Question

How does the resistance of a wire change when:

(i) its length is tripled?
(ii) its diameter is tripled?
(iii) its material is changed to one whose resistivity is three times?

Answer 1

Resistance,
$R=\frac{\mathrm{\rho }l}{A}$
where l = length of wire
A= area of cross-section of the wire

(i) When the length is tripled, l' = 3l

New resistance,
$R\text{'}=\frac{\mathrm{\rho }l\text{'}}{A}$

$R\text{'}=\frac{\mathrm{\rho }\left(3l\right)}{A}=\frac{3\mathrm{\rho }l}{A}$

$R\text{'}=3R$

Thus, the resistance will become three times the original resistance.

(ii) When the diameter is tripled, d' = 3d

Radius, r' = 3r

Area of cross-section, $A\text{'}=\mathrm{\pi }r{\text{'}}^2=\mathrm{\pi }(3r{)}^2=9\mathrm{\pi }{r}^2=9A$

Thus, the area of cross-section will become nine times.

Then the new resistance,
$R\text{'}=\frac{\mathrm{\rho }l}{A}$

$R\text{'}=\frac{\mathrm{\rho }l}{9A}$

$R\text{'}=\frac{R}{9}$

Thus, the resistance will be reduced by nine times.

(iii) When the material is changed to one whose resistivity is three times:

New resistivity, $\mathrm{\rho }\text{'}=3\mathrm{\rho }$

The new resistance,
$R\text{'}=\frac{\mathrm{\rho }\text{'}l}{A}$
​

$R\text{'}=\frac{3\mathrm{\rho }l}{A}=3R$

Thus, the resistance will become three times the original resistance.