Physics Solution Manual for 1100 and 2101

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Unformatted text preview: terial. In order to rank the resistances, we need to evaluate L and A for each configuration in terms of L0, the unit of length. Resistance a b c 2 = ρ L L0 × 2 L0 0 1 L0 R=ρ = ρ 8L 2 L0 × 4 L0 0 1 2 L0 R=ρ = ρ 2L L0 × 4 L0 0 R=ρ 4 L0 Rank 1 3 2 Therefore, we expect that a has the largest resistance, followed by c, and then by b. b. Equation 20.2 states that the current I is equal to the voltage V divided by the resistance, I = V / R . Since the current is inversely proportional to the resistance, the largest current arises when the resistance is smallest, and vice versa. Thus, we expect that b has the largest current, followed by c, and then by a. SOLUTION a. The resistances can be found by using the results from the REASONING: Chapter 20 Problems a b c 1061 2 2 −2 R = ρ = 1.50 × 10 Ω ⋅ m = 0.600 Ω −2 L 5.00 × 10 m 0 1 1 −2 R = ρ = 1.50 × 10 Ω ⋅ m = 0.0375 Ω −2 8L 8 × 5.00 × 10 m 0 1 1 −2 R = ρ = 1.50 × 10 Ω ⋅ m = 0.150 Ω −2 2L 0 2 × 5.00 × 10 m ( ) ( ) ( ) b. The current in each case is given by Equation 20.2, where the value of the resistance is obtained from part (a): 3.00 V V I= = = 5.00 A a R 0.600 Ω 3.00 V V I= = = 80.0 A b R 0.0375 Ω 3.00 V V I= = = 20.0 A c R 0.150 Ω ______________________________________________________________________________ 11. REASONING The resistance R of a wire that has a length L and a cross-sectional area A is L given by Equation 20.3 as R = ρ . Both wires have the same length and cross-sectional A area. Only the resistivity ρ of the wire differs, and Table 20.1 gives the following values: ρAluminum = 2.82 × 10−8 ·m and ρCopper = 1.72 × 10−8 ·m. Applying Equation 20.3 to both wires and dividing the two equations will allow us to eliminate the unknown length and cross-sectional area algebraically and solve for the resistance of the copper wire. SOLUTION Applying Equation 20.3 to both wires gives RCopper = ρCopper L A and RAluminum = ρ Aluminum L A Dividing these two equations, eliminating L and A algebraically, and solving the result for RCopper give RCopper ρ Copper L / A ρ Copper = = RAluminum ρ Aluminum L / A ρ Aluminum ρCopper 1.72 ×10−8 Ω ⋅ m RCopper = RAluminum = ( 0.20 Ω ) = 0.12 Ω −8 2.82 ×10 Ω ⋅ m ρ Aluminum 1062 ELECTRIC CIRCUITS 12. REASONING AND SOLUTION The resistance of the cable is R= V ρL = I A Since A = π r2, the radius of the cable is (1.72 × 10–8 Ω ⋅ m ) ( 0.24 m )(1200 A ) = π (1.6 × 10 –2 V ) ρ LI = r= πV 9.9 × 10 –3 m ______________________________________________________________________________ 13. REASONING AND SOLUTION From Equation 20.5 we have that R = R0[1 + α (T – T0)]. Solving for T gives 99.6 Ω −1 125 Ω = 20.0 °C + = −34.6 °C T = T0 + α 3.72 × 10−3 (C°) −1 ______________________________________________________________________________ R −1 R0 14. REASONING AND SOLUTION Using Equation 20.3 and the resistivity of aluminum from Table 20.1, we find R= ρL ( 2.82 × 10–8 Ω ⋅ m ) (10.0 × 103 m ) = = 0.58 Ω A 4.9 × 10 –4 m 2 ______________________________________________________________________________ 15. 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 . Solving for A, we have A = ρ L / R . We can use this expression to find the ratio of the cross-sectional area of the aluminum wire to that of the copper wire. SOLUTION Forming the ratio of the areas and using resistivity values from Table 20.1, we have Aaluminum ρ aluminum L / R ρ aluminum 2.82 × 10 –8 Ω ⋅ m = = = = 1.64 ρcopper L / R ρ copper Acopper 1.72 × 10 –8 Ω ⋅ m ______________________________________________________________________________ Chapter 20 Problems 1063 16. REASONING The resistance R of the spooled wire decreases as its length L decreases, L according to R = ρ (Equation 20.3). The resistivity ρ and cross-sectional area A of the A wire do not change. Because the same battery is used, the potential difference V across the V wire is the same in both cases. Therefore, R = (Equation 20.2) explains why the current I I increases as the resistance R of the wire decreases. RA L (Equation 20.3) for L, we obtain L = . Thus, the initial A ρ length L0 and final length Lf of the wire are given by SOLUTION Solving R = ρ L0 = R0 A ρ and Lf = Rf A ρ (1) where R0 is the initial resistance, and Rf the final resistance, of the wire. Taking the ratio of Equations (1) eliminates the unknown quantities A and ρ, allowing us to solve for Lf in terms of L0 and the initial and final resistances: Rf A ρ Lf R = =f L0 R0 A R0 or ρ From R = R Lf = f L0 R 0 (2) V (Equation 20.2), the initial resistance R0 and final resistance Rf of the spooled I wire are R0 = V I0 and Rf = V If (3) Substituting Equations (3) into Equation (2) yields V I Lf = f V I 0 L = I 0 L = 2.4 A 75 m = 58 m ) ( 0 I f 0 3.1 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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