# s31 - Solutions to Homework#3 Chapter 21 6 Picture the...

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Solutions to Homework #3 Chapter 21 6. Picture the Problem : Electric current is delivered to a television set at a specified voltage. Strategy: The power delivered to the television is the energy per charge (voltage) multiplied by the charge per time (current) as given by equation 21-4. For part (b) we can use the definition of current (equation 21-1) and the charge on an electron to find the required time for 10 million electrons to pass through the circuit. Solution: 1. (a) Solve equation 21-4 for I : 78 W 0.65 A 120 V P I ε = = = 2. (b) Solve equation 21-1 for t : ( 29 ( 29 7 19 12 e 1 10 e 1.6 10 C/e ) 2.5 10 s 2.5 ps 0.65 A N e Q t I I - - - - ∆ = = = = = Insight: Ten million electrons in 2.5 picoseconds! It’s a good thing electric companies don’t charge by the electron! 16. Picture the Problem : The four conducting cylinders shown in the figure are all made of the same material, though they differ in length and/or diameter. They are connected to four different batteries, which supply the necessary voltages to give the circuits the same current, I . Strategy: Use Ohm’s law, V I R = (equation 21-2) together with the expression for resistance, R L A ρ = (equation 21-3) to determine the ranking of the voltages. Solution: Combining equations 21-2 and 21-3 we find that 2 2 4 I L I L L V B A D D π = = = , where B is a constant. We may then calculate the voltages: ( 29 1 2 2 3 3 , 4 2 L L V B B D D = = 2 2 2 2 2 , L L V B B D D = = 3 2 , L V B D = and ( 29 4 2 2 1 . 4 2 L L V B B D D = = From these calculations we can determine the ranking, V 4 < V 1 < V 3 < V 2 . Insight: The smallest resistance (case 4) needs the smallest voltage in order to flow current I . 20. Picture the Problem : A wire conducts current when a potential difference is applied between the two ends. Strategy: Ohm’s Law (equation 21-2) gives the relationship between potential difference, current, and resistance for any circuit element. To solve this problem we can combine equation 21-3 with Ohm’s Law to find the resistivity of the material from which the wire was made. Solution: Combine equations 21-2 and 21-3 to find : ( 29 ( 29 ( 29 ( 29 2 3 2 8 2 1 4 0.33 10 m 12 V 7.1 10 m 4 4 2.1 A 6.9 m L L D V V IR I I A I L D - - = = = = = = Insight: The resistivity of this material is a little higher than tungsten but lower than iron. If it is a pure metal we might suspect it is made out of one of the transition metals. 32. Picture the Problem : A light bulb consumes electric power by drawing current at a certain voltage. Strategy:

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## This note was uploaded on 02/10/2012 for the course PHY 102 taught by Professor Korotkova during the Spring '09 term at University of Miami.

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s31 - Solutions to Homework#3 Chapter 21 6 Picture the...

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