Question

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.

Answer 1

Given,

Number of turns $(n)=820$

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

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

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

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

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

We know that,

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

Now, substituting the values,

$\Rightarrow R = \dfrac{820\times 1.72\times 10^{-8}\times 0.2286}{5.18\times 10^{-7}}\Omega$

$\Rightarrow R = 6.2\Omega$