samp05 - Since the fuse can handle 15 A, the number allowed...

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SAMPLE PROBLEM (Class 5) Since the lamps are in parallel, each lamp has the same applied emf, E = 120 V. Then, for each lamp, P = i  E , so i = P /  E . i = (500 W)/(120 V) = 4.17 A for each lamp. Since they are in parallel, the total current is the sum of the currents through the resistors.
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Unformatted text preview: Since the fuse can handle 15 A, the number allowed is, N = (15 A)/(4.17 A) = 3.60. Since a fraction of a lamp is not possible, the answer is, N = 3 lamps...
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This note was uploaded on 09/17/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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