ism_chapter_25

# ism_chapter_25 - Chapter 25 Electric Current and...

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333 Chapter 25 Electric Current and Direct-Current Circuits Conceptual Problems *1 Determine the Concept When current flows, the charges are not in equilibrium. In that case, the electric field provides the force needed for the charge flow. 2 Determine the Concept Water, regarded as a viscous liquid flowing from a water tower through a pipe to ground is another mechanical analog of a simple circuit. 3 Picture the Problem The resistances of the wires are given by , A L R ρ = where L is the length of the wire and A is its cross-sectional area. We can express the ratio of the resistances and use our knowledge of their lengths and diameters to find the resistance of wire A. Express the resistance of wire A: A A A A L R = where is the resistivity of the wire. Express the resistance of wire B: B B A L R = Divide the first of these equations by the second to obtain: A B B A B B A A A A A L L A L A L R R = = or, because L A = L B , R A A R A B A = (1) Express the area of wire A in terms of its diameter: 2 A 4 1 A d A π =

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Chapter 25 334 Express the area of wire B in terms of its diameter: 2 B 4 1 B d A π = Substitute in equation (1) to obtain: R d d R 2 A 2 B A = or, because d A = 2 d B , () R R d d R 4 1 2 B 2 B A 2 = = and correct. is ) ( e 4 •• Determine the Concept An emf is a source of energy that gives rise to a potential difference between two points and may result in current flow if there is a conducting path whereas a potential difference is the consequence of two points in space being at different potentials. *5 •• Picture the Problem The resistance of the metal bar varies directly with its length and inversely with its cross-sectional area. Hence, to minimize the resistance of the bar, we should connect to the surface for which the ratio of the length to the contact area is least. Denoting the surfaces as a , b , and c , complete the table to the right: Surface L A L / A a 10 8 0.8 b 4 20 0.2 c 2 40 0.05 Because connecting to surface c minimizes R : correct. is ) ( c 6 •• Picture the Problem The resistances of the wires are given by , A L R ρ = where L is the length of the wire and A is its cross-sectional area. We can express the ratio of the resistances and use the definition of density to eliminate the cross-sectional areas of the wires in favor of the ratio of their lengths.
`Electric Current and Direct-Current Circuits 335 Express the resistance of wire A: A A A A L R ρ = where is the resistivity of copper. Express the resistance of wire B: B B B A L R = Divide the first of these equations by the second to obtain: A B B A B B A A B A A A L L A L A L R R = = or, because L A = 2 L B , B A B A 2 R A A R = (1) Using the definition of density, express the mass of wire A: A A A A A 'L 'V m = = where is the density of copper. Express the mass of wire B B B B B A 'L 'V m = = Because the masses of the wires are equal: B B A A A 'L A 'L = or B A A B L L A A = Substitute in equation (1) to obtain: () B B B B A A 4 2 2 2 R R R L L R = = = and correct. is ) ( b 7 Picture the Problem The power dissipated in the resistor is given by P = I 2 R . We

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ism_chapter_25 - Chapter 25 Electric Current and...

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