741_PartUniversity Physics Solution

# 741_PartUniversity Physics Solution - Current Resistance...

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Current, Resistance, and Electromotive Force 25-5 EXECUTE: The ratio of the current at 20 C ° to that at the higher temperature is (0.860 A) (0.220 A) 3.909. = 0 20 1 ( ) 3.909 T R TT R α =+ − = , where 0 20 C. T = ° 20 0 31 / 1 3.909 1 20 C 666 C. 4.5 10 (C ) T RR −− =°+ = ° × ° EVALUATE: As the temperature increases, the resistance increases and for constant applied voltage the current decreases. The resistance increases by nearly a factor of four. 25.23. IDENTIFY: Relate resistance to resistivity. SET UP: R = ρ L/A EXECUTE: (a) R = L/A = (0.60 m)(0.25 m)/(0.12 m) 2 = 10.4 (b) R = L/A = (0.60 m)(0.12 m)/(0.12 m)(0.25 m) = 2.4 EVALUATE: The resistance is greater for the faces that are farther apart. 25.24. IDENTIFY: Apply L R A = and VI R = . SET UP: 2 Ar π = EXECUTE: 42 7 (4.50 V) (6.54 10 m) 1.37 10 m. (17.6 A)(2.50 m) RA VA LI L × === = × Ω EVALUATE: Our result for shows that the wire is made of a metal with resistivity greater than that of good metallic conductors such as copper and aluminum. 25.25. IDENTIFY and SET UP: Eq. (25.5) relates the electric field that is given to the current density. V = EL gives the potential difference across a length L of wire and Eq. (25.11) allows us to calculate R. EXECUTE: (a) Eq. (25.5): / so / EJ J E ρρ == From Table 25.1 the resistivity for gold is 8 2.44 10 m. × Ω⋅ 72 8 0.49 V/m 2.008 10 A/m m E J = × ×Ω () ( ) 2 27 23 0.41 10 m 11 A IJ AJ r ππ = × × = (b) ( ) 0.49 V/m 6.4 m 3.1 V VE L = (c) We can use Ohm’s law (Eq. (25.11)): . R = 3.1 V 0.28 11 A V R I = Ω EVALUATE: We can also calculate R from the resistivity and the dimensions of the wire (Eq. 25.10): ( ) 8 2 2 3 m 6.4 m 0.28 , 0.42 10 m LL R = = Ω × which checks. 25.26. IDENTIFY and SET UP: Use V = EL to calculate E and then / E J = to calculate . EXECUTE: (a) 0.938 V 1.25 V/m 0.750 m V E L = (b) 8 1.25 V/m so 2.84 10 m 4.40 10 A/m E J = = × Ω × EVALUATE: This value of is similar to that for the good metallic conductors in Table 25.1. 25.27. IDENTIFY: Apply 00 1 R RT T =+− ⎡⎤ ⎣⎦ to calculate the resistance at the second temperature. (a) SET UP: 1 0.0004 C (Table 25.1). Let 0 be 0.0 C and be 11.5 C. °° EXECUTE: ( ) 0 1 0 100.0 99.54 1 1+ 0.0004 C 11.5 C R R Ω = Ω +− (b) SET UP: 1 0.0005 C =− ° (Table 25.2). Let 0 0.0 C and 25.8 C. = °= ° EXECUTE: ( ) ( ) 1 1 0.0160 1+ 0.0005 C 25.8 C 0.0158 =+−= Ω ° ° = Ω ⎢⎥ EVALUATE: Nichrome, like most metallic conductors, has a positive and its resistance increases with temperature. For carbon, is negative and its resistance decreases as T increases.

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25-6 Chapter 25 25.28. IDENTIFY: 00 [1 ( )] T R RT T α =+− SET UP: 0 217.3 R . 215.8 T R . For carbon, 1 0.00050 (C ) . =− ° EXECUTE: 0 0 1 ( / ) 1 (215.8 / 217.3 ) 1 13.8 C 0.00050 (C ) T RR TT −Ω Ω −= = = ° ° . 13.8 C 4.0 C 17.8 C. T =+ = °° ° EVALUATE: For carbon, is negative so R decreases as T increases. 25.29. IDENTIFY and SET UP: Apply L R A ρ = to determine the effect of increasing A and L .
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741_PartUniversity Physics Solution - Current Resistance...

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