Fall 2011 Homework 8 Solutions

# Fall 2011 Homework 8 Solutions - Physics 21 Fall 2011...

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Unformatted text preview: Physics 21 Fall, 2011 Solution to HW-8 25-44 If a “75 W” bulb (75 W are dissipated when con- nected across 120V) is connected across a 220 V potential difference (as is used in Europe), how much power does it dissipate? (a) The power dissipation P and potential difference V across the bulb are variables, but the resistance R across the light bulb is a physical constant whether you are in America or Europe. From the equation sheet we know that P = IV and V = IR . We can combine these formulas to eliminate the current I , which we don’t know, and get a relation be- tween power, voltage, and resistance: P = IV = V R V = V 2 R ⇒ R = V 2 P . Using the US values P = 75 W and V = 120 V, we find R = (120 V) 2 75 W = 192 Ω . We can then use this resistance to solve for the power dis- sipated when the bulb is connected to the new potential difference: P = V 2 R = (220 V) 2 192 Ω = 252 W . 25-46 A battery-powered global positioning system (GPS) receiver operating on a voltage of 9.1 V draws a current of 0.20 A. a) How much electrical energy does it consume during a time of 2.0 h? (a) We can calculate the energy use over a period of time if we know the rate that the device uses electrical energy (power). The power can be determined from: P = IV = 9 . 1 V × . 2 A = 1 . 82 J / s = 1 . 82 W The total energy used over a time of two hours is: E tot =...
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## This note was uploaded on 12/18/2011 for the course PHYS 021 taught by Professor Hickman during the Spring '08 term at Lehigh University .

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Fall 2011 Homework 8 Solutions - Physics 21 Fall 2011...

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