b) True by charge conservation. c) True. V 1 = V 2 ,soI 1 R d) False. P = IV.V 1 = V 2 , but I 1 I 2 ,soP 1 P 2 . e) False. P = IV = V 2 R . Since R 2 > R 1 ,P 2 < P 1 f) True. Potential is independent of path. g) True. Charges lose potential energy (as heat) in . 1 R h) False. See answer to (g). i) False. They are at the same potential.
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26.5: a) R eq = 1 2.4 + 1 1.6 + 1 4.8 1 = 0.8 . b) I 2.4 = ε R = (28 V) (2.4 ) = 11.67 A; I 1.6 = R 1.6 = (28 V) (1.6 ) = 17.5 A; I 4.8 = R = (28 V) (4.8 ) = 5.83 A. c) I total = R total = (28 V) (0.8 ) = 35 A. d) When in parallel, all resistors have the same potential difference over them, so here all have V = 28 V. e) P = I 2 R = (1 1 . 6 7 A ) 2 (2.4 ) = 327 W; P 1.6 = I 2 R = (17.5 A) 2 (1.6 ) = 490 W; P 4.8 = I 2 R 4.8 = (5.83A) 2 (4.8 ) = 163 W. f) For resistors in parallel, the most power is dissipated through the resistor with the least resistance since P = I 2 R = V 2 R , withV = constant.
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26.6: a) R eq R i = 2.4 Ω+ 1.6 Ω + 4.8 Ω = 8.8 . b) The current in each resistor is the same and is I = ε R eq = 28 V 8.8 = 3.18 A. c) The current through the battery equals the current of (b), 3.18 A. d) V 2.4 = IR 2.4 = (3.18 A)(2.4 ) = 7.64 V;V 1.6 = IR = (3.18 A)(1.6 ) = 5.09 V;V 4.8 = IR 4.8 = (3.18 A)(4.8 ) = 15.3V. e) P = I 2 R = (3.18 A) 2 (2.4 ) = 24.3 W; P = I 2 R = (3.18 A) 2 (1.6 ) = 16.2 W; P 4.8 = I 2 R = (3.18 A) 2 (4.8 ) = 48.5 W. f) For resistors in series, the most power is dissipated by the resistor with the greatest resistance since P = I 2 R with I constant.
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26.7: a) P = V 2 R V = PR = (5.0 W)(15,000 ) = 274 V. b) P = V 2 R = (120 V) 2 9,000 = 1.6 W.
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26.8: R eq = 1 3.00 + 1 6.00 1 + 1 12.0 + 1 4.00 1 = 5.00 . I total = ε R total = (60.0 V) (5.00 ) = 12.0 A I 12 = 4 12 + 4 (12.0) = 3.00 A; I 4 = 12 12 + 4 = 9.00 A; I 3 = 6 3 + 6 = 8.00 A; I 6 = 3 3 + 6 = 4.00 A.
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26.9: R eq = 1 3.00 Ω+ 1.00 + 1 5.00 7.00 1 = 3.00 . I total = ε R total = (48.0 V) (3.00 ) = 16.0 A . I 5 = I 7 = 4 4 + 12 (16.0) = 4.00 A; I 1 = I 3 = 12 4 + 12 = 12.0 A .
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26.10: a) The three resistors R are in parallel, so: 2 ,R 3 and R 4 R 234 = 1 R 2 + 1 R 3 + 1 R 4 1 = 1 8.20 + 1 1.50 + 1 4.50 1 = 0.99 R eq = R 1 + R 234 = 3.50 Ω + 0.99 Ω = 4.49 . b) I 1 = ε R eq = 6.0 V 4.49 = 1.34 A V 1 = I 1 R 1 = (1.34 A) (3.50 ) = 4.69 V. V R 234 = I 1 R 234 = (1.34 A) (0.99 ) = 1.33 V I 2 = V R 234 R 2 = 1.33 V 8.20 = 0.162 A, I 3 = V R 234 R 3 = 1.33 V 1.50 = 0.887 A and I 4 = V R 234 R 4 = 1.33 V 4.50 = 0.296 A.
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26.11: Using the same circuit as in Problem 27.10, with all resistances the same: R eq = R 1 + R 234 = R 1 + 1 R 2 + 1 R 3 + 1 R 4 1 = 4.50 Ω+ 3 4.50 1 = 6.00 .
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Chapter_26_Solutions - 1 1 26.1: a) Req = + = 12.3 . 32 20...

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