Cambridge Physics Problems 23: resistor of constant resistance
Physics and electronics discussion, revision, exam and homework help.
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Re: Cambridge Physics Problems 23: resistor of constant resistanceWe are supposed to find the proportion of the two metals, instead of starting from "14/44 constantan and 30/44 parts manganin".(Original post by Orphan)
Err, don't you just need a wire that is 14/44 constantan and 30/44 parts manganin.
Use this to work out the average resistance of the wire and hence the length you need?
I don't know... Wouldn't that make the working more complex?(Original post by Joinedup)
Would working out the lengths ratio required to achieve 0 temp coefficient first help?
Caution - still on my first coffee. -
Re: Cambridge Physics Problems 23: resistor of constant resistance
Err, But the question asks for the two lengths - not the proportions. "14/44 constantan and 30/44 parts manganin" is both intuative and correct and is required to generate any constant resistance wire. After that the problem is trivially easy surely?
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Re: Cambridge Physics Problems 23: resistor of constant resistanceYour formula is for temperature zero but you also need to express the equation for total R = 5 for the case where the temperature is any temperature theta.(Original post by johnconnor92)


How do I make out a ratio of Lc/Lm from the final working? Simple algebra but I can't make head or tails of it. Please help!!!
Thank you!
If you do that you will get an equation where theta cancels and you obtain a formula for the ratio of the lengths.
Combine that with the formula you already have to get actual values for the two lengths. -
Re: Cambridge Physics Problems 23: resistor of constant resistance
Setting theta to zero gives you the resistance when there is no temperature change, not when there is no resistance change with temperature. You need to keep those terms and make the constant in front of the theta that is dependent on Lc and Lm equal to zero.
edit: nvm Stonebridge said it