Physics Solution Manual for 1100 and 2101

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Unformatted text preview: _________________ 1064 ELECTRIC CIRCUITS 17. REASONING Assuming that the resistance is R at a temperature T and R0 at a temperature T0, we can write the percentage change p in resistance as p= R − R0 R0 × 100 Equation 20.5, on the other hand, gives the resistance as a function of temperature as follows: R = R0 1 + α (T − T0 ) where α is the temperature coefficient of resistivity. Substituting this expression into the expression for the percentage change in resistance gives p= R − R0 R0 × 100 = R0 + R0α (T − T0 ) − R0 R0 ×100 = α (T − T0 ) 100 (1) The change in temperature is unknown, but it is the same for both wires. Therefore, we will apply Equation (1) to each wire and divide the two expressions to eliminate the unknown change in temperature. From the result we will be able to calculate the percentage change in the resistance of the tungsten wire. SOLUTION Applying Equation (1) to both wires gives pTungsten = α Tungsten (T − T0 )100 and pGold = α Gold (T − T0 )100 Dividing these two expressions, eliminating (T − T0) algebraically, and solving for pTungsten give pTungsten α Tungsten (T − T0 )100 α Tungsten = = α Gold pGold α Gold (T − T0 )100 pTungsten α Tungsten = pGold α Gold −1 0.0045 ( C° ) = 9.3% = ( 7.0% ) −1 0.0034 ( C° ) 18. REASONING AND SOLUTION Suppose that when the initial temperature of the wire is T0 the resistance is R0, and when the temperature rises to T the resistance is R. The relation between temperature and resistance is given by Equation 20.5 as R = R0[1 + α (T – T0)], where α is the temperature coefficient of resistivity. The initial and final resistances are related to the voltage and current as R0 = V/I0 and R = V/I, where the voltage V across the Chapter 20 Problems 1065 wire is the same in both cases. Substituting these values for R0 and R into Equation 20.5 and solving for T, we arrive at T = T0 + I0 − 1 I = 360 ° C 4.5 × 10 −4 (C° ) −1 ______________________________________________________________________________ 19. α = 20 ° C + 1.50 A − 1 1.30 A SSM WWW REASONING The resistance of a metal wire of length L, cross-sectional area A and resistivity ρ is given by Equation 20.3: R = ρ L / A . The volume V2 of the new wire will be the same as the original volume V1 of the wire, where volume is the product of length and cross-sectional area. Thus, V1 = V2 or A1L1 = A2L2. Since the new wire is three times longer than the first wire, we can write A1L1 = A2 L2 = A2 (3L1) or A2 = A1 / 3 We can form the ratio of the resistances, use this expression for the area A2 , and find the new resistance. SOLUTION The resistance of the new wire is determined as follows: R2 R1 = ρ L2 / A2 L2 A1 ( 3L1 ) A1 = = =9 ρ L1 / A1 L1 A2 L1 ( A1 / 3) Solving for R2 , we find that R2 = 9 R1 = 9(21.0 Ω ) = 189 Ω ______________________________________________________________________________ 20. REASONING The length L of the wire is related to its resistance R and cross-sectional area A by L = A R / ρ (see Equation 20.3), where ρ is the resistivity of tungsten. The resistivity is known (see Table 20.1), and the cross-sectional area can be determined since the radius of the wire is given. The resistance can be obtained from Ohm’s law as the voltage divided by the current. SOLUTION The length L of the wire is L= AR ρ (1) 1066 ELECTRIC CIRCUITS Since the cross-section of the wire is circular, its area is A = π r 2 , where r is the radius of the wire. According to Ohm’s law (Equation 20.2), the resistance R is related to the voltage V and current I by R = V / I . Substituting the expressions for A and R into Equation (1) gives (π r 2 ) V 2 120 V ( −3 π 0.0030 ×10 m ) AR I= 1.24 A = 0.049 m L= = ρ ρ 5.6 ×10−8 Ω ⋅ m ______________________________________________________________________________ 21. REASONING AND SOLUTION The mass m of the aluminum wire is equal to the density d of aluminum times the volume of the wire. The wire is cylindrical, so its volume is equal to the cross-sectional area A times the length L; m = dAL. The cross-sectional area of the wire is related to its resistance R and length L by Equation 20.3; R = ρL/A, where ρ is the resistivity of aluminum (see Table 20.1). Therefore, the mass of the aluminum wire can be written as ρL L m = dAL = d R The resistance R is given by Ohm’s law as R = V/I, so the mass of the wire becomes d ρ L2 I ρL m=d L= V R (2700 kg/m3 )(2.82 × 10−8 Ω ⋅ m)(175 m) 2 (125 A) = 9.7 × 102 kg 0.300 V ______________________________________________________________________________ m= 22. REASONING AND SOLUTION According to Equation 20.6c, the power delivered to the iron is 2 V 2 (120 V) P= = = 6.0 × 10 2 W R 24 Ω ______________________________________________________________________________ 23. REASONING AND SOLUTION a. According to Equation 20.6c, the resistance is 2 V 2 (12 V ) R= = = 4.4 Ω P 33 W Chapter 20 Problems 1067 b. According to Equation 20.6a, the current is P 33 W = = 2.8 A...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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