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Unformatted text preview: CHAPTER 20 ELECTRIC CIRCUITS CONCEPTUAL QUESTIONS ____________________________________________________________________________________________ 1. REASONING AND SOLUTION a. When S 1 is set to position A , current can flow from the generator only along the path that requires S 2 to be set to position A . Hence, the light will be on when S 2 is in position A . b. When S 1 is set to position B , current can flow from the generator only along the path that requires S 2 to be set to position B . Hence, the light will be on when S 2 is in position B . ______________________________________________________________________________ 2. REASONING AND SOLUTION When an incandescent light bulb is turned on, the tungsten filament becomes white hot. Since the voltage is constant, the power delivered to the light bulb is given by Equation 20.6c: . From Equation 20.5, P = V 2 / R [1 ( )] R R T T α = + − , where a is the temperature coefficient of resistivity and is a positive number. Thus, as the filament temperature increases, the resistance of the wire increases, and as the filament heats up, the power delivered to the bulb decreases. ____________________________________________________________________________________________ 3. SSM REASONING AND SOLUTION Two materials have different resistivities. Two wires of the same length are made, one from each of the materials. The resistance of each wire is given by Equation 20.3: / R L A ρ = , where ρ is the resistivity of the wire material, and L and A are, respectively, the length and crosssectional area of the wire. Even when the wires have the same length, they may have the same resistance, if the crosssectional areas of the wires are chosen so that the ratio / L A ρ is the same for each. ____________________________________________________________________________________________ 4. REASONING AND SOLUTION The resistance of a wire is given by Equation 20.3: / R L A ρ = , where ρ is the resistivity of the wire material, L is the length of the wire, and A is its crosssectional area. Since the crosssectional area is proportional to the square of the diameter, a doubling of the diameter causes the crosssectional area to be increased four fold. From Equation 20.3, we see that doubling both the diameter and length causes the resistance of the wire to be reduced by a factor of 2. ____________________________________________________________________________________________ 5. REASONING AND SOLUTION One electrical appliance operates with a voltage of 120 V, while another operates with a voltage of 240 V. The power used by either appliance is given by Equation 20.6c: . Without knowing the resistance R of each appliance, no conclusion can be reached as to which appliance, if either, uses more power....
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This note was uploaded on 05/09/2008 for the course PHYS 23 taught by Professor Holland during the Spring '08 term at Pacific.
 Spring '08
 holland
 Current, Light

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