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Zorluk: OrtaResistance and Resistivity

A technician tests a solid aluminum ground wire and measures its electrical resistance as 1.2 Ω1.2\ \Omega. If the wire is replaced with an aluminum wire of the same alloy that is three times as long and has twice the cross-sectional area, what is the electrical resistance of the replacement wire?

  1. A
    0.8 Ω0.8\ \Omega
  2. 1.8 Ω1.8\ \OmegaCevap
  3. C
    7.2 Ω7.2\ \Omega
  4. D
    0.2 Ω0.2\ \Omega

Cevap

The resistance of the replacement wire is 1.8 Ω1.8\ \Omega.
The resistance of a uniform conductor is directly proportional to its length and inversely proportional to its cross-sectional area (R=ρL/AR = \rho L / A). Increasing the length by a factor of 33 increases resistance by 3×3\times, while doubling the cross-sectional area reduces resistance by half (/2/2). Combining these effects yields a net multiplier of 3/2=1.53/2 = 1.5. Multiplying the initial 1.2 Ω1.2\ \Omega by 1.51.5 results in 1.8 Ω1.8\ \Omega.

Adım Adım Çözüm

1
Identify the relationship governing conductor resistance based on material and physical geometry.
The resistance formula is R=ρLAR = \rho \frac{L}{A}, where ρ\rho is resistivity, LL is wire length, and AA is cross-sectional area.
Electrical resistance is directly proportional to conductor length and inversely proportional to cross-sectional area.
2
Apply the scaling factors for the new conductor dimensions.
Length changes to L′=3LL' = 3L and area changes to A′=2AA' = 2A. The new resistance is R′=ρ3L2A=32(ρLA)=1.5RR' = \rho \frac{3L}{2A} = \frac{3}{2} \left(\rho \frac{L}{A}\right) = 1.5 R.
Since the material remains identical, resistivity ρ\rho is unchanged.
3
Calculate the numerical value of the new resistance.
R′=1.5×1.2 Ω=1.8 ΩR' = 1.5 \times 1.2\ \Omega = 1.8\ \Omega.
Multiplying the baseline resistance by the scaling factor gives the final resistance.

Anahtar Kavram

Conductor resistance scales directly with length and inversely with cross-sectional area (R=ρL/AR = \rho L / A).
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Resistance and Resistivity

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