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Homework Help: Measuring the resistance value of some bulbs.

  1. Jun 6, 2017 #1
    1. The problem statement, all variables and given/known data
    A small generator of 5kW, 240V and a constant frequency is used to light up 35 bulbs in parallel, each one has (240 V, 60 W) printed on it. Calculate the resistance of each light bulb.

    2. Relevant equations

    3. The attempt at a solution
    Attempt 1

    Pw = V²/R ----> R = V²/Pw ------> R = 240²/60 = 960Ω

    Attempt 2

    i (passing thrrough the circuit) = 5*10³ /240 = 20.83 A
    i (passing through each light bulb) = 20.83/35 = 0.6 A

    Pw (in each bulb) = i²R -----> R = Pw/i² ----> R = 60/ 0.6² = 166.667Ω

    Why do I get different results?
    Last edited: Jun 6, 2017
  2. jcsd
  3. Jun 6, 2017 #2


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    Because your second method doesn't make any sense. You are treating the MAXIMUM POSSIBLE power output as though it were all being used.
  4. Jun 6, 2017 #3


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    The 5 kW only appears on the name plate of the generator: I can deliver 21 A. In the case of 35 bulbs x 60 W it only has to deliver just over 2 kW.
  5. Jun 6, 2017 #4

    Charles Link

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    Just because it is a generator capable of putting out 5kWatts does not mean that is the actual total power it puts out. 35 bulbs x60 watts/bulb=2100 watts. (I see 2 others, @phinds and @BvU gave essentially the same answer, just ahead of me.)
  6. Jun 6, 2017 #5
    The 5kW rating of a generator is the maximum rating. It will only provide 5kW if the resistance you use is low enough.
    The same thing is usually true of current ratings of power supplies. The voltage is constant, as long as no more than the maximum power or current is drawn.
    If you would use 140 lamps, for instance, they would have a resistance of 240Ω, and the generator would have to produce more than 5kw and one or more of the following would likely happen:
    - The output voltage of the generator would become lower than 240 V.
    - a fuse will blow
    - The generator would get damaged
    - Whatever the power source of the generator is would get damaged
  7. Jun 6, 2017 #6
    Ok, guys, I will correct this. Here we are:
    i = 2100/240 = 8.75 A
    i (in each bulb) = 8.75/ 35 = 0.25 A
    Pw = i²/R -----> R = Pw/i² = 60/0.25² = 960 Ω

    Ok, I got the same result, thank you!
  8. Jun 6, 2017 #7


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    Sometimes a question setter will provide unnecessary information. Part of the challenge is to figure out which information is relevant. (I like that because that is what happens in the real world.)
    Can you see how you could have gone straight to the answer without knowing anything about the generator?
  9. Jun 6, 2017 #8
    That's what I did in Attempt 1 in #1, I used the information printed on the bulb itself (the power and the voltage).
  10. Jun 6, 2017 #9


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    Ok, sorry for the noise.
  11. Jun 7, 2017 #10
    No, no, you made me aware that it is sometimes will give me unnecessary information. I should know what to use and what to not.
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