A copper wire has a diameter of 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Asked by Vishal kumar | 1 year ago |  211

Solution :-

The resistance of the copper wire of length in meters and area of cross-section m2 is given by the formula

$$R=p\frac{t}{A}$$

The area of cross-section of the wire can be calculated as follows

$$A=n(\frac{Diameter}{2})^2$$

Substituting the values in the formula, we get

$$l = \frac{RA}{P}=\frac{10\,\times\,3.14\,\times(\frac{0.0005^2}{2})}{(1.6\,\times\,10^{-8})}=\frac{10\,\times\,3.14\,\times\,25}{4\,\times\,1.6}=122.72\,m$$

If the diameter of the wire is doubled, then the new diameter will be 1 mm or 0.001 m

Therefore, the resistance can be calculated as follows:

$$R= p\frac{l}{A}=1.6\,\times\,10^{-8}\times\frac{122.72\,m}{m(\frac{0.001}{2})^2}=250.2\,\times10^{-2}=2.5\,\Omega$$

The length of the wire is 122.72 m and the new resistance is 2.5 Ω.

Answered by Shivani Kumari | 1 year ago

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