qz4sp09sol - February 17, 2009 PHY2054 Discussion-Spring 09...

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February 17, 2009 PHY2054 Discussion-Spring ‘09 Quiz 4 (Chapter 18.1-18.7) Name: UFID: *1. (2.5pts) A battery having an emf of 8.00 V delivers 125. mA when connected to a 60.0-Ω resistor. What is the internal resistance of the battery? The internal resistor is connected in series with the external resistor. Thus we have ε = I(R+r) r = ε/I-R = 8/0.125-60 = 4.00 Ω **2. (2.5pts) Three 40.0-W, 120-V lightbulbs are connected across a 24.0-V power source. Find the total power delivered to the three bulbs. The resistance of the lightbulb is P = ΔV 2 /R R = ΔV 2 /P = 120 2 /40 = 360 Ω R 2 and R 3 are connected in parallel. The equivalent resistance of these two lightbulbs is R’ = (1/R 1 +1/R 2 ) -1 = (1/360+1/360) -1 = 180 Ω R 1 and R’ are connected in series. Therefore, the equivalent resistance of the combination is R eq = R 1 +R’ = 360+180 = 540 Ω Since the potential difference of 24 V is applied to the combination, the total power is P tot = ΔV’ 2 /R eq = 24 2 /540 = 1.07 W
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This note was uploaded on 12/13/2011 for the course PHY 2054 taught by Professor Avery during the Fall '08 term at University of Florida.

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qz4sp09sol - February 17, 2009 PHY2054 Discussion-Spring 09...

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