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. IDENTIFY: / IQ t = . Use d I nqvA = to calculate the drift velocity d . v SET UP: 28 3 5.8 10 m . n 19 1.60 10 C q . EXECUTE: (a) 2 420 C 8.75 10 A. 80(60 s) Q I t == = × (b) d . = 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 −− × = × ×× × EVALUATE: d v is smaller than in Example 25.1, because I is smaller in this problem. 25.8. IDENTIFY: / 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. SET UP: + Na and Cl each have magnitude of charge qe = + EXECUTE: (a) 16 16 19 total Cl Na ( ) (3.92 10 2.68 10 )(1.60 10 C) 0.0106 C. Qn 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. EVALUATE: The Cl ions have negative charge and move in the direction opposite to the conventional current direction. 25.12. IDENTIFY: E J ρ = , where / J IA = . The drift velocity is given by d . = SET UP: For copper, 8 1.72 10 m =×Ω . 28 3 8.5 10 / m . n EXECUTE: (a) 52 32 3.6 A 6.81 10 A/m . (2.3 10 m) I J A × (b) 85 2 (1.72 10 m)(6.81 10 A/m ) 0.012 V m. EJ × × = (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 vI × = 1333 min 22 hrs! t =≈ EVALUATE: The currents propagate very quickly along the wire but the individual electrons travel very slowly. 25.16. IDENTIFY: Apply L R A = and solve for L . SET UP: 2 /4 AD = , where 0.462 mm D = . EXECUTE: 8 (1.00 )( 4)(0.462 10 m) 9.75 m. 1.72 10 m RA L Ω× = ×Ω EVALUATE: The resistance is proportional to the length of the wire. 25.36. IDENTIFY: 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. SET UP: The current is counterclockwise, because the 16 V battery determines the direction of current flow. 25

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25-2 Chapter 25 EXECUTE: 16.0 V 8.0 V (1.6 5.0 1.4 9.0 ) 0 I +− + + + = 16.0 V 8.0 V 0.47 A 1.6 5.0 1.4 9.0 I == Ω+ (b) 16.0 V (1.6 ) ba VI V = , so 16.0 V (1.6 )(0.47 A) 15.2 V. ab a b VVV = (c) 8.0 V (1.4 5.0 ) ca V ++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. EVALUATE: (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. IDENTIFY and SET UP: By definition . P p LA = Use , and PV IEV L I J A = to rewrite this expression in terms of the specified variables. EXECUTE: (a) E is related to V and J is related to I, so use P = VI. This gives VI p LA = and so E JpE J LA = (b) J is related to I and ρ is related to R, so use 2 .
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## This note was uploaded on 09/22/2011 for the course PHY 09C 9c taught by Professor Zhu during the Spring '11 term at UC Davis.

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Physics9C_A_HW6_Soln - CURRENT RESISTANCE AND ELECTROMOTIVE...

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