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Light bulb Circuit problem

  1. Jun 14, 2006 #1
    [​IMG]
    "Light bulbs with the wattage ratings shown in the figure above are connected in a circuit. (a) What current does the voltage source deliver to the circuit? (b) Find the power dissipated by each bulb. (Asusume normal operating resistances of the bulbs to be constant at any voltage.)"

    I know P = IV = V2/R = I2R, and R = V/I. I know the current is the same for the 15W, the 40W and the combination in parallel of the 60W and 100W.

    I tried expressing the power of each bulb as a ratio of the power of the first bulb, 15W:
    P1 = P1, P2 = (8/3)P1, P3 = 4P1, P4 = (20/3)P1
    then since the currents are equal, P1/R1 = (8/3)P2/R2
    R1 = (3/8) R2

    but I don't know where to go from here...especially with the parallel part. I don't know how to combine the powers in the parallel part...
     
  2. jcsd
  3. Jun 15, 2006 #2

    nrqed

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    The simplest way is to remember that the power ratings are given for a given voltage (usually 120 volts). Therefore, using P= V^2/R, you get that the relation between two light bulb resistances is simply [itex] R_B = R_A {P_B \over P_A} [/itex]. For example, this gives [itex] R_1 = R_2 {P_1 \over P_2} = R_2 { 3 \over 8} [/itex] as you had. Applying this to all the other light bulbs, you will be able to relate all of them to R_1.
     
  4. Jun 15, 2006 #3

    andrevdh

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    means that the resistances of the bulbs should be considered to stay the same, irrespective of the operating conditions.

    So for example for the 40W bulb:

    [tex]R = \frac{V^2}{P} = \frac{120 \times 120}{40} = 360 \varOmega[/tex]
     
    Last edited: Jun 15, 2006
  5. Jun 15, 2006 #4

    berkeman

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    Boy, that's a very bad question for your prof/textbook to be asking. On the one hand it's good to get you thinking about working backwards from power to resistances, but in reality, that circuit is a joke and very misleading. If they wanted to get you thinking about power, they should have used something other than than light bulbs. Light bulbs only work at their rated voltage, and their cold resistance is nothing like their hot/working resistance. If you connected light bulbs together as shown in the problem diagram, they would not light up, and they would not consume the powers indicated. I hate it when some dork instructor or author (without real world experience usually) puts forth a non-realistic problem like this. It ends up putting wrong concepts into half of the students' heads.

    Rant off, down off soap box. :mad:
     
  6. Jun 15, 2006 #5
    yeah, it's one of the problems with the highest difficulty in my textbook.
     
  7. Jun 15, 2006 #6

    nrqed

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    I understand what you are saying (about real life vs idealized questions) but, playing devil's advocate, I would say that, imho, physics should be taught in a layered fashion. I think that one should be taught to solve idealized situations first, and then the situations should be made more and more realistic by adding layers of complexity (mimicking the way science progresses, after all).

    Now, I agree that some idealizations are really way off from real life. Assuming that the resistance of a lightbulb is constant, independent of the current flowing through it is farfetched. Granted. But doing projectile motion without taking into account air resistance is also crazy since air friction has such a large effect. Does that mean that we should do projectile motion with air friction included even in introductory classe? Maybe. It is a matter of opinion and of pedagogy.

    Personally, having taught at the lower college level for 10 years, I prefer to show the maths and the concepts of projectile motion in the context of a constant gravitational force first (and have the students build confidence in their algebra skills and problem solving in that context) before later pointing out that the resulst we get do not work in real life because air friction has a huge effect and then describing these effects qualitatively.

    Maybe others would feel preferable to include air drag from the start.

    I personally think that teaching involves a difficult balance between describing real life physics and helping the students build confidence in both their algebra skills and their conceptual understanding. What is the most pedagogical approach is a matter of debate.

    Just my personal opinion, obviously...

    Best regards

    Patrick
     
  8. Jun 15, 2006 #7

    berkeman

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    Your points are well taken, nrqed. I just get frustrated when I see things that are really being mis-taught, especially things that can put incorrect concepts into the student's brain. My favorite basic electronics book, "The Art of Electronics" by Horowitz and Hill has a section at the end of each chapter called "Bad Circuits". It shows common errors made by people regarding the subject of each chapter, like incorrect biasing of opamps or transistor circuits, comparators without hysteresis feedback, incorrectly using digital logic gates as analog components, forgetting base drive current limit resistors, etc. A lot of those errors come from oversimplified teaching, with the resulting mis-concepts being imparted to the students' brains and held there for years until they make the mistake in real life and have to figure out what they did wrong.

    Simplification for the sake of teaching initial concepts is okay if done correctly. As in your projectile motion example, I'm sure that you mention explicitly that "for the sake of simplicity, we will first neglect air resistance in our calculations". That way everybody knows that air resistance makes a difference, but they'll get to the details later. In the case of the OP's problem, the author of the problem should have at least put in a disclaimer about how real light bulbs work, or just have called the loads "constant power loads". The problem's author did a disservice to the students IMO by implying that real light bulbs can be hooked together that way.

    Dang, how'd I get back up on this soap box again?
     
  9. Jun 15, 2006 #8
    It's kind of like frictionless pains and baseballs in vacuums.
     
  10. Jun 15, 2006 #9

    berkeman

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    Funny typo :rofl: Frictionless planes can definitely be painful. I'm guessing you've seen me try to ice skate!:biggrin:
     
    Last edited: Jun 15, 2006
  11. Jun 17, 2006 #10

    mukundpa

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    The only difficulty is this, once you know the voltage at which the power is rated you will be able to find the resistance of the bulb and the things will become too easy. Try with the usual voltage 120 volts and I think you will get the correct answer.

    best of luck.

    MP
     
  12. Jun 17, 2006 #11
    I got the right answers by finding the resistances with V = 120V. Thanks
     
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