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If I hook up an L.E.D. to a 9V, is the L.E.D. in series or parallel?

  1. Mar 5, 2014 #1
    I was helping my friend wire up some pretty lights for his automobile. I walked him through series and parallel circuits. He asked me if a single L.E.D. connected to a battery was in series or parallel with the battery. A said series! But then I thought a little more ...isn't it in parallel too?
     
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  3. Mar 5, 2014 #2

    berkeman

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    It is a series circuit. The current flows continuously around the circuit, not dividing anywhere. When you have a parallel circuit, the current divides between the branches.

    BTW, you will need a series resistor to limit the current the LED draws. A typical LED is about 2V, so you will need to drop 7V across the resistor. If you want to supply 10mA to the LED, what value should the resistor be?
     
  4. Mar 6, 2014 #3
    Key word, divide.


    Supplied voltage: 9V

    The L.E.D. is going to "use" 2V of the supplied voltage, so now there's going to be a potential difference of 7V. The relationship between voltage, current, and resistance is as follows, E=IR. So, if the supplied voltage is 7V, then what resistor value do you need in order to get .01 amperes of current (see pic below)?

    Screenshot_2014-03-06-14-51-44~2.jpg

    In understand the above, but I did run into a few hrmmm moments.

    A typical 9V puts out around a half of an amp. If this 9V put out 1 amp, would all of this^ still hold true?

    We solve for 7V. That makes me want to think the diode is going to see 7V. That's not true...

    We say the diode uses 2V, so we subtract it from our supplied voltage. The diode also uses 10mA, yet we don't subtract this figure from the supplied current.
     
    Last edited by a moderator: Mar 6, 2014
  5. Mar 6, 2014 #4

    davenn

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    the current capability of the power source is irrelevant

    you could have a 9V, 10A power source, it doesn't matter.
    The resistor value is still calculated for the voltage drop and the current required to light the LED


    Ohh and type your maths in as text, that image is very difficult to read

    Dave
     
  6. Mar 6, 2014 #5

    meBigGuy

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    Echoing davenn:
    The voltage across the resistor determines the current in the circuit. PERIOD Doesn't matter whether your supply has 10000 A capacity or 20ma capacity.

    The voltage across the resistor is Vsupply minus Vdiode. As you increase the current (reduce the resistor) you will see more drop across the diode which affects the result. Usually you assume a constant voltage across the diode from the data sheet and it is close enough.
     
  7. Mar 9, 2014 #6
    To address your original question a bit more. In a sense, it's a bit of a trick question. If you have only two components and you connect them, they are wired both series and parallel! It's when you have more than two components that the circuit becomes series or parallel. If the circuit is complicated enough, portions will be series and portions will be parallel, hence the term series/parallel.

    In the case of your circuit however, you actually have three components - the LED, the battery, and a resistor (needed to prevent the LED from self-destructing and pointed out in previous posts), and these three are definitely wired in series. In some cases (like automotive LED tail lights), the resistor is built into the LED assembly, but it is there even if it isn't seen (alternately it might be a current limiting circuit which allows the LED assembly to work over a range of applied voltages).

    This conundrum occurs when you have a single component (like a buzzer) and a power source (e.g., a battery). The same current flows through both and the same voltage appears across each, hence they are in series and in parallel.

    The voltage/current test is a good one to keep in mind: If the same (not value-wise but the same actual current) flows through any number of components, they are in series. Likewise, if the same (again, actual - not value) voltage appears across any number of components, they are in parallel.
     
    Last edited: Mar 9, 2014
  8. Mar 9, 2014 #7

    Averagesupernova

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    It gets tricky. Is the internal resistance of a battery in series or parallel with the load? Our measurements tell us it is in series and it is ALWAYS modeled this way. It also is in series with the battery itself (internal ideal). This tells us that ALL loads are in series with a battery even when they are in parallel with other loads.
     
  9. Mar 9, 2014 #8

    sophiecentaur

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    The distinction between the names 'series and parallel' is pretty much a red herring, in most situations. The circuit is the circuit and the sums apply, whatever you choose to call it. (see my present signature) None of the formulae used in circuit calculations explicitly use the terms series or parallel. Any circuit is merely a set of components joined by nodes.
     
