Question

If diameter of conductor is doubled, then its resistance will be reduced to

Answer 1

Given:

Diameter of conductor is doubled

To Find:

Change in resistance

Solution:

We know that,

• $\orange{\boxed{\bf{Diameter=2\times radius}}}$

• Resistance of cylindrical conductor R is given by

$\pink{\boxed{\bf{R=\dfrac{\rho l}{\pi r^{2}}}}}$

where,

ρ is resistivity of conductor

l is length of conductor

r is radius of conductor

$\rule{190}{1}$

Considering conductor to be cylindrical

Let the original resistance of conductor be R, resistivity be ρ, length of conductor be l, original diameter of conductor be D and r be the original radius of conductor

So,

$\longrightarrow\rm{R=\dfrac{\rho l}{\pi r^{2}}}-----(1)$

Also,

$\longrightarrow\rm{D=2r}$

$\rule{190}{1}$

Now, after doubling the diameter, new diameter D' will be

$\longrightarrow\rm{D'=2D}$

$\longrightarrow\rm{D'=2(2r)}$

$\longrightarrow\rm{D'=4r}-----(2)$

Let the new radius be r' and new resistance be R'

So, from (2), we get

$\longrightarrow\rm{2r'=4r}$

$\longrightarrow\rm{r'=2r}$

Also,

$\longrightarrow\rm{R'=\dfrac{\rho l}{\pi(r')^{2}}}$

$\longrightarrow\rm{R'=\dfrac{\rho l}{\pi(2r)^{2}}}$

$\longrightarrow\rm{R'=\dfrac{\rho l}{\pi4r^{2}}}$

$\longrightarrow\rm{R'=\dfrac{\rho l}{4\pi r^{2}}}$

$\longrightarrow\rm{R'=\dfrac{1}{4}\bigg(\dfrac{\rho l}{\pi r^{2}}\bigg)}$

$\longrightarrow\rm{R'=\dfrac{1}{4}\bigg(R\bigg)}$

$\longrightarrow\rm\green{R'=\dfrac{R}{4}}$

Hence, the new resistance is 1/4 times of original resistance.