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. π = 4π₯106π ππππππ
What will be the conductivity of the cable if its length is (a) reduced by 1β4 of its original length (b) doubled?
Let's first find the resistivity of the cable:
Finally, we can calculate the conductivity of the cable:
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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