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ch20 - CHAPTER 20 ELECTRIC CIRCUITS CONCEPTUAL QUESTIONS 1...

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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 0 0 [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 cross-sectional area of the wire. Even when the wires have the same length, they may have the same resistance, if the cross-sectional 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 cross-sectional area. Since the cross-sectional area is proportional to the square of the diameter, a doubling of the diameter causes the cross-sectional 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. P = V 2 / R ____________________________________________________________________________________________
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Chapter 20 Conceptual Questions 1031 6. REASONING AND SOLUTION Two light bulbs are designed for use at 120 V and are rated at 75 W and 150 W. The power used by either bulb is given by Equation 20.6c: . We see from Equation 20.6c that, at constant voltage, the power used by a bulb is inversely proportional to the resistance of the filament. Therefore, the filament resistance is greater for the 75-W bulb. P = V 2 / R ____________________________________________________________________________________________ 7. REASONING AND SOLUTION Rather than state how many watts of power an appliance uses, appliance instructions often give statements such as "10 A, 120 V" instead. This statement gives the current used by the appliance for a specific voltage. It provides all the information necessary to determine the power usage of the appliance from Equation 20.6a: P = IV .
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