How to differentiate between single and multiloop circuits

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To differentiate between single and multiloop circuits, one must consider both the number of loops and the spatial arrangement of components rather than just the number of voltage sources. A circuit can appear multiloop without actually being one, and simplification is key to understanding its structure. The placement of the power supply does not determine the circuit type; rather, it's about how components are connected. Using Kirchhoff's law can help analyze the circuit, ensuring that the voltage leaving the battery equals the voltage returning. Ultimately, recognizing the arrangement and simplifying the circuit will clarify whether it is a single or multiloop circuit.
novafx
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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. :rolleyes: Thanks for the help!:smile:
 
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The term "loop" should be big hint. Does "loop" make you think about batteries or some sort of spatial arrangement?
 
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"?
 
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?
 
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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novafx said:
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?

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.
 
Ah ok thanks. :smile:
 
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