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Five identical light bulbs are wired in series to a 12-volt car battery. The circuit has 0.12 A of current...?

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  • Five identical light bulbs are wired in series to a 12-volt car battery. The circuit has 0.12 A of current...?


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Answer #1 | 14/01 2014 11:49
Total power = V * I = 12 * 0.12 = 1.44 w dissipated by each light bulb = total power / 5 = 1.44 / 5 = 0.288 w Goodbye
Answer #2 | 15/01 2014 07:23
You can do this two different ways: determine the power out of the battery and divide that power by 5 to determine the consumption of one bulb: 12*12/100 = 144/100 = 1.44W for 5 bulbs so 1.44W/5 = 0.288W per bulb = 288mW OR we could figure the resistance of each bulb by determining the resistance seen by the battery Req = V/I = 12/0.12 = 100Ω so that means each bulb has resistance 100/5 = 20Ω = Rb The power consumed by each bulb is then I²*Rb = 0.12²*20 = 0.288W = 288mW same answer.
Answer #3 | 15/01 2014 10:36
Total Volts will divided equally across the five identical bulbs connected in series and same current must flow through each bulb. Power per bulb = (Volts/bulb)*(Amps) = (12Volt/5 bulb)*(.12A) = .288 Watts = 288mW

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