- Given:
l = 3 m,
d = 0.005 m,
ρ=1.7⋅10−8Ohm⋅m.
Required:
R - ?
Solution:
R=ρSl=ρπd24l=1.7⋅10−8⋅3.14⋅0.00524⋅3=0.0026 Ohm.
2.
Given:
R0=10 Ohm,
α=−0.0005 1/°C,
t0=20°C,
t1=30°C,
t2=10°C.
Required:
R1 - ?
R2 - ?
Solution:
R1=R0(1+α(t1−t0))=10(1+(−0.0005)(30−20))=9.95 Ohm.
R2=R0(1+α(t2−t0))=10(1+(−0.0005)(10−20))=10.05 Ohm.
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