Question #171536

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.

Expert's answer

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 mm^{2}) / (4 A × 10 m) = (220 V × 10^{-5} m^{2}) / (4 A × 10 m) = **5.5 × 10**^{-5}** Ω⋅m**

ii) According to the equation:

R = R_{ref} [1 + α(T - T_{ref})]

where R - resistance, R_{ref} - resistance at reference temperature, α - temperature coefficient, T - temperature, T_{ref }- reference temperature.

As R = V/I, the equation is:

V/I = (V_{ref} / I_{ref}) [1 + α(T - T_{ref})]

From here:

I = V / (V_{ref} / I_{ref}) [1 + α(T - T_{ref})] = 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.**

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