Question

A volltage drop of 0.00039 V is measured across a 1.2 m long piece of copper wire when 10

mA of current passes through it.

a. What is the resistivity of copper?

b. What is the radius of the wire?

c. What is the power dissipated in the wire?'

b is r=4.08 x 10-4 ^m and c is 3.9 x 10^6 J

If someone could show me step by step how to get to these answer that would be great.

Answer 1

Resistance of 1.2 m wire = voltage drop / current = 0.00039/(10*0.001) = 0.039 ohms

a) resistivity of copper = 1.698*10^-8 ohm/m

b) We have resistance = Resistivity*length / cross sectional area

so, 0.039 = 1.698*10^-8 * 1.2 / cross sectional area

cross sectional area = 0.0000005224615385 m^2

we have pi*r^2 = 0.0000005224615385

3.14*r^2 = 0.0000005224615385

r = 0.000407908 m = 4.079 *10^-4 m

radius = 4.079 *10^-4 m

c) Power dissipated =current ^2 * Resistance = i^2 * R = (10*0.001)^2 * 0.039 = 0.0000039 J = 3.9 x 10^6 J

Answer 2

Resistance = V/I = 0.00039/10*10^(-3) = 0.039

Resistance = Resistivity*L/A

Hence resistivity comes out to be 1.7*10^(-8)

b) Resistance = Resistivity*L/A

A = 1.7*10^(-8)*1.2/0.039 = 52.04*10^(-8)

A = pi*r^2

52.04*10^(-8) = 3.14*r^2

hence r = 4.07*10^(-4) m

c) P = VI

= 0.00039*10*10^(-3) = 3.9*10^(-6)J

Answer 3

fg

Answer 4

step1:

V= IR

V= voltage, I = current R = resistance

R = V/I = 0.00039/0.001 = 0.39 ohm

R = p * L/A

p = resistivity of copper it is 1.68x10^-8

given L = 1.2m

find A

Power = VI = V/R2 = IR2 we can use any formula

Answer 5