Mean temperature of winding when current falls

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SUMMARY

The discussion centers on calculating the mean temperature of winding when current falls, utilizing the resistance-temperature relationship defined by the equation R2=R1(1+alpha(t2-t1)). The user initially calculates R1 as 50 ohms and R2 as 63.94 ohms, but encounters discrepancies in solving for t2. The correct manipulation of the equation reveals an error in the sequence of operations, leading to the correct temperature of 16.274 degrees Celsius. The source of the equation is confirmed to be from B.L. Theraja.

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DevonZA
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Homework Statement


upload_2016-5-1_12-0-27.png


Homework Equations


R2=R1(1+alpha(t2-t1))

The Attempt at a Solution


R1=250/5=50ohms
R2=250/3.91=63.94ohms

R2=R1(1+alpha15degrees(t2-t1))
63.94=50(1+1/254.5(t2-15))
t2=

Now I found this online but the answers provided still don't match, 84.25 being the closest.

upload_2016-5-1_12-12-12.png


When I manipulate the equation I get a totally different answer:
63.94=50(1+1/254.5(t2-15))
63.94/50=(1+1/254.5(t2-15))
1.2788/1+1/254.5=t2-15
1.274=t2-15
t2=1.274+15=16.274 degrees

Please help me understand where I am going wrong.
 
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DevonZA said:
where I am going wrong.
DevonZA said:
63.94/50=(1+1/254.5(t2-15))
1.2788/1+1/254.5=t2-15
upload_2016-5-1_12-12-12-png.100061.png

Is this from B.L. Theraja?
 
Not sure I found it online
 
DevonZA said:
Not sure I found it online
It is from B.L. Theraja. The page looks familiar. The red line in the above post is your error.
 
What have I done wrong though? Every way I enter it into my calculator yields an incorrect answer
 
DevonZA said:
63.94/50=(1+1/254.5(t2-15))
1.2788/1+1/254.5=t2-15[/\QUOTE]
This step is wrong. Check the sequence of operations.
 
Ah I got it :-) thank you
 

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