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Unformatted text preview: Assignment 4 Solutions Chpt 26 1. (a) The charge that passes through any cross section is the product of the current and time. Since t = 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 × 10 3 C. (b) The number of electrons N is given by q = Ne , where e is the magnitude of the charge on an electron. Thus, N = q / e = (1200 C)/(1.60 × 10 –19 C) = 7.5 × 10 21 . 23. The resistance of conductor A is given by 2 A A r L R π ρ = , where r A is the radius of the conductor. If r o is the outside diameter of conductor B and r i is its inside diameter, then its crosssectional area is π ( r o 2 – r i 2 ), and its resistance is ) ( 2 2 i B r r L R − = π ρ The ratio is . 3 ) 50 . ( ) 50 . ( ) . 1 ( 2 2 2 2 2 2 = − = − = mm mm mm r r r R R A i B A . 26. Let be the radius of the kite string and 2.00 mm r = 0.50 mm t = be the thickness of the water layer. The crosssectional area of the layer of water is ....
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 Spring '08
 Parsons
 Charge, Current, Electric charge, Electrical resistance, Ri, crosssectional area, Chpt

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