Resistivity

Resistvity is the total resistance of a material

Equation for calculating resistance

If we know the:

• area of the cross-section of the wire
• length of the wire
• resistivity of the material

we can calculate the resistance of the wire using the equation:

R = \frac{\rho L}{A}

where:

• R is the resistance (in ohms, \Omega)
• \rho is the resistivity of the material (in ohm metres, \Omega m)
• L is the length of the wire (in metres, m)
• A is the area of the cross-section of the wire (in square metres, m^2)

The resistivity of copper is 1.8\times10{-8}\Omega m. Calculate the resistance of a copper wire of length 0.6m and cross-sectional area of 1mm^2

• R=\frac{\rho L}{A}
• \rho=1.8\times10^{-8}\Omega m
• L=0.6m
• A=1mm^2=1\times10^{-6}m^2
• R=\frac{1.8\times10^{-8}\times0.6}{1\times10^{-6}}=1.08\times10^{-2}\Omega
• Answer: 1.08\times10^{-2}\Omega

Two of these wires are used to connect a lamp to a power supply of negligible internal resistance. The p.d. across the lamp is 12V and its power is 36W. Calculate the potential difference across each wire.

• power, P=\frac{V^2}{R}
• Rearranging for R: R=\frac{V^2}{P}
• V=12V
• P=36W
• R=\frac{12^2}{36}=4\Omega
• Each wire has resistance of 4\Omega
• Total resistance of wires = 4\Omega + 4\Omega = 8\Omega
• Total resistance in circuit = 8\Omega + resistance of lamp
• Using V=IR to find current, I=\frac{V}{R}
• I=\frac{12V}{8\Omega}=1.5A
• Potential difference across each wire, V=IR
• V=1.5A \times 4\Omega = 6V
• Answer: 6V across each wire

What length of copper wire of diameter 0.1mm is required to make a coil with resistance 0.5 \Omega? (Resistivity of copper = 1.7 \times 10^{-8} \ \Omega m)

• R=\frac{\rho L}{A}
• Rearranging for L: L=\frac{RA}{\rho}
• A=0.05^2 \pi \approx 7.85 \times 10^{-9} \ m^2
• R=0.5 \ \Omega
• \rho=1.7 \times 10^{-8} \ \Omega m
• L=\frac{0.5 \times 7.85 \times 10^{-9}}{1.7 \times 10^{-8}} \approx 0.23 \ m
• Answer: 0.23 \ m

If the resistivity of copper is 1.7\times10^{-8}\Omega m calculate the resistance of 1cm^3 of copper, when in the form of a wire of diameter 0.02cm

• R=\frac{\rho L}{A}
• V=1cm^3=1\times10^{-6}m^3
• A=\pi r^2=\pi(0.01\times10^{-2})^2=\pi\times10^{-8}m^2
• L=\frac{V}{A}=\frac{1\times10^{-6}}{\pi\times10^{-8}}=\frac{100}{\pi}m
• \rho=1.7\times10^{-8}\Omega m
• R=\frac{1.7\times10^{-8}\times\frac{100}{\pi}}{\\pi\times10^{-8}}=\frac{1700}{\pi^2}\Omega
• Answer: \frac{1700}{\pi^2}\Omega \approx 172\Omega

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