A current-carrying gold wire has a diameter of 0.90 mm . The electric field in t

Question

A current-carrying gold wire has a diameter of 0.90 mm . The electric field in the wire is 0.48 V/m . Part A What is the current carried by the wire? Part C What is the resistance of a 6.5-m length of this wire? 25.70 A person with body resistance between his hands of 10 k accidentally grasps the terminals of a 16-kV power supply. Part B What is the power dissipated in his body? Part C If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be Imax=1.00mA or less?

Explanation / Answer

A)The resistivity of gold wire is approximately = 2.44*10^-8 m

A=pi*(0.45*10-³)²=6.36*10^-7 m²

We knows that current density j = i/A

Or i =EA/ = 0.48*6.36*10^-7/2.44*10^-8

i = 12. 51 A

B)V = E L = 0. 48*6.5 = 3.12 v

R = V/i = 3.12/12.51 = 0.25 ohms

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