Draw a circuit diagram to measure resistance of wire

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Homework Help Overview

The discussion revolves around drawing a circuit diagram to measure the resistance of a wire, focusing on the inclusion of a rheostat in series with the power supply. Participants explore the implications of circuit design choices in relation to safety and measurement accuracy.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the necessity of a rheostat in the circuit, questioning the consequences of omitting it. There are considerations about controlling voltage and current to prevent damage to the wire and power supply.

Discussion Status

Guidance has been offered regarding the importance of controlling the voltage applied to the unknown resistance and the potential risks of short-circuiting the power supply. Multiple interpretations of the circuit setup are being explored, particularly concerning the role of the rheostat.

Contextual Notes

Participants note that the wire has a very low resistance, which raises concerns about the safety of the circuit configuration without additional resistance. There is an emphasis on practical considerations versus theoretical reasoning in circuit design.

Janiceleong26
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Homework Statement


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Homework Equations

The Attempt at a Solution


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This is how I drew it (in pencil) , but there should be a rheostat connected in series with the power supply, why?i can already calculate the resistance straight away by V/I
 
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There may be additional reasons, but a very basic one is that if you connected that circuit without additional resistance, such as from a rheostat, bad things would happen because you'd have a short circuit.
 
A piece of wire has a very low resistance. That means that you are in effect short circuiting the power supply with the wire. The reostat adds additional resistance in the circuit to keep the current drawn from the power supply within the limits of the power supply, so it is more for practical than theoretical reasons that it is needed.
 
Janiceleong26 said:
This is how I drew it (in pencil) , but there should be a rheostat connected in series with the power supply, why?i can already calculate the resistance straight away by V/I
You need control over the voltage applied to the unknown resistance, else it is likely to be destroyed before you can take your measurements. Your circuit shows a fixed voltage source connected to the unknown R, and with R being unknown you can't know ahead of time whether this voltage will be safe for the unknown resistor. A rheostat set to maximum can offer protection, and its inclusion shows you have considered this angle.

Had you shown the voltage source to be variable, then the rheostat would probably not be considered a necessary inclusion.

The function highlighted by andrewkirk is also appropriate, because every voltage source has a limit to the current it can supply.
 
You can vary the current in the wire by changing the reostat setting and thereby obtain a range of I, V values from which you could draw a graph and relate the gradient of the graph to the wire's resistance.
 
andrevdh said:
You can vary the current in the wire by changing the reostat setting and thereby obtain a range of I, V values from which you could draw a graph and relate the gradient of the graph to the wire's resistance.

NascentOxygen said:
You need control over the voltage applied to the unknown resistance, else it is likely to be destroyed before you can take your measurements. Your circuit shows a fixed voltage source connected to the unknown R, and with R being unknown you can't know ahead of time whether this voltage will be safe for the unknown resistor. A rheostat set to maximum can offer protection, and its inclusion shows you have considered this angle.

Had you shown the voltage source to be variable, then the rheostat would probably not be considered a necessary inclusion.

The function highlighted by andrewkirk is also appropriate, because every voltage source has a limit to the current it can supply.
andrewkirk said:
There may be additional reasons, but a very basic one is that if you connected that circuit without additional resistance, such as from a rheostat, bad things would happen because you'd have a short circuit.
Thanks! Got it
 

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