Question

A wire has a resistivity of 6.80 x 10-8 ohm . meter when the temperature is 75.0 degrees Farenheit. If the temperature coefficient of resistivity of the material that the wire is made from is 5.0 x 10-3 (1/Kelvin), what will be the resistivity of the wire when it is cooled to 45.0 degrees Farenheit? Note: before working this problem you must convert the temperature from the Farenheit temperature scale to the Kelvin scale. Give your answer in the form "a.bc" x 10-8 Ω.m.

(2) A heating element is made by maintaining a potential difference of 13.5 V along the length of a certain wire with a 6.80 x 10^-6 m² cross section area and a resistivity of 4.68 x 10^-6 ohm.m. If the element dissipates 5.0 W, what is its length? Put your answer in the form of "a.bc x 10^(x) m"

(3)a potential difference of 12.8 V is maintained between the ends of a 1500 cm length of wire whose diameter is 0.56 mm. The conductivity of the wire is 4.80 x 10⁷(ohm.m)^-1. Determine the rate at which energy in the wire is transformed from kinetic to thermal energy. Give your answer in the form "a.bc x 10^(x) unit".

(4)A current of 126 mA exists in a wire for 5.60 minutes. What is the rate that charge passes a point in the wire? Give your answer in the form "a.bc x 10^(x)" C/s. Continuing the previous question, what is the rate of electrons passing through a cross section perpendicular to the wire's axis? Give your answer in the form "a.bc x 10^(x)" electrons/second. Assume that the charge moves in a straight line from one end of the wire to the other, how many electrons are moved through the cross section during 38.0 ms? Give your answer in the form "a.bc x 10^(x)" electrons

(5) An electrical current of 8.50 mA exists in a solid cylindrical wire whose diameter is 1.50 mm. Calculate the magnitude of the current density in the wire. Put your answer in the form of "a.bc x 10^(x)" A/m².  Assume that electrons are the charge carriers and the conduction electron density is 8.55 x 10²⁹/m³. (Note that this is the charge carrier density and could also be stated as 8.55 x 10²⁹ charge carriers/m³ or electrons/m³.) Calculate the electron drift speed in the wire. Put your answer in the form of "a.bc x 10^(x) m/s".

(6.)A wire with a resistance of 6.80 mΩ is drawn out through a die so that its new length is 4 times its original length. Find the resistance of the longer wire, assuming that the resistivity and density of the material are not changed during the drawing process. Put your answer in the form of "a.bc x 10^(x) ohm".

Answer 1

(a) Given that :

resitivity of wire, ₀ = 6.8 x 10^-8.m

initial temperature, T₀ = 75 ⁰F = 297.04 K

temperature coefficient of resistivity, = 5 x 10^-3 K^-1

The resistivity of the wire when it is cooled to 45 ⁰F which will be given as :

using an equation,    = ₀ + ₀ (T -T₀)    { eq.1 }

where, T = 45 ⁰F = 280.37 K

inserting the values in eq.1,

= (6.8 x 10^-8.m) + (5 x 10^-3 K^-1) (6.8 x 10^-8.m) [(280.37 K) - (297.04 K)]

= (6.8 x 10^-8.m) + (34 x 10^-11 K^-1.m) (-16.67 K)

= (6.8 x 10^-8.m) - (566.7 x 10^-11.m)

= (6.8 x 10^-8.m) - (0.566 x 10^-8.m)

= 6.23 x 10^-8.m

(b) Given that :

potential difference, V = 13.5 V

cross-section area of wire, A = 6.8 x 10^-6 m²

resistivity, = 4.68 x 10^-6.m

power dissipated, P = 5 W

using an equation,      R = (L / A)                                                              { eq.2 }

and   P = V² / R

Or R = V² / P                                                                                           { eq.3 }

equating eq.2 & 3, we get

V² / P = (L / A)  

inserting the values in above relations,

(13.5 V)² / (5 W) = (4.68 x 10^-6.m) [L / (6.8 x 10^-6 m²)]

L = (36.45 ) (6.8 x 10^-6 m²) / (4.68 x 10^-6.m)

L = (247.86 x 10^-6.m²) / (4.68 x 10^-6.m)

L = 52.9 m

(c) Given that :

potential difference, V = 12.8 V

length of the wire, L = 1500 cm = 15 m

diameter of the wire, d = 0.56 x 10^-3 m

radius of the wire, r = 0.28 x 10^-3 m

conductivity of wire, = 4.8 x 10⁷ (.m)^-1

using eq.2,   R = (L / A)                

where, = resistivity = 1 / 1 / [4.8 x 10⁷ (.m)^-1]

= 2.08 x 10^-8.m

and   A = area of wire = r² (3.14) (0.28 x 10^-3 m)²

A = 2.46 x 10^-7 m²

Now, inserting the values in eq.2 -

R = (2.08 x 10^-8.m) [(15 m) / (2.46 x 10^-7 m²)]

R = 12.6 x 10^-1

R = 1.26

The rate at which energy in the wire is transformed from kinetic to thermal energy which is given as :

using eq.3,         P = V² / R

inserting the values in eq.3,

P = (12.8 V)² / (1.26 )

P = 130.1 W