P27_069.PDF

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Chapter 27 - 27.69
69. (Third problem of Cluster )
(a) We use Eq. 27-17 with ρ = 1 8 ρ 0 (we are neglecting any thermal expansion of the material) and
T
−
T 0 = 100 K in order to obtain α =2 . 5
×
10 − 3 / K. Now with this value of α but T = 600 K (so
R =1 . 75(8 . 0Ω)=14Ω.
(b) We are assuming the wires have unknown but equal length (not the lengths shown in Figure 27-33).
With α D =5 . 0
−
T 0 = 300 K) we find ρ =1 . 75 ρ 0 →
T 0 = 300 K. With the same assumptions as in
part (a), this implies R =2 . 5 R 0 where R 0 = 16 Ω (that the resistance of D is twice that of C at
300 K is evident in part (a) of the previous solution. Therefore, R =2 . 5(16 Ω) = 40 Ω for wire D
at T = 600 K.
×
10 − 3 / K, we find ρ =2 . 5 ρ 0 for T
−
T

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