Answer to Question #133068 in Electric Circuits for louie

Question #133068
1. A coil of annealed copper wire has 820 turns, the average length of which is 9 in. If the diameter of the wire is 32 mils, calculate the total resistance of the coil at 200C.
1
Expert's answer
2020-09-15T08:20:38-0400

Given,

Number of turns (n)=820(n)=820

Length of the coil (l)=9inch=0.0254×9m=0.2286m(l)=9 inch =0.0254\times 9 m =0.2286m

Diameter of the wire (d)=32mils=32×2.54×105m=81.28×105m(d)=32mils = 32\times 2.54\times 10^{-5}m=81.28\times 10^{-5}m

Let total resistance of the coil =(R)=(R)

We know that the resistivity of the annealed copper at 20oC(ρ)=1.72×108Ωm20^oC (\rho)=1.72\times 10^{-8} \Omega m

Cross sectional area of the wire (A)=πd24m2=3.14×(81.28×105)24=5.18×107m2(A)=\dfrac{\pi d^2}{4}m^2 =\dfrac{3.14\times (81.28\times 10^{-5})^2}{4}=5.18\times 10^{-7}m^2

We know that,

Resistance of the coil (R)=nρlA(R)= \dfrac{n\rho l}{A}

Now, substituting the values,

R=820×1.72×108×0.22865.18×107Ω\Rightarrow R = \dfrac{820\times 1.72\times 10^{-8}\times 0.2286}{5.18\times 10^{-7}}\Omega

R=6.2Ω\Rightarrow R = 6.2\Omega


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