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Power consumed by lightbulb

  1. Feb 16, 2016 #1
    1. The problem statement, all variables and given/known data

    Figure 1 shows the relationship between voltage and current when voltage is applied to a certain light bulb. Next, as shown in Figure 2, the bulb is connected in series with a 12-V battery and a 5 ohm resistor.
    https://scontent-kul1-1.xx.fbcdn.net/hphotos-xap1/v/t34.0-12/12721826_1254566191224449_288507912_n.jpg?oh=d4c85c21eeaf583d3caae6daabab282b&oe=56C532FB

    https://scontent-kul1-1.xx.fbcdn.net/hphotos-xlp1/v/t34.0-12/12659760_1254566187891116_1651482263_n.jpg?oh=d855cbc09ea2be59fc688fdd3b258da0&oe=56C58981
    What is the amount of power consumed by the light bulb?

    2. Relevant equations
    Ohm law

    3. The attempt at a solution
    I tried both P=VI and P=V^2/R but the answers are wrong. May I know the solution for this?
     
  2. jcsd
  3. Feb 16, 2016 #2

    gneill

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    How did you try them? Show your attempts. Why do you think the attempts failed? Sorry, but we don't solve homework for you here; we can provide help and advice, but not provide answers in place of your own efforts.

    Have you covered the concept of a load line in your course? :wink:
     
  4. Feb 16, 2016 #3
    My first attempt:
    P = VI
    According to the graph, when V = 12, I = 3.5
    P = 12 x 3.5
    = 42

    My second attempt:
    P=V^2 / R
    = 12^2 / 5
    = 144/5
    = 28.8

    It's wrong because I compared my attempts with the answer ofc. The answer is 4.
     
  5. Feb 16, 2016 #4

    cnh1995

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    ontent-kul1-1.xx.fbcdn.net%2Fhphotos-xap1%2Fv%2Ft34.0-12%2F12721826_1254566191224449_288507912_n.jpg
    As gneill said earlier, this should be solved using the conept of load line. Are you familiar with the method?
     
  6. Feb 16, 2016 #5

    gneill

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    Did you think about why your attempts did not work?

    Without worrying about specific values, can you describe what you would expect the behavior of the circuit will be in terms of current and potential drops?
     
  7. Feb 16, 2016 #6

    cnh1995

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    Since you can't write the equation for the given i-v curve of the bulb, I believe you need to use the graphical approach. Since this is a series circuit, same current will flow through the bulb and 5Ω resistor. Can you write another equation for voltage across the bulb and current through the bulb using this?
     
  8. Feb 16, 2016 #7
    My apologize, I have never learn the concept of load line before, that's why I thought this question is simple and didn't put in much effort. I'm sorry.

    Did you mean the concept of in series circuit, the current throughout the circuit is constant, and the voltage will drops when lose in electric potential?
    So I have to calculate the potential difference of the light bulb?

    This is my attempt:
    V = IR
    12= I (5)
    I = 2.4 A

    I'm not sure if I have to deal with the i-v curve to find out the voltage across the bulb, if so, the voltage should be 3 V? Or did I miss out any formula?
     
  9. Feb 16, 2016 #8

    cnh1995

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    I meant a generalized equation..You can't decide the voltage across the bulb using only the given graph.. Are you familiar with KVL?
     
  10. Feb 16, 2016 #9

    cnh1995

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    ul1-1.xx.fbcdn.net%252Fhphotos-xap1%252Fv%252Ft34.0-12%252F12721826_1254566191224449_288507912_n.jpg
    This is the characteristic of the bulb. This bulb is placed in a circuit with a resistance 5Ω in series with it. What you will write using KVL will bè the generalized characteristic of the whole circuit. In that characteristic, you are putting your bulb with its own characteristic. Hence, the operating point of the circuit will be the intersection of these two characteristic curves. Once you get the generalized equation using KVL, plot it on the same graph and see..
     
