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does resistance in a "ideal wire" not change due to temperature?

https://oxfordpat.wordpress.com/oxford-pat-2016-question-13/ surely resistance increases with temperature?
Original post by alvan15
https://oxfordpat.wordpress.com/oxford-pat-2016-question-13/ surely resistance increases with temperature?


Yes, you are correct. But be careful with the terminology.

The word 'ideal' is another way of saying a resistance follows Ohms law without needing to invoke thermal considerations or any other parameter which would cause a deviation form Ohms law such as inductance, capacitance etc.

EDIT: For an ideal conductor, there is no resistance and hence no energy losses.
(edited 6 years ago)
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alvan15
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Original post by uberteknik
Yes, you are correct. But be careful with the terminology.

The word 'ideal' is another way of saying the conductor follows Ohms law without needing to invoke thermal considerations or any other parameter which would cause a deviation form Ohms law such as inductance, capacitance etc.


thank you , that explains it perfectly :smile:
Original post by alvan15
https://oxfordpat.wordpress.com/oxford-pat-2016-question-13/ surely resistance increases with temperature?


Ideal wires have zero resistance. That's why they're ideal, lol.
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alvan15
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Original post by Physics Enemy
Ideal wires have zero resistance. That's why they're ideal, lol.


thanks :P
Original post by uberteknik
Yes, you are correct. But be careful with the terminology.

The word 'ideal' is another way of saying the conductor follows Ohms law without needing to invoke thermal considerations or any other parameter which would cause a deviation form Ohms law such as inductance, capacitance etc.


It has virtually no resistance to begin with, it's not like there's a significant resistance that's maintained despite the temperature increase.
Original post by alvan15
thank you , that explains it perfectly :smile:


Following on from Physics Enemy, yes, that does imply that an 'ideal resistance' follows Ohms law perfectly and an 'ideal conductor' is perfect with no resistance and hence no energy losses. For calculation purposes it can therefore be ignored.
They're testing whether you can suss out what makes a wire, or any component, ideal. You have to consider its purpose. We want a wire to conduct optimally.
(edited 6 years ago)

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