A copper wire has diameter 0.5 mm and resistivity of $1.6\times10^{-8}$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?

Given: Resistivity $(\rho)=1.6 × 10-8\ \omega m$, Resistance $(R)=10\ \Omega$, Diameter $(d)=0.5\ mm=5\times10^{-4}\ m$

To do: To find the length of this wire to make its resistance $10\ Ω$. And to calculate the change in resistance if the diameter is doubled?

Solution:

So, Radius of the wire $r=0.25\ mm=2.5\times10^{-4}\ m$

So, the area of cross-section, $A=\pi r^2$

$A=(\frac{22}{7})(2.5\times10^{-4})^2$

$A=(\frac{22}{7})(6.25\times10^{-8})$

$A=1.964\times10^{-7}\ m^2$

Let $l$ be the length of the wire.

We know that

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

Or $l=\frac{R\times A}{\rho}$

Substituting the values in the above equation we get

$l=\frac{(10\times1.964\times10^{-7})}{1.6\times10^{-8}\ m}$

$l=\frac{1.964\times10^{-6}}{1.6\times10^{-8}}$

$l=122.72\ m$

If the diameter of the wire is doubled, the new diameter$=2\times0.5=1\ mm=0.001\ m$

Let $R'$ be the new resistance.

$R'=\rho \frac{l}{4A}$

Or $R'=\rho(l)\times \frac{1}{4A}$

Hence, if the diameter doubles, resistance becomes 1/4 times.

So, new resistance $R'=2.5\ \Omega$

Therefore, the length of the wire is $122.7\ m$ and the new resistance becomes $\frac{1}{4}$ times.

Updated on: 10-Oct-2022

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