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**Unformatted text preview: **Chapter 17 Current and Resistance Problem Solutions 17.1 The charge that moves past the cross section is ( ) Q I t Δ = Δ , and the number of electrons is ( ) ( ) ( ) ( ) 3 20 19 80.0 10 C s 10.0 min 60.0 s min 3.00 10 electrons 1.60 10 C I t Q n e e-- Δ Δ = = × = = × × The negatively charged electrons move in the direction opposite to the conventional current flow. 17.3 The current is Q V I t R Δ Δ = = Δ . Thus, the change that passes is ( ) V Q t R Δ Δ = Δ , giving ( ) ( )( ) 1.00 V 0.100 A 20.0 s 2.00 C 10.0 Q t Δ = Δ = = Ω 69 70 &amp;#1; &amp;#1; &amp;#1; &amp;#1; &amp;#1; &amp;#1; CHAPTER 17 17.8 Assuming that, on average, each aluminum atom contributes one electron, the density of charge carriers is the same as the number of atoms per cubic meter. This is A A density N n mass per atom M N M ρ ρ = = = , or ( ) ( ) ( ) 23 3 6 3 3 28 3 6.02 10 mol 2.7 g cm 10 cm 1 m 6.0 10 m 26.98 g mol n × = = × The drift speed of the electrons in the wire is then ( )( )( ) 4...

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