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Quiz 2 Solution Sect - 7B.S11.C Quiz 2 Last 6 digits of student ID DL SECTION Grade Last name first name First three letters of your last name No

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No books or notes. Calculators OK. Show all your work below. Answers alone receive no credit! 1. You hook up a light bulb (unknown resistance) to a battery (unknown voltage), and measure the current using an ammeter, as shown in Circuit 1. The ammeter has no internal resistance (i.e. it has no effect on the circuit other than to measure the current). You find that the current in Circuit 1 is 0.6 A . You then add an identical light bulb as shown in Circuit 2. a) Using just the information given above, what current will you measure in Circuit 2? Use the energy density equation to justify your answer. Going around the Circuit 1 in a complete loop, we get ± ² ³ ´ µ · ² ¸ , which leads to I 1 ²¹³/R . Since the current in circuit 1 is .6 A, we now know that ³ /R = .6 A. Going around Circuit 2 in a complete loop, remembering that the battery is the same, and that we have added another identical resistor, we get ± ² ³ ´ µ º · ´ µ º · ² ³ ´ »µ º · ² ¸ . So now we have I 2 = ³ /2R = (.6 A)/2 = .3 A. b) You now use a voltmeter to measure the voltage difference between points 1 and 2 and find that ± ² ´¼½¹± . Using your answer to part a), and the appropriate energy density equation, find the resistance of the light bulb in Ohms (Ω). Between points 1 and 2 in Circuit 2, there is one resistor, so ΔV = V
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This note was uploaded on 10/24/2011 for the course PHYSICS 7B taught by Professor Johnconway during the Spring '07 term at UC Davis.

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Quiz 2 Solution Sect - 7B.S11.C Quiz 2 Last 6 digits of student ID DL SECTION Grade Last name first name First three letters of your last name No

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