A resistance element having cross sectional area of 10 mm2 and
a length of 10 meters. This resistance takes a 4 A current from a 220
V supply at ambient temperature of 20 oC. Calculate, i) The resistivity
of material and 2) current it will take when the temperature rises to 60
oC. Assume α20=0.0003/oC.
i) According to the equation:
R = ρL / A
where R - resistance, ρ - resistivity, L - length, A - cross-sectional area.
From here:
ρ = RA / L = VA / IL
where V - voltage, I - current.
As a result, resistivity equals:
ρ = (220 V × 10 mm2) / (4 A × 10 m) = (220 V × 10-5 m2) / (4 A × 10 m) = 5.5 × 10-5 Ω⋅m
ii) According to the equation:
R = Rref [1 + α(T - Tref)]
where R - resistance, Rref - resistance at reference temperature, α - temperature coefficient, T - temperature, Tref - reference temperature.
As R = V/I, the equation is:
V/I = (Vref / Iref) [1 + α(T - Tref)]
From here:
I = V / (Vref / Iref) [1 + α(T - Tref)] = 220 V / {(220 V / 4 A) × [1 + 0.0003 °C(60 °C - 20 °C)]} = 4 A
Answer: 5.5 × 10-5 Ω⋅m; 4 A.
Comments
Leave a comment