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What diameter should you make a copper wire so its resistance is 0.015 Ω per meter of length?
physics
Guest Feb 25, 2015

Best Answer 

 #1
avatar+20553 
+5

What diameter should you make a copper wire so its resistance is 0.015 $$\Omega$$ per meter of length ? 

The resistance R of a copper wire with the length l can be calculated with the following formula:

 

where

R = resistance (ohms, Ω)

ρ = resistivity (ohm meter, Ω m)  Resistivity Copper: 1.7 x 10-8 Ω m

l = length of conductor (m)

A = cross-sectional area of conductor (m2)

d  = the nominal diameter of the wire

 

We have:

$$\frac{R}{l}= 0.015\ \frac{\Omega}{m}$$

$$\boxed{\dfrac{R}{l}= \dfrac{4\cdot \rho}{\pi \cdot d^2}} = 0.015\ \frac{\Omega}{m}\\\\
\dfrac{4\cdot \rho}{\pi \cdot d^2}} = 0.015\ \frac{\Omega}{m}\\\\
d^2 = \dfrac{4\cdot \rho}{\pi \cdot0.015\ \frac{\Omega}{m}}} \quad | \quad \rho_{copper} = 1.7 \cdot 10^{-8}\ \Omega\ m\\\\\\
d^2 = \dfrac{4\cdot 1.7 \cdot 10^{-8}\ \Omega\ m}{\pi \cdot0.015\ \frac{\Omega}{m}}} \quad | \quad \sqrt{} \\\\\\
d = 2 \cdot 10^{-4} \sqrt{ \dfrac{1.7}{\pi\cdot 0.015} }\ m \\\\
d = 0.00120125135\ \mathrm{m} \\
d = 1.2\ \mathrm{mm}$$


 

 

heureka  Feb 25, 2015
 #1
avatar+20553 
+5
Best Answer

What diameter should you make a copper wire so its resistance is 0.015 $$\Omega$$ per meter of length ? 

The resistance R of a copper wire with the length l can be calculated with the following formula:

 

where

R = resistance (ohms, Ω)

ρ = resistivity (ohm meter, Ω m)  Resistivity Copper: 1.7 x 10-8 Ω m

l = length of conductor (m)

A = cross-sectional area of conductor (m2)

d  = the nominal diameter of the wire

 

We have:

$$\frac{R}{l}= 0.015\ \frac{\Omega}{m}$$

$$\boxed{\dfrac{R}{l}= \dfrac{4\cdot \rho}{\pi \cdot d^2}} = 0.015\ \frac{\Omega}{m}\\\\
\dfrac{4\cdot \rho}{\pi \cdot d^2}} = 0.015\ \frac{\Omega}{m}\\\\
d^2 = \dfrac{4\cdot \rho}{\pi \cdot0.015\ \frac{\Omega}{m}}} \quad | \quad \rho_{copper} = 1.7 \cdot 10^{-8}\ \Omega\ m\\\\\\
d^2 = \dfrac{4\cdot 1.7 \cdot 10^{-8}\ \Omega\ m}{\pi \cdot0.015\ \frac{\Omega}{m}}} \quad | \quad \sqrt{} \\\\\\
d = 2 \cdot 10^{-4} \sqrt{ \dfrac{1.7}{\pi\cdot 0.015} }\ m \\\\
d = 0.00120125135\ \mathrm{m} \\
d = 1.2\ \mathrm{mm}$$


 

 

heureka  Feb 25, 2015

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