Question

What is the resistance of

a. $2.0m-$long gold wire that is $0.20\mathrm{mm}$in diameter?

b. A $10\mathrm{cm}$long piece of carbon with a $1.0\mathrm{mm}\times 1.0\mathrm{mm}$square cross-section?

Short answer

a. The resistance of a $2.0\mathrm{m}$long gold wire that is $0.20\mathrm{mm}$in diameter is$1.53\mathrm{\Omega }.$

b. The resistance of a long piece of carbon with a $1.0\mathrm{mm}\times 1.0\mathrm{mm}$square cross-section is$3.5\mathrm{\Omega }.$

Step-by-step solution

Given Information (Part a) 

Gold wire length$=2.0m$

Diameter$=0.20mm$

Explanation (Part a)

The resistance in the first case, since the cross-section area is a circle, will be given by

$R=\rho \frac{L}{A}$

Modify the expression,

$=\rho \frac{4L}{\pi D^2}$

Substitute the values,

$R=\left(2.4\times {10}^{-8}\Omega m\times \frac{4\times 2m}{\pi \times 0.0002m{m}^2}\right)$

$R=1.53\Omega$

Final Answer (Part a) 

Hence, the resistance of a $2.0\mathrm{m}$long gold wire that is $0.20\mathrm{mm}$in diameter is $1.53\Omega$.

Given Information (Part a) 

Carbon Cross-section dimension$=1.0mm\times 1.0mm$

Explanation (Part b)

The resistance in the second case, since the cross-section area is a square, will be given by

$R=\rho \frac{L}{A}$

$=\rho \frac{L}{a^2}$

Substitute the values,

$R=\left(3.5\times {10}^{-5}\Omega m\times \frac{2}{0.001m{m}^2}\right)$

$=3.5\Omega$

Final Answer (Part a) 

Hence, the resistance of a long piece of carbon with a $1.0mm\times 1.0mm$is $3.5\mathrm{\Omega }.$