Physics Current Electricity

R =? *L/A

This is the formula of calculating a wire's resistance. If you look closely at the formula,

A= cross sectional area

More the surface area, more will be the area for passage of free electrons.

Hence, we can easily conclude that thickness of a wire plays an jmportant role in determining it's conductivity.

This happens due to the lack of electrons. Metals have a solid flow of electric current in their bodies due to the presence of free electrons moving throughout. These electrons are responsible for carrying electric charges which makes them a good conductor of electricity. however, the lack of these electrons in non-metals makes them an insulator.

As per electric current, the key difference between an electron and a proton is that electrons are directly responsible for carrying electric current. On the other hand, protons do not contribute in the passage of current since they are held tightly by the nucleus in the centre of the atom.

$\Rightarrow \frac{R}{100}=\frac{40}{60}$

$\Rightarrow \Delta R=\left(\frac{\Delta l}{l}+\frac{\Delta l}{\left(100-l\right)}\right)R$

$=\left(\frac{0.1}{40}+\frac{0.1}{60}\right)*66.7=0.27=0.3$

When wire is stretched uniformly

$R\propto \mathcal{R}$

$\frac{R}{R^{\text{'}}}={\left(\frac{l}{nl}\right)}^2=\frac{1}{n^2}$

$R^{\text{'}}=n^{2}R$

When this elongated wire is cut into 5 parts, resistance of each part $=\frac{n^{2}R}{5}$  

Effective resistance between A and C $=\frac{n^{2}R}{5}$  

$01*10^4\Omega \pm 5\mathrm{\%}=10*10^3\Omega \pm 5\mathrm{\%}$

Resistance of whole wire is

$R=r^{\text{'}}*l$

$20=1*l$

$l=20m$

Let length of the shorter section = a

$\frac{1}{R}=\frac{1}{a}+\frac{1}{\left(20-a\right)}$

$\frac{1}{1.8}=\frac{20-a+a}{a\left(20-a\right)}?20a-a^2=36$

$a=2and18$

$?a=2m$

  $R_i=\frac{\rho \mathcal{l}}{A}$
$R_f=\frac{\rho \left(1.25\mathcal{l}\right)}{\left(A/1.25\right)}={\left(1.25\right)}^2*\frac{\rho \mathcal{l}}{A}$

$\Rightarrow R_f=1.5625*R_i$

$\Rightarrow \frac{R_f-R_l}{R_l}*100=56.25\mathrm{\%}$