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Difficulty: MediumElectric Current and Resistance

An electric heating element made of a wire with resistivity 1.0×10−6 Ω⋅m1.0 \times 10^{-6}\,\Omega\cdot\text{m} and a uniform cross-sectional area of 5.0×10−7 m25.0 \times 10^{-7}\,\text{m}^2 carries a steady current of 4.0 A4.0\,\text{A} when connected to a 240 V240\,\text{V} direct-current supply. What is the length of the wire?

  1. A
    15 m15\,\text{m}
  2. 30 m30\,\text{m}Answer
  3. C
    60 m60\,\text{m}
  4. D
    120 m120\,\text{m}

Answer

30 m30\,\text{m}
By applying Ohm's Law (R=VIR = \frac{V}{I}), the total resistance of the heating element is found to be 60 Ω60\,\Omega. Substituting this along with resistivity ρ=1.0×10−6 Ω⋅m\rho = 1.0 \times 10^{-6}\,\Omega\cdot\text{m} and cross-sectional area A=5.0×10−7 m2A = 5.0 \times 10^{-7}\,\text{m}^2 into the resistivity equation R=ρLAR = \frac{\rho L}{A} yields L=RAρ=30 mL = \frac{R A}{\rho} = 30\,\text{m}.

Step-by-Step Solution

1
Calculate the electrical resistance of the wire using Ohm's Law.
R=VI=240 V4.0 A=60 ΩR = \frac{V}{I} = \frac{240\,\text{V}}{4.0\,\text{A}} = 60\,\Omega
The resistance must be determined from the operational potential difference and current before finding the geometric dimensions.
2
Relate resistance to length, cross-sectional area, and resistivity using R=ρLAR = \frac{\rho L}{A}.
60 Ω=(1.0×10−6 Ω⋅m)×L5.0×10−7 m260\,\Omega = \frac{(1.0 \times 10^{-6}\,\Omega\cdot\text{m}) \times L}{5.0 \times 10^{-7}\,\text{m}^2}
The resistance of a uniform conductor is directly proportional to its length and inversely proportional to its cross-sectional area.
3
Solve the equation for the wire length LL.
L=60×5.0×10−71.0×10−6=30 mL = \frac{60 \times 5.0 \times 10^{-7}}{1.0 \times 10^{-6}} = 30\,\text{m}
Rearranging the formula gives L=RAρL = \frac{R A}{\rho} to isolate the required length.

Key Concept

Relationship between potential difference, current, resistance, and wire dimensions (R=VI=ρLAR = \frac{V}{I} = \frac{\rho L}{A})
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Topics

Electric Current and Resistance

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