Question

A thin gold wire (resistivity 2.44×10−8Ωm) with a diameter of 2.05mm and a length of 60m is connected to a small voltage source. When connected, a current of 10A flows through the wire.  

a) What is the voltage supplied by the voltage source?

b) If we replace the gold wire with a platinum wire (resistivity 1.06×10−7Ωm) that has the same diameter and length, what would the current through the platinum wire be?

Answer 1

resistance of wire

R = rho * L / A

R = 2.44* 10^-8* 60 / ( 3.14 / 4* ( 2.05* 10^-3)^2)

R = 0.4437 Ohm

a)

applied voltage is given by

V = iR = 10* 0.4437 = 4.437 V

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b)

new resistance

R' = 1.928 ohm

i = V/R' = 4.437 / 1.928 = 2.3 A

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