41_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

41_Ch 18 College Physics ProblemCH18 Direct-Current...

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Direct-Current Circuits 139 (c) The total resistance of the single branch of the series circuit (Case 1) is 2R. Thus, the current in this case is 1 2 IV R = ∆ . Note that this is one half of the current that flows through each bulb in the parallel circuit (Case 2). Since the power supplied is proportional to the square of the current, the power supplied to each bulb in Case 2 is four times that supplied to each bulb in Case 1. Thus, the bulbs in 2 I Case 2 are much brighter than those in Case 1. (d) If either bulb goes out in Case 1, the only conducting path of the circuit is broken
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

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