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E&M: AC Circuit Light Bulb

  1. Apr 12, 2015 #1
    1. The problem statement, all variables and given/known data
    How large an inductance should be connected in series with a 120 volt (rms), 60 watt light bulb if it is to operate normally when the combination is connected across a 240 volt, 60 Hz line? (First determine the inductive reactance required. You may neglect the resistance of the inductor and the inductance of the light bulb.)

    2. Relevant equations
    z = iωL
    V = L(dI/dt)
    P = V2/z
    I(t) = I0cos(ωt)
    More?

    3. The attempt at a solution
    I'm essentially stuck at the beginning of this problem. All I've done so far is calculate the inductive reactance = 120πL, but I have no idea where to go from there. All help is greatly appreciated!
     
  2. jcsd
  3. Apr 12, 2015 #2

    collinsmark

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    Hello NoSniping,

    Welcome to Physics Forums! :smile:

    That's a good start! :smile:

    From here, there exist two approaches you can take that lead to the same answer.

    What is the rms current in the light bulb when it operating normally from the 120 V (rms) source (without the inductor)? What does the reactance (and ultimately inductance of the inductor) have to be such that the current in the overall circuit is the same, except when connected to a 240 V (rms) source?

    If you treat the light bulb and inductor as an rms voltage divider, what does the reactance (and ultimately inductance of the inductor) have to be such that voltage across the bulb is 120 V (rms), even though the entire circuit spans across a 240 V (rms) source?

    [Edit: Hint: be careful when calculating the total impedance in whichever method you choose. One can't simply just add components together. Pythagorean's theorem is involved.]
     
    Last edited: Apr 12, 2015
  4. Apr 12, 2015 #3
    Thank you so so much for replying!!

    So rms current would be P/V = 120/60 = 2 A. Then I'm not sure what to do...I know there's V = L(dI/dt), but will this current vary in time?

    Or should I use Vrms = z*Irms?
     
  5. Apr 12, 2015 #4

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    Try that one again. :wink: I think you have your numbers mixed around.

    Invoking V = L(dI/dt) makes things more complicated than it needs to be. [At least for this problem.]

    Look up in your textbook (or even on the Internet) how one finds the magnitude of the impedance of two components in parallel series (particularly one purely resistive component and another purely reactive component). [Edit: Hint, in case you missed my edit in the last post: the Pythagorean theorem is involved.]

    Once you are able to work with impedances, this problem reduces to just using the complex version of Ohm's law (i.e., Ohm's law with complex numbers).
     
    Last edited: Apr 12, 2015
  6. Apr 12, 2015 #5
    Oh whoops Irms = 1/2 A. According to the internet, impedances in series just add...maybe I'm missing something here? I don't understand how this works. I thought I had it with Vrms = Irms*z...I don't see where the other impedance comes in
     
  7. Apr 12, 2015 #6

    SammyS

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    Impedances add however, one is purely real the other imaginary. They behave as if they are out of phase with each other.
     
  8. Apr 12, 2015 #7
    Could you help me understand where the other impedance comes from? I see that there's one from the inductor but that's all
     
  9. Apr 12, 2015 #8

    SammyS

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    The resistance of the light bulb is the other.
     
  10. Apr 12, 2015 #9

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    Yes, impedances add so long as you treat them as complex.

    For example, if
    [itex] Z_1 = R + j0 [/itex]
    and
    [itex] Z_2 = 0 + j\omega L [/itex]

    Then combining them in series gives you

    [itex] Z_{series} = Z_1 + Z_2 = R + j \omega L [/itex]

    where '[itex] j [/itex] represents [itex] \sqrt{-1} [/itex], the unit imaginary number.

    Now, what is the magnitude of that series impedance?
    (If you didn't catch my edits in previous posts, I'll give you the hint again: The Pythagorean theorem is involved.)
     
  11. Apr 13, 2015 #10
    I honestly have no idea how the pythagorean theorem comes in here. Could you show me please?
     
  12. Apr 13, 2015 #11

    SammyS

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    What is the expression you have learned for calculating impedance of resistor, inductor, and/or capacitor in series?
     
  13. Apr 13, 2015 #12

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    Your textbook surely has the explanation in there somewhere.

    To point you in the right direction, a complex number has two components: an imaginary component and a real component.

    The figure below shows two complex numbers on the complex plane.
    220px-Complex_conjugate_picture.svg.png
    [Image taken from Wikipedia]

    In the image, both numbers are different, although they share the same magnitude. (The figure shows what are called "complex conjugate" numbers, but that's more detail than I wish to get into here.)

    My point is that although a complex number can be represented by its real and imaginary components, it can also be specified in terms of its magnitude (r in the figure) and phase angle ([itex] \varphi [/itex] in the figure). [Edit: to be really accurate, that bottom complex number should have had its angle specified as [itex] - \varphi [/itex], but again, that's more detail than I want to get into here.]

    For the problem in this thread, the phase angle [itex] \varphi [/itex] isn't very important, but the magnitude is.

    -----------------------
    Edit: Actually, this is probably a better, less cluttered figure also taken from Wikipedia. So what is the magnitude or "length" (if you will) of that number?:
    220px-Complex_number_illustration.svg.png

    ---
    Yet another edit: In my earlier description, I used "[itex] j [/itex]" to represent the unit imaginary number instead of "[itex] i [/itex]". This is common when discussing electricity to avoid confusion with the current, which is also denoted by the letter "[itex] i [/itex]". When discussing pure mathematics, the unit imaginary number is usually "[itex] i [/itex]", but in electrical engineering (and physics discussing electricity) it's "[itex] j [/itex]".
     
    Last edited: Apr 13, 2015
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