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Physics9C_A_HW6_Soln

Physics9C_A_HW6_Soln - CURRENT RESISTANCE AND ELECTROMOTIVE...

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25-1 C URRENT , R ESISTANCE , AND E LECTROMOTIVE F ORCE 25.2. I DENTIFY : / I Q t = . Use d I n q v A = to calculate the drift velocity d . v S ET U P : 28 3 5.8 10 m . n = × 19 1.60 10 C q = × . E XECUTE : (a) 2 420 C 8.75 10 A. 80(60 s) Q I t = = = × (b) d . I n q v A = This gives 2 6 d 28 19 3 2 8.75 10 A 1.78 10 m s. (5.8 10 )(1.60 10 C)( (1.3 10 m) ) I v nqA π × = = = × × × × E VALUATE : d v is smaller than in Example 25.1, because I is smaller in this problem. 25.8. I DENTIFY : / I Q t = . Positive charge flowing in one direction is equivalent to negative charge flowing in the opposite direction, so the two currents due to Cl and + Na are in the same direction and add. S ET U P : + Na and Cl each have magnitude of charge q e = + E XECUTE : (a) 16 16 19 total Cl Na ( ) (3.92 10 2.68 10 )(1.60 10 C) 0.0106 C. Q n n e = + = × + × × = Then total 0.0106 C 0.0106A 10.6 mA. 1.00 s Q I t = = = = (b) Current flows, by convention, in the direction of positive charge. Thus, current flows with Na + toward the negative electrode. E VALUATE : The Cl ions have negative charge and move in the direction opposite to the conventional current direction. 25.12. I DENTIFY : E J ρ = , where / J I A = . The drift velocity is given by d . I n q v A = S ET U P : For copper, 8 1.72 10 m ρ = × Ω⋅ . 28 3 8.5 10 /m . n = × E XECUTE : (a) 5 2 3 2 3.6 A 6.81 10 A/m . (2.3 10 m) I J A = = = × × (b) 8 5 2 (1.72 10 m)(6.81 10 A/m ) 0.012 V m. E J ρ = = × Ω⋅ × = (c) The time to travel the wire’s length l is 28 3 19 3 2 4 d (4.0 m)(8.5 10 m )(1.6 10 C)(2.3 10 m) 8.0 10 s. 3.6 A ln q A l t v I × × × = = = = × 1333 min 22 hrs! t = E VALUATE : The currents propagate very quickly along the wire but the individual electrons travel very slowly. 25.16. I DENTIFY : Apply L R A ρ = and solve for L . S ET U P : 2 /4 A D π = , where 0.462 mm D = . E XECUTE : 3 2 8 (1.00 )( 4)(0.462 10 m) 9.75 m. 1.72 10 m RA L π ρ × = = = × Ω⋅ E VALUATE : The resistance is proportional to the length of the wire. 25.36. I DENTIFY : The sum of the potential changes around the circuit loop is zero. Potential decreases by IR when going through a resistor in the direction of the current and increases by E when passing through an emf in the direction from the to + terminal. S ET U P : The current is counterclockwise, because the 16 V battery determines the direction of current flow. 25
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25-2 Chapter 25 E XECUTE : 16.0 V 8.0 V (1.6 5.0 1.4 9.0 ) 0 I + Ω + Ω + Ω + Ω = 16.0 V 8.0 V 0.47 A 1.6 5.0 1.4 9.0 I = = Ω + Ω + Ω + (b) 16.0 V (1.6 ) b a V I V + Ω = , so 16.0 V (1.6 )(0.47 A) 15.2 V. a b ab V V V = = = (c) 8.0 V (1.4 5.0 ) c a V I V + + Ω + Ω = so (5.0 )(0.47 A) (1.4 )(0.47 A) 8.0 V 11.0 V. ac V = + + = (d) The graph is sketched in Figure 25.36. E VALUATE : (0.47 A)(9.0 ) 4.2 V. cb V = Ω = The potential at point b is 15.2 V below the potential at point a and the potential at point c is 11.0 V below the potential at point a , so the potential of point c is 15.2 V 11.0 V 4.2 V = above the potential of point b . Figure 25.36 25.47. I DENTIFY and S ET U P : By definition . P p LA = Use , and P VI E VL I JA = = = to rewrite this expression in terms of the specified variables.
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