Question

a conductor has a cross section of 15cm^2 and specific Resistance of 7.6mu.ohm.cm at 0°c. if the temperature coefficient of Resistance of the material is 0.005°c^-1 . estimate its Resistance in ohm per 2km, when its temperature is 50°c​

Answer 1

Explanation:

i hope you understand

ok ..

Answer 2

Complete Question: A conductor has a cross sectional area of 15 cm² and specific resistance of 7.6 μ Ω cm at 0°C. If the temperature coefficient of resistance of the material is 0.005 °C⁻¹. Estimate its resistance in ohm per 2 km, when its temperature is equal to 50°C.

Answer: The resistance of the conductor when its temperature is 50°C is equal to 0.127 ohm per 2 km

Given:

Cross-sectional area of conductor (A) = 15 cm²

Specific Resistance of the conductor at 0°C (ρ₀) = 7.6 μ Ω cm

Temperature coefficient of resistance of the material (α) = 0.005 °C⁻¹

The temperature of the resistance (t) = 50°C

Length of Conductor (L) = 2 km

To Find:

The resistance of 2 km length of conductor in Ohm at a temperature of 50°C

Solution:

Cross-sectional area of the conductor (A) = 15 cm²

Length of the conductor (L) = 2 km = 2 × 10³ m = 2 × 10⁵ cm

The specific resistance at 0°C (ρ₀) = 7.6 μ Ω cm = 7.6 × 10⁻⁶ Ω cm

→ The relationship between the specific resistance at 0°C (ρ₀) and the specific resistance at t°C is given as:

                                   $\rho_{t} =\rho_{0} (1+\alpha t)$

→ where 'α' is temperature coefficient of resistance of the material.

→ Therefore the specific resistance at 50°C will be:

                                   $\rho_{50} =\rho_{0} (1+\alpha t)\\\\\rho_{50} =(7.6\times10^{-6} )(1+0.005\times50)\\\\\rho_{50} =9.5\times10^{-6} \:\Omega\:cm$

→ So the specific resistance at 50°C: ρ₅₀ = 9.5 × 10⁻⁶ Ω cm

→ The resistance of a cylindrical metallic wire is directly proportional to its length (L) and inversely proportional to its cross-sectional area (A). The resistance (R) of a metallic conductor is given as:

                                   $Resistance(R)=\rho\frac{L}{A}$

where 'ρ' is the specific resistance of the material of the conductor.

→ Hence the resistance of the 2 km length of the conductor would be:

 $Length\:of\:Conductor(L)=2\:km=2\times10^{5}cm\\\\ Cross-sectional\:area\:of\;Conductor(A)=15\:cm^{2} \\\\Specific\:resistance \:of \:Conductor(\rho)=9.5\times10^{-6}\: \Omega \:cm$

 $\therefore Resistance(R)=\rho(\frac{L}{A}) =9.5\times10^{-6}\times(\frac{2\times10^{5} }{15} )\\\\\therefore R=\frac{19}{150} =0.127\:\Omega$

Hence the resistance of the conductor when its temperature is 50°C is equal to 0.127 ohm per 2 km.

#SPJ2