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To what value would you have to raise the temperature of a copper wire (originally at 20 ∘C) to increase its resistance by 17 % ?

User Arva
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1 Answer

2 votes

Answer:

The temperature is
T_2=  63.59 ^oC

Step-by-step explanation:

Generally the equation that represents the variation of resistance with respect to temperature is mathematically represented as


R = R_i (1 +  \alpha *  \Delta  T  )

=>
R = R_i (1 +  \alpha *  [T_2 -  T_1] )

Here
T_1 is the original temperature with value
T_1 = 20^oC


R_i is the initial resistance of the copper

R is the resistance at temperature
\Delta  T


\alpha is the temperature coefficient of resistivity with a value


\alpha =3.9 *10^(-3) \  ^oC^(-1)

From the question we are told that


R =  17\% \  of  R_i + R_i

So


R =  0.17R_i + R_i


R =  1.17R_i

Hence


1.17 R_i = R_i (1 + 3.9 *10^(-3)  *  [T_2 -  20] )


1.17  =  (1 + 3.9 *10^(-3)  *  [T_2 -  20] )

=>
0.17 = 3.9 *10^(-3)  *  [T_2 -  20] )

=>
[T_2 -  20]  = 43.5897

=>
T_2=  63.59 ^oC

User Ocrdu
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