Please help me with all 4 parts! Course Contents.» Homework 06 » Wire Properties

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Course Contents.» Homework 06 » Wire Properties Timer Notes d Evalu A potential difference of 1.80 V will be applied to a 39.00 m length of 18-gauge aluminum wire (diameter 0.0400 in). Calculate the current. 6.721 A Submit Answer Incorrect. Tries 1/10 Previous Tries Calculate the magnitude of the current density. Submit Answer Tries 0/10 Calculate the magnitude of the electric field within the wire. Submit Answer Tries 0/10 Calculate the rate at which thermal energy will appear in the wire. Submit Answer Tries 0/10

Explanation / Answer

Resistance is R = rho*l/A

rho is the resistivity of aluminium wire = 2.65*10^-8 ohm-m

l = 39 m

0.04 in = 0.001016 m

r = d/2 = 0.001016/2 m

A = pi*r^2 = 3.142*(0.000508^2) = 8.11*10^-7 m^2

then

R = rho*l/A = (2.65*10^-8*39)/(8.11*10^-7) = 1.27 ohm

then

current i = V/R = 1.8/1.27 = 1.42 A

current denisty j = i/A = 1.42/(8.11*10^-7) = 1.75*10^6 A/m^2

Electric field is E = V/l = 1.8/39 = 0.046 V/m

rate at which thermal energy will appear is P =i^2*R = 1.42^2*1.27 = 2.56 W

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