Part 1: A number 12 copper wire has a diameter of 2.053 mm. Calculate the resist

Question

Part 1:
A number 12 copper wire has a diameter of 2.053 mm. Calculate the resistance of a 34.0 m long piece of such wire. (Use 1.72×10-8 m for the resistivity of copper.)

Part 2:

Part 3:

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

Part 4:

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.)

Part 5:

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.)

Explanation / Answer

Diameter d = 2.053 mm
Radius r = d/ 2
             = 1.0265 mm
              = 1.0265 x 10 -3 m
Area of cross section A = r 2
                                  =3.3103 x 10 -6 m 2

Length L = 34 m

Resistivity = 1.72 x 10 -8 m

Resistance R = L / A

Substitute values we get  R =  0.1766

(b). Current i = 20 A

Power P = i 2 R

            = 70.66 watt

(c)voltage  V = iR

                   = 3.532 volt

(d)Current density J = i / A

                              = 6.041 x 10 6 A / m 2

(e).Drift velocity  v = J / nq

Where  n = The density of electrons in copper is 8.47×1028 m-3

            q = charge of electron

            = 1.6 x 10 -19 C

So, v = 4.458 x 10 -4 m/ s

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