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Question: List three factors on which the resistance of a conductor depends. The resistivities of some substan...

List three factors on which the resistance of a conductor depends. The resistivities of some substances are given below

MaterialsABCDE
Resistivity1.6×1081.6 \times {10^{ - 8}}6.4×1086.4 \times {10^{ - 8}}10×10810 \times {10^{ - 8}}96×10896 \times {10^{ - 8}}100×104100 \times {10^{ - 4}}

Answer the following questions in relation to them giving justification for each:

  1. Which material is best for making connecting cords?
  2. Which material do you suggest to be used in heating devices?
  3. You have two wires of the same length and the same thickness. One is made of material A and another of material D. If the resistance of wire made of A is2Ω2\Omega , what is the resistance of the other wire?
Explanation

Solution

Resistance depends on the length, type, thickness, and temperature of the material. Resistivity which is also known as specific resistance of the material is the measure of the resistance of the material. We prefer materials with low resistance for making connecting cords and materials with high resistance to make heating devices.
Formula used:
RlAR \propto \dfrac{l}{A} where R stands for the resistance of the material, l stands for the length of the material, and A stands for the area of the material
R=ρlAR = \rho \dfrac{l}{A} where ρ\rho stands for the resistivity of the material.
WhenlA=1\dfrac{l}{A} = 1 ,
\Rightarrow R=ρR = \rho

Step by step Solution:
The resistance of a material is directly proportional to the length of the material and it is inversely proportional to the area of cross-section. There are other factors also which affect the resistance which is the type of material and the temperature of the material.

List of factors that affect the resistance of a conductor

  1. Type of the material
  2. Length of the material
  3. Thickness/ Area of the material
  4. Temperature of the material.
    -Material A is the best for making connecting cords.
    For making connecting cords materials with low resistance is the best, this will avoid loss of current due to heating.
    -Material D is best for making heating devices.
    For making heating devices we need materials with very high resistance since it will induce more heat in the material. Since material E has a resistivity in the range of104Ωm{10^{ - 4}}\Omega - m, it falls under the category of semiconductors hence material E cannot be used for making heating devices. This makes material D best among the given materials to make a heating device.
    -Let ρ1{\rho _1} be the resistivity of material A and ρ2{\rho _2} be the resistivity of the material D.
    The resistance of the material A is given by R1{R_1}.
    We have to find the resistance of the material D (R2){(R_2)} . To find the resistance of the material D, we should use the formula for resistance which involves resistivity. The formula is given by
    R=ρlAR = \rho \dfrac{l}{A}
    Using the formula for two resistances, R1{R_1} and R2{R_2}, and then taking their ratio gives us,
    R1R2=ρ1lAρ2lA\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{\rho _1}\dfrac{l}{A}}}{{{\rho _2}\dfrac{l}{A}}}
    We know that the length and thickness are same for the material. Hence we get lAlA=1\dfrac{{\dfrac{l}{A}}}{{\dfrac{l}{A}}} = 1
    Therefore, R1R2=ρ1ρ2\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{\rho _1}}}{{{\rho _2}}} (Since lAlA=1\dfrac{{\dfrac{l}{A}}}{{\dfrac{l}{A}}} = 1 )
    R2=R1ρ2ρ1{R_2} = \dfrac{{{R_1}{\rho _2}}}{{{\rho _1}}} Ω\Omega
    From the equation of resistance that includes resistivity in it, we have deduced an expression for the resistance of the material. That is shown in the above steps. From the ratio of resistances, we have to write an equation for the resistance to be found which in this case isR2{R_2}. After finding the equation substitute the values ofR1,ρ1,ρ2{R_1},{\rho _1},{\rho _2}
    R1=2Ω{R_1} = 2\Omega (Given in the question)
    ρ1=1.6×108Ωm{\rho _1} = 1.6 \times {10^{ - 8}}\Omega - m (Given in the question)
    ρ2=96×108Ωm{\rho _2} = 96 \times {10^{ - 8}}\Omega - m (Given in the question)
    R2=2×96×1081.6×108Ω{R_2} = \dfrac{{2 \times 96 \times {{10}^{ - 8}}}}{{1.6 \times {{10}^{ - 8}}}}\Omega
    R2=120Ω{R_2} = 120\Omega

Note:
Keep in mind that the order of resistivity of conductors is in the range 108{10^{ - 8}} Ωm\Omega m.The resistivity of semiconductors (example: Silicon, Germanium, Carbon, Graphite etc.) are in the range 105103Ωm{10^{ - 5}} - {10^3}\Omega m and the resistivity of bad conductors are in the range 10141016Ωm{10^{14}} - {10^{16}}\Omega m. Pay attention to power while solving problems.