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Zorluk: OrtaResistance and Resistivity

A specialized alloy heating wire has a length of 40 meters40\text{ meters}, a uniform cross-sectional area of 0.8 mm20.8\text{ mm}^2, and a material resistivity of 2.5×10−7 Ω⋅m2.5 \times 10^{-7}\ \Omega\cdot\text{m}. What is the total electrical resistance of this wire in ohms (Ω\Omega)?

Cevap: 12.5 Ω

Cevap

12.5 Ω
The electrical resistance of a uniform conductor is determined by R=ρLAR = \rho \frac{L}{A}. Converting 0.8 mm20.8\text{ mm}^2 to 8×10−7 m28 \times 10^{-7}\text{ m}^2 and substituting L=40 mL = 40\text{ m} along with ρ=2.5×10−7 Ω⋅m\rho = 2.5 \times 10^{-7}\ \Omega\cdot\text{m} yields (2.5×10−7)(40)8×10−7=12.5 Ω\frac{(2.5 \times 10^{-7})(40)}{8 \times 10^{-7}} = 12.5\ \Omega.

Adım Adım Çözüm

1
Convert the cross-sectional area from square millimeters to square meters.
A=0.8 mm2=0.8×10−6 m2=8×10−7 m2A = 0.8\text{ mm}^2 = 0.8 \times 10^{-6}\text{ m}^2 = 8 \times 10^{-7}\text{ m}^2
Resistivity is given in ohm-meters (Ω⋅m\Omega\cdot\text{m}), requiring all geometric dimensions to be in standard SI units (meters and square meters).
2
Apply the fundamental resistance formula relating resistivity, length, and cross-sectional area.
R=ρLAR = \rho \frac{L}{A}
The total resistance of a uniform conductor is directly proportional to its resistivity and length, and inversely proportional to its cross-sectional area.
3
Substitute the known values into the equation and solve for resistance.
R=(2.5×10−7 Ω⋅m)(40 m)8×10−7 m2=100×10−78×10−7=12.5 ΩR = \frac{(2.5 \times 10^{-7}\ \Omega\cdot\text{m})(40\text{ m})}{8 \times 10^{-7}\text{ m}^2} = \frac{100 \times 10^{-7}}{8 \times 10^{-7}} = 12.5\ \Omega
Canceling the common power factor 10−710^{-7} simplifies the expression to 100/8=12.5 Ω100 / 8 = 12.5\ \Omega.

Anahtar Kavram

Calculating conductor resistance from material resistivity, length, and cross-sectional area using R=ρL/AR = \rho L / A.
Tahmini Süre:1m 15s
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Konular

Resistance and Resistivity

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