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Question: Calculate the resistance of an aluminium cable of length \[10\,{\text{km}}\] and diameter \[2.0\,{\t...

Calculate the resistance of an aluminium cable of length 10km10\,{\text{km}} and diameter 2.0mm2.0\,{\text{mm}} if the resistivity of aluminium is 2.7×108Ωm2.7 \times {10^{ - 8}}\,\Omega \,{\text{m}} .

Explanation

Solution

First of all, we will convert the length and diameter into an S.I system of units. We know, resistance varies directly with length and inversely with the area of cross section. We will use this technique to find the resistance.

Complete step by step answer:
In the given problem, we are supplied the following data:
There is an aluminium cable whose length is 10km10\,{\text{km}} .The diameter of the wire is 2.0mm2.0\,{\text{mm}} .The given resistivity of aluminium is 2.7×108Ωm2.7 \times {10^{ - 8}}\,\Omega \,{\text{m}} . We are asked to find the resistance that the aluminium cable offers to the flow of current.

To begin with, we have to understand the situation.
Here, the length of the cable along with diameter is given. The resistivity is also provided.We know that the resistance depends on the length and area of the cross section of a wire. But however, the resistance varies directly with the length and varies inversely with the cross-sectional area of the wire.

We will now convert the length and diameter into the S.I system of units.
l=10km l=10×1000m l=10000m d=2.0mm d=2×103m r=d2 r=2×103m2 r=1×103ml = 10\,{\text{km}} \\\ \Rightarrow l = 10 \times 1000\,{\text{m}} \\\ \Rightarrow l = 10000\,{\text{m}} \\\ \Rightarrow d = 2.0\,{\text{mm}} \\\ \Rightarrow d = 2 \times {10^{ - 3}}\,{\text{m}} \\\ \Rightarrow r = \dfrac{d}{2} \\\ \Rightarrow r = \dfrac{{2 \times {{10}^{ - 3}}\,{\text{m}}}}{2} \\\ \Rightarrow r = 1 \times {10^{ - 3}}\,{\text{m}}

We know, the expression which gives the resistance of wire is given below:
R=ρ×lAR = \rho \times \dfrac{l}{A} …… (1)
Where,
RR indicates the resistance of the cable.
ρ\rho indicates the resistivity of the material.
ll indicates the length of the cable.
AA indicates the area of the cross section of the wire.

We substitute the required values in the equation (1) we get:
R=ρ×lA R=2.7×108×10000π×r2 R=2.7×104×13.14×(1×103)2 R=85.98ΩR = \rho \times \dfrac{l}{A} \\\ \Rightarrow R = 2.7 \times {10^{ - 8}} \times \dfrac{{10000}}{{\pi \times {r^2}}} \\\ \Rightarrow R = 2.7 \times {10^{ - 4}} \times \dfrac{1}{{3.14 \times {{\left( {1 \times {{10}^{ - 3}}} \right)}^2}}} \\\ \therefore R = 85.98\,\Omega

Hence, the resistance of the aluminium cable is 85.98Ω85.98\,\Omega .

Note: It is important to remember that resistance is not a fixed quantity that is present in a conductor. If we take a thin and a thick wire of the same material, both the wires are of the same length. Then the thin wire will offer more resistance to the flow of current. It is also important to remember that resistivity is the property of a material. The value of resistivity remains the same even if the material is small or big, but the temperature must remain constant.