  11. Feb 16, 2016 #10

    gneill

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    In a way, yes. You need to realize that the resistor is not the only resistance in the circuit; the bulb will also have a resistance. But that resistance changes with the amount of current flowing through it (the I-V graph is not a straight line).

    The actual current that flows in the circuit depends upon the total resistance, and the total resistance depends on the current... so a bit of a "catch-22" situation. This is where a graphical approach to "solve" the implied simultaneous equations comes in. That's the essential purpose of a load line. But as you say, if you haven't been introduced to the load line concept yet, then this is your introduction :smile:
    Suppose that the current was 2.4 A as you've calculated. It assumes that the full 12 V is dropped across the 5 Ohm resistor. Check the graph of the IV curve for the light bulb. What voltage would be dropped across the bulb if 2.4 A were flowing through it? How many volts would be left for the resistor? What "new" value of current does that imply for the resistor?
     
  12. Feb 16, 2016 #11
    I am not familiar with KVL too.. All I know about it is only the algebraic sum of all voltages in a enclosed circuit must be equal to 0.
    But based on the characteristic of the full circuit, I wonder if the graph will look like this?
    https://scontent-kul1-1.xx.fbcdn.net/hphotos-xlf1/v/t34.0-12/12713983_1254736014540800_1468100975_n.jpg?oh=ab29a3871fe8fd646623308ec1287df2&oe=56C546C7

    The intersection is the operating point, hence, the potential difference across the bulb = 2 V?

    If 2.4 A were flowing through the bulb, the potential drop will be 3V.
    12V - 3V = 9V
    There will be 9V left for the resistors.
    V = IR
    9 = I(5)
    I = 1.8 A

    Am I right? Can I use this current to calculate the power consumed? (Using P=VI)
     
  13. Feb 16, 2016 #12

    cnh1995

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    Bingo!:smile::smile:
    Now can you find the power consumed in the bulb?
     
  14. Feb 16, 2016 #13

    cnh1995

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    And that's what KVL is! You knew it already..:smile:
     
  15. Feb 16, 2016 #14
    Yea I think I got it already, since I = 1.8 A and V = 2 V, the power consumed should be P = 1.8 x 2 = 3.6 W
    And the best answer would be 4 W

    Thank you so much gneill and cnh1995! Really appreciated it!
     
  16. Feb 16, 2016 #15

    cnh1995

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    No. You assumed the current to be 2.4A first and after calculation, it came out to be 1.8 A. How is that possible in a series circuit? That way, your voltages too won't add up to 0..
    .fbcdn.net%25252Fhphotos-xap1%25252Fv%25252Ft34.0-12%25252F12721826_1254566191224449_288507912_n.jpg
    Look at the operating point. What is the current at the operating point?
     
    Last edited: Feb 16, 2016
  17. Feb 16, 2016 #16

    gneill

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    Yes! What you've drawn is the load line for the source and resistor. Where they intersect is the graphical solution to the problem at hand.
    Not yet (actually you've found the correct voltage and current above by graphical means).

    If you took that 1.8 A and once again determined the potential drop across the bulb, and then another "new" current for the circuit, and repeated this process until the solution converged, you would zero in on the intersection of the load line and the IV curve of the bulb:
    upload_2016-2-16_12-8-55.png

    This is why the load line concept is important. It afford a means to quickly solve, in a graphical manner, the simultaneous equations for a circuit when only the IV characteristic of some non-linear component (the bulb in this case) is given.
     
  18. Feb 16, 2016 #17
    Oh! My bad. So does it means that all the assumptions of current is to find out the operating point of the load line in the graph?
    The current throughout the series circuit must be constant, so the assumptions cannot be the right current.
    Thus, the answer is straightforward which is the operating point, the current of the circuit will be 2 A, and the voltage across the light bulb will be 2 V?
     
  19. Feb 16, 2016 #18

    gneill

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    Yup. That's it.
     
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