αt1 = Rt2-Rt1t2-t1×1Rt1
α20 = 60-4040-20×140
= 140 °C - 1
αt1 = 1t1+1/α0
α20 = 120+1/α0 = 140
20+ 1α0 = 40
1α0=20
α0= 120 °C - 1
αt1 = 1t1+1/α0
α100 = 1100+1/0.01
= 1100+100
=0.005°C-1
銅: Rt2Rt1 = 234.5+t2234.5+t1
9690 = 234.5+t2234.5+35.5
t2=53.5°C
Dt=t2-t1=53.5-35.5=18°C
Rt2=Rt1 [1+αt1(t2-t1)]
R75=100 [1+0.1%(75-25)]
=105Ω
銅: Rt2 = Rt1×234.5+t2234.5+t1
R65.5=25×234.5+65.5234.5+15.5
=25×300250
=30Ω
αt = 1| T0 |+T
αt | T0 | + αt T = 1
αt | T0 | = 1 - αt T
| T0 | = 1 - αt T αt
α0 = αt1-αtT
PS:α0 與 | T0 | 互為導數
直徑為 D 毫米,圓面積 = πD24 × 10-6 (m2)
R = ρ lA = 4ρlπD2 (MΩ)
若導線被剪掉四分之一長度,則長度變成原來 3/4,R = 4ρlπD2 ×3/4 = 3ρlπD2 (MΩ)
若導線被均勻拉長為原來的N 倍( 體積不變),則電阻變成原來的 N2 倍 R = 4ρlN2πD2 (MΩ)
若室溫上升為t2℃
R2R1 = 234.5+t2234.5+t1 ⟹ R2 = R1 × 234.5+t2234.5+t1 = R1 ×
Rt2 = 4ρ𝓁πD2 × =
α30 = 6 - 3150 - 30 × 13
= 1120 °C-1