19_InstSolManual_PDF_Part12

# 19_InstSolManual_PDF_Part12 - 19-12 Chapter 19 V2 5 I 2R...

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power. Use the rated power to find the resistance of each bulb. In parallel, the full line voltage is applied to each bulb. Solve: The resistance of each bulb is: and (a) In series, The current in each bulb is the current in the equivalent circuit, The power consumed by each bulb is and (b) In parallel, for each bulb, so and (c) P is proportional to and now for each bulb, so they would each consume 4 times their rated power: and The power rating for each bulb would be exceeded and they would quickly burn out. Reflect: In series, the bulb with the greatest resistance dissipates the greatest power. When the bulbs are connected individually across the 120 V outlet, the bulb with the least resistance dissipates the greatest power. 19.56. Set Up: For resistors in series the equivalent resistance is The current is the same in each. so (b) The 20.0 resistor would dissipate more than 2.00 W and would fail.
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