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Voltage Divider Circuits. Set Me Straight.

  1. Dec 16, 2008 #1
    Hello all.

    I wonder if anybody could clear this up for a beginner.

    In the circuit attached: Is V OUT 1 of a higher voltage than V OUT 2?

    Does the current travel at 9V, until it drops at R1? Does it drop again at R2 and give an even lower voltage to V OUT 2? Or does the presence of two resistors apply the total resistance to the entire circuit, giving both V OUTs equally lowered voltage?

    Any help would be much appreciated. Thanks.

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  3. Dec 16, 2008 #2


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    The voltage drops across every resistor.
    The current remains the same through every circuit element.
    9V > Vout1 > Vout2 > 0

    This looks like an LTSpice circuit; have you tried just reading the values for Vout1 and Vout2?

    p.s Welcome to PF :smile:
  4. Dec 16, 2008 #3
    I downloaded the program just for this post. I don't yet know how to use it.
    So, if I attach LEDs to both V OUTs, they will shine with equal brightness?


    Thanks for the welcome! Sorry if this question should be obvious.
    Last edited: Dec 16, 2008
  5. Dec 16, 2008 #4
    Vout2 = 0V. i don't like the voltage source symbol. i can't tell if Vout1 is -3V or +3V.
  6. Dec 16, 2008 #5
    The source is +9, if that helps. But I think I get the point guys. Thanks for the help!
    Last edited: Dec 16, 2008
  7. Dec 17, 2008 #6


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    Aaaack, how did I miss this? There's no "circuit" here, no path for current to return to the voltage source.

    So it's zero current through the resistors. Then all voltages are zero except at the +terminal of the source which is at +9V.

    In principle, you'd want to connect the voltage source back to ground somehow.
  8. Dec 17, 2008 #7
    The symbols are perhaps slightly deceptive. The triangle at the bottom represents the ground. The Voltage source is meant to be a source of positive voltage - must be conventional flow. It's an unusual shape for a "circuit" but irrelevant to my question.

    Which was whether each V OUT receives equal (0) voltage or if 1 receives more (1/3) than 2 (0).

    Good thinking though.
  9. Dec 17, 2008 #8
    Meaningless circuit because there is no circuit so no current flow and no volt drop so as it stands V1 = V2 = -9 v wrt the +ve of the source.

    Does the arrow down mean 'earth' or 'ground'. It si not the right symbol. A arrow like that usually means that circuit connects with another.
  10. Dec 17, 2008 #9
    It does mean ground yes. It's an odd symbol; I know. However, that's what LTspice called a ground. Keep in mind I've accumulated a mere five minutes experience with the program, just today.

    Say there is a circuit.


    In such a situation, where the total resistance leaves 0 voltage, will neither LED shine, or will one shine dimly while the other doesn't at all?

    That's the best way in which I can ask this question. If that still doesn't make sense, I clearly understand too little and need to experiment some.

    Thanks everyone for so much response!
  11. Dec 17, 2008 #10


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    I've used LTspice myself, but don't have it installed yet on this computer that I started using a month or 2 ago. Keep playing around with the program and the circuit, you'll learn the more you do it.

    To make this circuit "meaningful" you'll need to connect the battery + back to ground. From what I remember, you click on the icon that looks like a line or wire along the top row of the LTspice window. Then you can draw the wire to make the connection. Hit <esc> (I think) to stop drawing the wire.

    Since the elements are all in series, and it's one big loop, both LED's shine equally (assuming they are identical to each other). Also, you could combine the two resistors into a single 330 ohms.

    Circuit elements in series always have the same current. So this means the LED's, resistors, and battery all have the same current.
  12. Dec 17, 2008 #11
    Brilliant. Thanks!

    Obviously quite an active community here. I look forward to next time. Thanks everyone!
  13. Dec 17, 2008 #12
    LEDs have a non-linear V-A curve. I will test some on a power supply later.
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