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751_PartUniversity Physics Solution

# 751_PartUniversity Physics Solution - Current Resistance...

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Current, Resistance, and Electromotive Force 25-15 But 2 1 r h r β + = 1 2 2 1 1 2 1 2 1 2 1 1 h r r h R r r r r rr rr ρ ρ ρ πβ π π ⎞⎛ = = = ⎟⎜ ⎠⎝ (b) E VALUATE : Let 1 2 . r r r = = Then 2 2 / / where and . R h r L A A r L h ρ π ρ π = = = = This agrees with Eq. (25.10). 25.64. I DENTIFY : Divide the region into thin spherical shells of radius r and thickness dr . The total resistance is the sum of the resistances of the thin shells and can be obtained by integration. S ET U P : / I V R = and 2 /4 J I r π = , where 2 4 r π is the surface area of a shell of radius r . E XECUTE : (a) 2 2 1 1 1 . 4 4 4 4 4 b b a a dr dr b a dR R r r r a b ab ρ ρ ρ ρ ρ π π π π π = = = − = = (b) 4 ( ) ab ab V V ab I R b a π ρ = = and 2 2 4 . ( )4 ( ) ab ab I V ab V ab J A b a r b a r π ρ π ρ = = = (c) If the thickness of the shells is small, then 2 4 4 ab a π π is the surface area of the conducting material. 2 1 1 ( ) 4 4 4 b a L L R a b ab a A ρ ρ ρ ρ π π π = = = , where . L b a = E VALUATE : The current density in the material is proportional to 2 1/ . r 25.65. I DENTIFY and S ET U P : Use E J ρ = to calculate the current density between the plates. Let A be the area of each plate; then I = JA. E XECUTE : 0 0 and E Q J E K KA σ ρ = = = P P Thus 0 0 and , Q Q J I JA KA K ρ ρ = = = P P as was to be shown. E VALUATE : 0 0 / and / / C K A d V Q C Qd K A = = = P P so the result can also be written as / . I VA d ρ = The resistance of the dielectric is / / , R V I d A ρ = = which agrees with Eq. (25.10). 25.66. I DENTIFY : As the resistance R varies, the current in the circuit also varies, which causes the potential drop across the internal resistance of the battery to vary. S ET U P : The largest current will occur when R = 0, and the smallest current will occur when R . The largest terminal voltage will occur when the current is zero ( R ) and the smallest terminal voltage will be when the current is a maximum ( R = 0). E XECUTE : (a) As R , I 0, so V ab E = 15.0 V, which is the largest reading of the voltmeter. When R = 0, the current is largest at (15.0 V)/(4.00 ) = 3.75 A, so the smallest terminal voltage is V ab = E – rI = 15.0 V – (4.00 )(3.75 A) = 0. (b) From part (a), the maximum current is 3.75 A when R = 0, and the minimum current is 0.00 A when R . (c) The graphs are sketched in Figure 25.66. E VALUATE : Increasing the resistance R increases the terminal voltage, but at the same time it decreases the current in the circuit. Figure 25.66 25.67. I DENTIFY : Apply . L R A ρ = S ET U P : For mercury at 20 C ° , 7 9.5 10 m ρ = × Ω⋅ , 1 0.00088 (C ) α = ° and 5 1 18 10 (C ) . β = × ° E XECUTE : (a) 7 2 (9.5 10 m)(0.12 m) 0.057 . ( 4)(0.0016 m) L R A ρ π × Ω⋅ = = = Ω

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25-16 Chapter 25 (b) 0 ( ) (1 ) T T ρ ρ α = + Δ gives 7 1 7 (60 C) (9.5 10 m)(1 (0.00088 (C ) )(40 C ) 9.83 10 m, ρ ° = × Ω⋅ + ° = × Ω⋅ ° so 8 3.34 10 m. ρ Δ = × Ω⋅ (c) 0 V V T β Δ = Δ gives 0 ( ) A L A L T β Δ = Δ . Therefore 5 1 4 0 (18 10 (C ) )(0.12 m)(40 C ) 8.64 10 m 0.86 mm. L L T β Δ = Δ = × ° ° = × = The cross sectional area of the mercury remains constant because the diameter of the glass tube doesn't change. All of the change in volume of the
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