Physics - Current Electricity Question with Solution | TestHub

Potential difference$= 5 \mathrm{V}$, length$=\text{0.1 m}=l$and the electron speed$=$drift velocity${\mathit{\text{v}}}_{\text{d}}=\text{2.5}\times 10^{-4} {\text{m s}}^{-1}$, electron density$\left(\text{n}\right)=8\times 10^{28}{\text{m}}^{-3}$and the charge on each electron(e)$=\text{1.6}\times 10^{-19} \text{C}$. We know$\mathit{\text{i}}=\mathit{\text{nAe}}{\mathit{v}}_{\text{d}}\mathit{ }\mathit{ }\mathit{.}\mathit{.}\mathit{.}\left(\mathrm{i}\right)$. And$V=iR ...\left(\mathrm{ii}\right)$



Resistance$R$is also equal to,$\frac{\rho l}{\text{A}}$$\text{⇒}\mathit{\text{R}}=\frac{\rho l}{\mathit{\text{A}}}$

$\rho =\frac{\mathit{\text{AR}}}{l}$$\left[\rho =\text{Resistivity}\right]$

$=\frac{\text{A}}{l}\times \frac{V}{\mathit{\text{i}}}$[from (ii)]$=\frac{\mathit{\text{Av}}}{l\times \mathit{\text{nA}}\text{ e}{\text{v}}_{\text{d}}}$& [from (i)]$=\frac{\mathit{\text{V}}}{l\times \mathit{\text{n}}\times \mathit{\text{e}}\times {\mathit{\text{v}}}_{\text{d}}}$

$=\frac{5}{\text{0.1}\times 8\times 10^{28}\times \text{1.6}\times 10^{-19}\times \text{2.5}\times 10^{-4}}$

$=\text{0.156}\times 10^{-4} \mathrm{\Omega } \text{m}=\text{1.56}\times 10^{-5} \mathrm{\Omega } \text{m}$