Answer to Question #172168 in Electric Circuits for JohnDoe

Question #172168

A 20 meter length of cable has a cross-sectional area of 1mm2Β and a resistance of 5 ohms. Calculate the conductivity of the cable. 𝜎 = 4π‘₯106π‘ π‘–π‘’π‘šπ‘’π‘›π‘ 


What will be the conductivity of the cable if its length is (a) reduced by 1⁄4 of its original length (b) doubled?


1
Expert's answer
2021-03-17T16:55:25-0400

Let's first find the resistivity of the cable:


"R=\\rho\\dfrac{l}{A},""\\rho=\\dfrac{RA}{l}."

Finally, we can calculate the conductivity of the cable:


"\\sigma=\\dfrac{1}{\\rho}=\\dfrac{l}{RA}=\\dfrac{20\\ m}{5\\ \\Omega\\cdot1\\ mm^2\\cdot(\\dfrac{1\\ m}{1000\\ mm})^2}=4\\cdot10^6\\ \\dfrac{S}{m}."

a)

"\\sigma_{new}=\\dfrac{l}{4RA}=\\dfrac{1}{4}\\cdot\\sigma=\\dfrac{1}{4}\\cdot4\\cdot10^6\\ \\dfrac{S}{m}=10^6\\ \\dfrac{S}{m}."

(b)

"\\sigma_{new}=\\dfrac{2l}{RA}=2\\sigma=2\\cdot4\\cdot10^6\\ \\dfrac{S}{m}=8\\cdot10^6\\ \\dfrac{S}{m}."

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