(a) The thickness of a wire is halved. What happens to its resistance? How will the resistivity
change.
Answers
Explanation:
The electric resistance of a material is a property that represents the opposition to the passage of the electric current. From the microscopic point of view the resistance depends on the atomic structure of the material. From the macroscopic point of view, the resistance depends on the linear dimensions of the material.
The value of the resistance of a wire can be calculated using the following expression:
R = ρL/s
Where:
ρ, is the specific resistance which is related to the material; also has a temperature dependence.
L, is the length of the wire.
S, is the area of cross section of the wire.
For a circular cross section wire, we have:
R1 = ρL/πr1^2 = (4ρL)/(πd1^2)
If the diameter is halved, we have for the resistance:
R2 = (4ρL)/[π(d1/2)^2] = (16ρL)/(πd1^2)
By comparing the values of resistance:
R2/R1 = (16ρL/πd1^2)/(4ρL/πd1^2)
R2/R1 = 16/4 = 4
R2 = 4R1
So, the resistance increases four times.
Answer:Explanation:
As the length of the conductor get doubled , the resistance of the conductor also increases. Doubling the length of the conductor will double the resistance of the conductor , but the thickness of the conductor must also get halved as it is stretched because it contain the same amount of metal in twice its length.