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Homework Help: How to differentiate between single and multiloop circuits

  1. Jul 16, 2009 #1
    This is just a general question that will help me start a problem, so I have no work to show. The question is: How can you tell that a problem is a multiloop circuit? Does it correspond to the number of voltage sources (batteries) in the circuit, or is it based off of how the components of the circuit are arranged? If it's the second one, could you tell me how to tell where the components would typically be for a multiloop circuit? I guess that's more than one question. :uhh: Thanks for the help!:smile:
  2. jcsd
  3. Jul 16, 2009 #2


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    The term "loop" should be big hint. Does "loop" make you think about batteries or some sort of spatial arrangement?
  4. Jul 16, 2009 #3
    Well I guess it has to do with the spatial arrangement. So would a double loop problem just need a single power source in the middle of two circuit "loops"?
  5. Jul 16, 2009 #4


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    Middle is a poor term to use when describing a loop (and probably just describes a poor rendition of a single circuit with some items in parallel). If there is only a single battery, then it is only a single loop circuit. This doesn't mean that more than one battery makes it a multiloop problem though.

    I'm afraid I failed at trying to get you to answer your own question simply. The answer to your original question isn't always a case of one or the other (which I initially took for granted). If you haven't learned already, then you will find that there are many ways that a circuit can be simplified.

    A circuit can (and will for class) appear quite complicated, or to be multiloop when it is not, or have multiple batteries in a single loop. You have to simplify the circuit before you can really decide whether it is multiloop or not. In the end, when it is as simple as it can be, then you can decide whether it is multiloop or not by the number of loops that it has.

    Does that make sense?
  6. Jul 16, 2009 #5
    Yeah I'm pretty sure I understand what you're saying. My problem is being able to look a circuit like the one attached, and trying to picture another, much simpler, circuit diagram. The fact that the power supply is in the center instead of the left or right is throwing me off. The circle represents an ideal multimeter, which I believe they mean that it does not have a resistance. The power supply is a "real" battery that has an internal resistance. I have find the resistance of the battery given the voltage when the circuit is open, and the voltage when the circuit is closed.

    I don't know how to look at this circuit in a simpler way. I need to use Kirchoff's law - I know that. The voltage leaving the battery equals the voltage going back to the battery, right? So 6 Volts = I(1/Rb) + [IRb + I(1000 Ohms)] when the circuit is closed?

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    Last edited: Jul 16, 2009
  7. Jul 16, 2009 #6
    Would it help you picture it if the battery and multimeter switched positions? They're both connected in parallel with each other by a junction, and the circuit diagram is the same whether the battery is in the middle or on the left.
  8. Jul 16, 2009 #7
    Ah ok thanks. :smile:
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