  10. Mar 9, 2014 #9
    Good point. In my reply, I was thinking in terms of an ideal source and not a real world battery. All batteries have internal resistance and that makes four components connected in series (LED, resistor, battery, resistor) in the case of the LED light. Actually, the LED itself has a small resistance but the overall characteristics of a diode predominate (that makes three resistors in the series string).

    Don't forget however that the battery can be modeled as a Norton equivalent which places the resistor in parallel with an ideal source:smile:
     
  11. Mar 9, 2014 #10

    Averagesupernova

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    An ideal current source.
     
  12. Mar 9, 2014 #11

    jim hardy

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    Where there's only one possible path for current, you can declare "Series" and "Parallel" either irrelevant or equivalent, your choice.

    That's along the same lines as my assertion "there exist only three numbers in the universe: none, one, and more than one.
    That's why artificial intelligence won't get truly smart until base three Boolean algebra progresses a bit further, based on three states True, False, and Indeterminate.."
    :biggrin::devil::wink:
     
  13. Mar 9, 2014 #12

    meBigGuy

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    a better collection of states are Yes , No , Unknown, Irrelevant, Sometimes, Maybe,
    Probably, Doubtful, Usually, Depends, Rarely, Partly

    These are use by the 20Q game. http://www.20q.net/ (the handheld version uses fewer choices)

    Worth trying --- works really well, even with just yes, no
     
  14. Mar 10, 2014 #13

    jim hardy

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    Thanks meBigGuy !
    Looks like a complex game.
    I still have to cheat at solitaire - if you hold ctrlZ and back up to start of game then exit it doesn't count as a loss.

    Here's an article about computing in bases other than 2. Those tristate logic chips made for wired-OR look natural for base 3 hardware.

    http://www.americanscientist.org/issues/issue.aspx?id=3268&y=0&no&content=true&page=5&css=print
    I'm still trying to figure out how to count in base e .
     
  15. Mar 10, 2014 #14

    meBigGuy

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    Interesting article.

    http://en.wikipedia.org/wiki/Non-integer_representation
    Basically base e would look like ternary (have 3 symbols) but the values represented would be different. You are stuck with an integer number of symbols. How do you count in binary? 0 1 ten eleven? or 0 1 one-zero one-one? I just say 0-1-2-3, so in base e I'd have to say 0, 1, 2, 2.71..., 3.71..., 4.71..., 7.38..., ... Could be pretty slow if I wanted to be precise. :)

    Let me think about the number base stuff. Multilevel logic can be implemented, but I don't think tristate really qualifies unless you have the tri-state propagate through the gates.

    20 questions is just answering yes and no to 20 questions by the computer and it will guess what you are thinking of. It's pretty amazing (well, my little handheld one surprises many people - I haven't used the online version). It does better if you use more answers. Think of a shovel, answer the questions, and it will probably guess shovel.
     
  16. Mar 10, 2014 #15

    sophiecentaur

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    You are probably referring to the Perhapsatron circuit. I sometimes think I have managed to built one but then it behaves differently and I'm not sure. :smile:
     
  17. Mar 10, 2014 #16

    berkeman

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    :rofl:
     
  18. Mar 10, 2014 #17

    jim hardy

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    :surprised



    sophie, i think you've just uncovered the secret of Microsoft Windows !

    :rofl:

    :thumbs:
     
  19. Mar 12, 2014 #18
    I do not see how I am solving for the voltage drop across the resistor, so the L.E.D. sees 2V.

    The resistor resists current. You reduce current, you reduce voltage, right? I get that. So from that 9V battery, the 700 ohm resistor is going to reduce current down to .01 amps after the resistor. The voltage should now be 2V after the resistor.

    I do not see that.

    Here, maybe it's best we forget about the diode. If I wire a 700 ohm resistor in series with a 9V, how do I know what the voltage drop is after the resistor?
     
  20. Mar 12, 2014 #19

    berkeman

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    V=IR.

    And it's "voltage drop across the resistor", not "voltage drop after the resistor". :smile:
     
  21. Mar 12, 2014 #20
    7=.01R

    R=700

    Nowhere in that equation do I see 2V
     
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