Ch20 - CHAPTER 20 ELECTRIC CIRCUITS CONCEPTUAL QUESTIONS 1 REASONING AND SOLUTION a When S 1 is set to position A current can flow from the

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 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....
View Full Document

This note was uploaded on 05/09/2008 for the course PHYS 23 taught by Professor Holland during the Spring '08 term at Pacific.

Page1 / 77

Ch20 - CHAPTER 20 ELECTRIC CIRCUITS CONCEPTUAL QUESTIONS 1 REASONING AND SOLUTION a When S 1 is set to position A current can flow from the

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online