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Why a lamp with X PD passes Y current

  1. Oct 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A string of electric lamps for decorating a Christmas tree consists of 20 12 V lamps connected in series across the 240 V mains. The power consumption of the whole string is 24 W.

    (a) What is the resistance of each lamp?
    (b) If one lamp becomes short-circuited, what is approximately the new power consumption of the string?
    (c) Explain why, when one of these lamps is tested by applying a potential difference of 0.1 V, it passes a current of 10 mA.

    Answers: (a) 120 Ω, (b) 25.3 W.

    2. The attempt at a solution
    (a) First find the power consuption of 1 lamp: 24 W / 20 = 1.2 W per 1 lamp. Then we find the resistance: P = V2 / R → R = V2 / P = 122 / 1.2 = 120 Ω (resistance of each lamp).

    (b) In that case we'll have the same voltage consumption (20 lamps on 12 V = 240 V), but we'll get only 19 lamps with resistance of 120 Ω (2280 Ω). So the new power consumption of the string is: P = V2 / R = 2402 / 2280 = 25.3 W.

    (c) This part I don't understand completely. The resistance should be R = V / I = 0.1 / 0.01 = 10 Ω. But what should I compare with what?
     
  2. jcsd
  3. Oct 4, 2016 #2

    gneill

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    What might make a resistance change value depending up on operating conditions?

    You recently worked on an exercise where a resistor in a circuit changed over time after power was applied. What was thought to be the cause of that change in resistance?
     
    Last edited: Oct 4, 2016
  4. Oct 4, 2016 #3
    You mean "ON operating conditions"?

    Well, we had a resistance of 120 Ohm and now we have a resistance of 10 Ohm. As we can see the PD was changed -- V = 0.1 V, though we don't know the original current. So the answer is "it passes a current of 10 mA because the PD has changed from 12 V to 0.1 V"?

    I just don't understand what the answer supposed to be. "Because X has changed (resistance / voltage / current) and therefore the lamp passes a current of 10 mA"?
     
  5. Oct 4, 2016 #4

    gneill

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    Yes. Or perhaps "upon" :smile:. Thanks for catching that.
    The PD is something external to the lamp that is applied for texting purposes. What you need to think about is what might be different about the lamp itself under those conditions. What's the difference between a lamp that's running with its normal operating PD of 12 V and one operating with just 0.1 V? Do you expect the lamp to be the same brightness in both cases? What makes a lamp glow?
     
  6. Oct 4, 2016 #5
    A lamp consumes 12 V, however, when only 0.1 V are applied to the lamp it is clearl that the lamp will pass less current and be less bright than under normal conditions.
     
  7. Oct 4, 2016 #6

    gneill

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    Again, what makes a lamp glow?
     
  8. Oct 4, 2016 #7
    When we put the cable into the wall. It gives it energy, current?
     
  9. Oct 4, 2016 #8

    gneill

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    What , physically, makes a lamp filament glow? What's the difference between a filament that's switched off and one that's giving off light?
     
  10. Oct 4, 2016 #9
    A bulb makes a lamp glow?
     
  11. Oct 4, 2016 #10

    gneill

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    No, the bulb is just a container for the filament.

    Please do a web search on "How do light bulbs work?". This is an important concept for this question (and the one you answered previously about why the potential across a resistor changed over time!).
     
    Last edited: Oct 4, 2016
  12. Oct 4, 2016 #11
    Electric energy is converted in light energy? The metal spring (filament) inside the glass shines.

    Isn't it just "if 0.1 V is applied the lamp will pass a current of 10 mA". Since the current is based on PD.
     
  13. Oct 4, 2016 #12

    gneill

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    How is electric energy converted to light energy? Why does the filament shine? What makes the filament different when it's passing a current from when it's not? What physical property of the material is different?
    (Hint: Why isn't it a good idea to touch a 100 W lightbulb while it's shining?) :wink:

    If that were true the current would be 0.1V / 120 Ω = 0.833 mA. Not the 10 mA that is claimed. What is different about the filament between these two cases?
     
  14. Oct 4, 2016 #13
    Current goes through the whole circuit.

    Hm, indeed it is 0.833 mA. In that case no idea what is different. It is only said that why if V = 0.1 V applied the current is 10 mA. Maybe it is short-circuited and because the current doesn't reach the lamp and uses the shortcut, the current speed increases and the cable heats up (I read about it in the short-circuit material).
     
  15. Oct 4, 2016 #14

    gneill

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    Current goes through all the wiring in the circuit. Why doesn't the whole circuit glow? What's different about the filament?

    Why/when do the elements on an electric stove glow? How about in an electric toaster? Or a space heater?
     
  16. Oct 4, 2016 #15
    It is made out of a different material.

    No idea actually. This is a far as I could go:
    I think this is logical.
     
  17. Oct 4, 2016 #16

    gneill

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    moenste, did you look up how a lightbulb works? What makes it shine?
     
  18. Oct 4, 2016 #17
    Yes I looked it up. Current flows through the components and filiament lights when temperature increases.
     
  19. Oct 4, 2016 #18

    gneill

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    [Emphasis in the quote is mine] The temperature increases. Very hot things give off visible light. Like fire, molten metal, the Sun, and so on.

    The filament gets hot because it has resistance and current flowing through a resistance causes power to be dissipated in the form of heat. When the temperature gets high enough it glows.

    Another property of metals such as filaments are made of is that as they get hot their electrical resistance goes up. The hotter they get the higher the resistance. So what might you say abut the resistance of a cold filament versus that of a hot filament?
     
  20. Oct 4, 2016 #19
    A hot filament has more resistance and a cold filament has less resistance?
     
  21. Oct 4, 2016 #20

    gneill

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    Right :approve:

    So can you see how to apply this to part (c)?
     
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