Chapter 26 - 26.1 a 3 12 20 1 32 1 1 eq Ω = = R b A 5 19...

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Unformatted text preview: 26.1: a) . 3 . 12 20 1 32 1 1 eq Ω = + =- R b) . A 5 . 19 3 . 12 V 240 eq = Ω = = R V I c) . A 12 20 V 240 ; A 5 . 7 32 V 240 20 32 = Ω = = = Ω = = Ω Ω R V I R V I 26.2: . 1 1 2 1 2 1 eq 1 2 1 2 1 1 2 1 eq R R R R R R R R R R R R + = ⇒ + = + =-- . and 2 2 1 1 2 eq 1 2 1 2 1 eq R R R R R R R R R R R R < + = < + = ⇒ 26.3: For resistors in series, the currents are the same and the voltages add. a) true. b) false. c) . 2 R I P = 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 ; c V V b false. h) true. 26.4: a) False, current divides at junction a . b) True by charge conservation. c) True. R I V V 1 so , 2 1 ∝ = d) False. . so , but , . 2 1 2 1 2 1 P P I I V V IV P ≠ ≠ = = e) False. . , Since . 1 2 1 2 2 P P R R IV P R V < = = 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. 26.5: a) . 8 . 8 . 4 1 6 . 1 1 4 . 2 1 1 eq Ω = Ω + Ω + Ω =- R b) ; A 5 . 17 ) 6 . 1 ( ) V 28 ( ; A 67 . 11 ) 4 . 2 ( ) V 28 ( 6 . 1 6 . 1 4 . 2 4 . 2 = Ω = = = Ω = = R ε I R ε I . A 83 . 5 ) 8 . 4 ( ) V 28 ( 8 . 4 8 . 4 = Ω = = R ε I c) . A 35 ) 8 . ( ) V 28 ( = Ω = = total total R ε I d) When in parallel, all resistors have the same potential difference over them, so here all have V = 28 V. e) = Ω = = = Ω = = ) 6 . 1 ( ) A 5 . 17 ( ; W 327 ) 4 . 2 ( ) A 67 . 11 ( 2 6 . 1 2 6 . 1 2 4 . 2 2 4 . 2 R I P R I P W. 163 ) 8 . 4 ( ) A 83 . 5 ( ; W 490 2 8 . 4 2 8 . 4 = Ω = = R I P f) For resistors in parallel, the most power is dissipated through the resistor with the least resistance since constant. with , 2 2 = = = V R V R I P 26.6: a) . 8 . 8 8 . 4 6 . 1 4 . 2 eq Ω = Ω + Ω + Ω = Σ = i R R b) The current in each resistor is the same and is . A 18 . 3 8 . 8 V 28 eq = Ω = = R ε I c) The current through the battery equals the current of (b), 3.18 A. d) = Ω = = = Ω = = ) 6 . 1 )( A 18 . 3 ( ; V 64 . 7 ) 4 . 2 )( A 18 . 3 ( 6 . 1 6 . 1 4 . 2 4 . 2 IR V IR V . V 3 . 15 ) 8 . 4 )( A 18 . 3 ( ; V 09 . 5 8 . 4 8 . 4 = Ω = = IR V e) = Ω = = = Ω = = ) 6 . 1 ( ) A 18 . 3 ( ; W 3 . 24 ) 4 . 2 ( ) A 18 . 3 ( 2 6 . 1 2 6 . 1 2 4 . 2 2 4 . 2 R I P R I P . W 5 . 48 ) 8 . 4 ( ) A 18 . 3 ( ; W 2 . 16 2 8 . 4 2 8 . 4 = Ω = = R I P f) For resistors in series, the most power is dissipated by the resistor with the greatest resistance since . constant with 2 I R I P = 26.7: a) . V 274 ) 000 , 15 )( W . 5 ( 2 = Ω = = ⇒ = PR V R V P b) . W 6 . 1 000 , 9 ) V 120 ( 2 2 = Ω = = R V P 26.8: Ω = Ω + Ω +...
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Chapter 26 - 26.1 a 3 12 20 1 32 1 1 eq Ω = = R b A 5 19...

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