Electrical Resistance of Wires of Different Cross Sections

Task number: 2113

Electrical resistance of a wire with a circular cross-section of a diameter of 0.5 mm is 2 Ω. The length of the wire is 8 m. How long should be a wire of the same material but with the radius twice as big, so that the resistance of the wire is also 2 Ω? Determine the resistivity of the material, of which the wire has been made.

  • Hint

    Look up the relation for calculating the resistance of a homogeneous wire based on its parameters (such as its length, cross-sectional area and material).

  • Analysis – analytical solution

    From relation \(R\,=\,\rho\,\frac{l}{A}\), where ρ is the resistivity of the material, we see that the resistance of the wire R is directly proportional to its length l and inversely proportional to its cross-sectional area A. It means that if we increase the cross section of the wire, its resistance will decrease linearly and if we increase the length of the wire, the resistance will increase too (also linearly).

    It implies that in order to maintain the same resistance of the wire, we have to lengthen the wire the same number of times as we have increased its cross section. Because of the increase of the cross section, the resistance of the wire was decreased, so we have to compensate this decrease by a proper increase of its length.

    The cross-sectional area of a wire is directly proportional to the square of its radius (A = πr2), and hence to the diameter of the wire as well. Therefore, if we increase the diameter of the wire two times, its cross-sectional area will increase four times. In order for the wire made of the same material to have the same resistance as the original “narrower” wire, its length has to be four times greater than the length of the original wire.

    Thus, if the original wire is 8 m long, then the “wider” wire described above will have the length of 32 m.

  • Calculating length of the wire

    Electrical resistance of a wire can be calculated by the following formula:

    \[R\,=\,\rho\,\frac{l}{A}\,,\]

    where ρ is the resistivity of the material, l length of the wire and A stands for its cross-sectional area.

    For two different wires of the same resistance it holds true:

    \[R_1\,=\,R_2.\]

    Therefrom:

    \[\rho_1\,\frac{l_1}{A_1}\,=\,\rho_2\,\frac{l_2}{A_2}\,.\]

    If both wires are made of the same material, then they have the same electrical resistivity:

    \[\rho_1\,=\,\rho_2\,.\]

    Then we can express the length of one wire through the parameters of the other wire:

    \[\frac{l_1}{A_1}\,=\,\frac{l_2}{A_2}\] \[l_2\,=\,\frac{A_2}{A_1}\,l_1\,.\tag{1}\]

    The cross-sectional area of the wire is the area of a circle of radius r:

    \[A\,=\,\pi r^2\,=\,\pi \left(\frac{d}{2}\right)^2\,,\]

    where d is the diameter of the wire.

    So we calculate the cross-sectional area of the wires by the following formulae:

    \[A_1\,=\,\pi \left(\frac{d_1}{2}\right)^2\,;\hspace{15px}A_2\,=\,\pi \left(\frac{d_2}{2}\right)^2.\]

    After that we substitute the cross-sectional area for these expressions in relation (1):

    \[l_2\,=\,\frac{\pi \left(\frac{d_2}{2}\right)^2}{\pi \left(\frac{d_1}{2}\right)^2}\,l_1\] \[l_2\,=\,\left(\frac{d_2}{d_1}\right)^2\,l_1.\]

    As we know from the task assignment, the diameter of the second wire is twice as big as the diameter of the first wire (d2 = 2d1). Thus:

    \[l_2\,=\,\left(\frac{2d_1}{d_1}\right)^2\,l_1\] \[l_2\,=\,2^2\,l_1\] \[l_2\,=\,4\,l_1.\]
  • Resistivity – solution

    First, we express the resistivity of the material from

    \[R\,=\,\rho\,\frac{l}{A}\,,\]

    where R is the resistance of a wire, l length of the wire and A stands for the cross-sectional area of the wire.

    Thus

    \[\rho\,=\,\frac{RA}{l}\,.\]

    The cross-sectional area of the wire A is the area of a circle of radius r, or of diameter d = 2r:

    \[A\,=\,\pi r^2\,=\,\pi \left(\frac{d}{2}\right)^2.\]

    Then we substitute the expressed cross-sectional area into the previous relation to obtain the final formula for calculating the resistivity of the wire:

    \[\rho\,=\,\frac{RA}{l}\,=\,\frac{R\pi \left(\frac{d}{2}\right)^2}{l}\] \[\rho\,=\,\frac{\pi R d^2}{4l}\,.\]

    As we know from the task assignment, the diameter of the wire is of 0.5 mm, its length is 8 m and the resistance is 2 Ω. Thus:

    d = 0.5 mm = 0.5·10−3 m
    l = 8 m
    R = 2 Ω

    Substitution for the given values:

    \[ \rho\,=\,\frac{\pi\cdot2\cdot\left(0{.}5{\cdot}10^{-3}\right)^2}{4{\cdot}8}\,\mathrm{\Omega m}\,=\,\frac{\pi\cdot2{\cdot} 0{.}5^2{\cdot} 10^{-6}}{32}\,\mathrm{\Omega m}\,\dot{=}\,4{.}9{\cdot}10^{-8}\,\mathrm{\Omega m}\,.\]
  • Answer

    The length of the wire should be 32 m. The resistivity of the material is 4.9·10-8 Ωm.

  • Links to the similar tasks

    See another problem dealing with calculating the resistance of a homogeneous wire based on its parameters: Electrical Resistances of Conductors of Different Lengths.

Difficulty level: Level 2 – Upper secondary level
Original source: Kohout, J. (2010). Study materials for the Electricity and Magnetism
seminar. Internal materials, Pilsen.
×Original source: Kohout, J. (2010). Study materials for the Electricity and Magnetism seminar. Internal materials, Pilsen.
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