A number 14 copper wire has a diameter of 1.628 mm. Calculate the resistance of

Question

A number 14 copper wire has a diameter of 1.628 mm. Calculate the resistance of a 37.0 m long piece of such wire. (Use 1.72×10-8 m for the resistivity of copper.)

For safety, the National Electrical Code limits the allowable amount of current which such a wire may carry. When used in indoor wiring, the limit is 15.0 A for rubber insulated wire of that size. How much power would be dissipated in the wire of the above problem when carrying the maximum allowable current?

What would be the voltage between the ends of the wire in the above problem?

What is the current density in the wire when it is carrying the maximum allowable current? (Current density is the current in the wire divided by the cross sectional area of the wire.)

What is the drift velocity of the electrons when the wire is carrying the maximum allowable current?
(The density of electrons in copper is 8.47×1028 m-3.)

Please answer all parts and show work

Explanation / Answer

From the given data:
1)
R = *L/A
with L = length in m,
A = cross section area in m^2
R = (1.72*10^-8*37)/(pi*(1.628*10^-3/2)^2) = 0.305

2)
P = U*I
with U = R*I
P = RI^2
P = 0.305*15^2 = 68.78 Watt
loss

3)
Voltage loss U = RI = 0.305*15= 4.575 V
4.575 V and 0 V at both ends.

4)
15/(pi*((1.628*10^-3)/2)^2) = 7.205*10^6 A/m^2

5)
v = I/(n*A*q)
with n = electrons/m^3, A = cross section area in m^2,
q = charge of an electron =- 1.6*10^-19 C
v = (15)/((8.47*10^28*(pi*((1.628*10^-3)/2)^2)*(1.6*10^-19))
v = 0.532*10^-3m/s

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