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Half of a coil wire of a heating device which is set in a socket, is chilled.How does...

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afrokaa | Student, College Freshman | (Level 2) Honors

Posted June 29, 2009 at 10:55 PM via web

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Half of a coil wire of a heating device which is set in a socket, is chilled.How does change the total released heat and the current in each half?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted June 30, 2009 at 3:17 AM (Answer #1)

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By cooling, the resistance is decreasing, then the current I=U/R is increasing and the total power is increasing also P=U^2/R

R1,2 are the resistance of the 2 halves and R1-r1 and R2+r2, their final resistance, because, due to the first half cooling, R1 is decreasing and R2 is increasing, because the current and temperature are increasing.

P1=R1*I1^2=R1[U/(R1+R2)]^2

P'1=R1-r1)[U/(R1-r1+R2+r2)^2

P1/P'1=[(R1-r1)/R1](R1+R2)^2/(R1+R2-r1+r2)^2=

=(1-r1/R1)/[1-(r1-r2)/(R1+R2)]^2=1-r1/R1<1

We consider that 2(r1-r2)/(R1+R2)<r1/R1, so the heat released in the chilled half is decreasing and in the unchilled half, is increasing.

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