Quiz6solutions

# Quiz6solutions - Quiz#6 Solutions 10 am Section(Section A...

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Quiz #6 Solutions 10 am Section (Section A) Version Version Version Version A B C D 1 B D E D 2 C E B E 3 D B E D 4 D D E A 5 E E D D 6 B B E B 11 am Section (Section B) Version Version Version Version A B C D 1 E C C D 2 B C C B 3 D A C E 4 D C C C 5 C E B B 6 B C A D Solutions for version A of each quiz are provided below. Any additional questions should be directed to: (Bryan) OR (Jason)

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1. As R decreases, the total series resistance through R2 and R decreases, thus allowing more current to pass. Assuming r (the resistance in series with V) to be small in comparison to the other resistances, then the circuit is modeled as parallel resistances – if the current through r2/R goes up, then the current through R1 must go down. 2. Need the resistances of the three bulbs, then reduce the circuit by parallel/series arguments. Bulb A: P = R*I^2 -> R = 24. Bulb B: P = V^2/R -> R = 160. Bulb C: V = I*R -> R = 240. B || C -> (1/240 + 1/160)^-1 -> 96. 96 in series with Bulb A gives 120 ohms. 3. The correct statement is that modern household plugs have 3 prongs, two of which are connected to ground wires. The problems with the other statements: A) Each device plugged into a receptacle must have the full voltage provided
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Quiz6solutions - Quiz#6 Solutions 10 am Section(Section A...

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