3.3 Limitations of ohm's law

The conductor which do not obey Ohm's law are called non-ohmic conductors.

Examples: Transistors, Diode, Insulators etc.

The $V-I$ graph for the non-ohmic conductor is not a straight line.

The phenomenon of complete loss of resistivity by certain metals and alloys when they are cooled below a certain temperature is called superconductivity.

The temperature at which substance undergoes a transition from normal conductor to superconductor in zero magnetic fields is called transition temperature or critical temperature.

If an electric current is set up in a superconductor, it can persist for a long time without any applied electromotive force (EMF).

Mercury $(Hg)$ becomes a superconductor at 4.2$K$.

Similarly, Lead $(Pb)$ becomes a superconductor at 7.0$K$.

(1) A superconductor is used to construct very strong magnets.

(2) It is used in long distance power transmission without any wastage of power.

Question 6. Calculate the electric field in a copper wire of cross-sectional area 4.0$m{m^2}$ carrying a current of $2A$. The conductivity of copper is $6.25 \times {10^7}S{m^{ - 1}}$.

Solution: Given $$\begin{equation} \begin{aligned} A = 4.0m{m^2} \\ I = 2A \\ \sigma = 6.25 \times {10^7}S{m^{ - 1}} \\\end{aligned} \end{equation} $$

As, $$\begin{equation} \begin{aligned} \Rightarrow j = \frac{I}{A} = \sigma E \\ \Rightarrow E = \frac{I}{{A\sigma }} \\ \Rightarrow \frac{2}{{4 \times {{10}^{ - 6}} \times 6.25 \times {{10}^7}}} \\ \Rightarrow 8 \times {10^{ - 3}}V{m^{ - 1}} \\\end{aligned} \end{equation} $$

Question 7. A wire of $5\Omega $ resistance is stretched to thrice its original length. What will be its,

(i) New resistivity

(ii) New resistance.

(i) Resistivity $\rho $ remains unchanged because it only depends only on nature of material and temperature. It is independent of dimensions.

(ii)

Given $L' = 3L$

We have, $$R = \rho \frac{l}{A}$$

Volume of wire is same, $$V = A'L' = AL$$

where,

$A$ & $A'$ is the intial and final area of cross-section of the conductor respectively.

$L$ & $L'$ is the initial and final length of the conductor respectively.

$$ \Rightarrow \frac{{A'}}{A} = \frac{l}{{l'}} = \frac{1}{3}$$

New resistance,$$ \Rightarrow R' = \rho \frac{{l'}}{{A'}}$$ So, $$\begin{equation} \begin{aligned} \Rightarrow \frac{{R'}}{R} = \frac{{\rho \frac{{l'}}{{A'}}}}{{\rho \frac{l}{A}}} \\ \Rightarrow \frac{{l'}}{l} \times \frac{A}{{A'}} \\ \Rightarrow \frac{1}{3} \times \frac{1}{3} = 9 \\\end{aligned} \end{equation} $$

Hence, $$R' = 9R$$

Question 8. At what temperature the resistance of a conductor is doubled of its resistance at $0^\circ C$ ? ($\alpha $ for copper is $3.9 \times {10^{ - 3}}^\circ {C^{ - 1}}$).

Solution: As we know, $$ \Rightarrow {R_2} = {R_1}(1 + \alpha \Delta T)$$ or, $$\begin{equation} \begin{aligned} \Rightarrow \alpha = \frac{{{R_2} - {R_1}}}{{{R_1}\left( {T - {T_1}} \right)}} \\ \Rightarrow \alpha = \frac{{2{R_0} - {R_0}}}{{{R_0}\left( {T - 0} \right)}} \\ \Rightarrow \alpha = \frac{1}{T} \\\end{aligned} \end{equation} $$ or,$$\begin{equation} \begin{aligned} \Rightarrow T = \frac{1}{\alpha } \\ \Rightarrow T = \frac{1}{{3.9 \times {{10}^{ - 3}}}} \\ \Rightarrow T = 256^\circ C \\\end{aligned} \end{equation} $$

Question 9. The resistance of a wire is $5\Omega $ at $50^\circ C$ and $6\Omega $ at $100^\circ C$. What is the resistance of wire at $0^\circ C$?

Solution: As we know, $$\begin{equation} \begin{aligned} \Rightarrow {R_T} = {R_0}(1 + \alpha \Delta T) \\ \Rightarrow {R_T} - {R_0} = {R_0}\alpha \Delta T \\\end{aligned} \end{equation} $$

At, $$\begin{equation} \begin{aligned} \Rightarrow T = 50^\circ C \\ \Rightarrow {R_{50}} - {R_0} = \alpha {R_0}(50) \\ \Rightarrow 5 - {R_0} = 50\alpha {R_0}\quad ...(i) \\\end{aligned} \end{equation} $$

At, $$\begin{equation} \begin{aligned} \Rightarrow T = 100^\circ C \\ \Rightarrow {R_{100}} - {R_0} = \alpha {R_0}(100) \\ \Rightarrow 6 - {R_0} = 100\alpha {R_0}\quad ...(ii) \\\end{aligned} \end{equation} $$

On dividing equation $(i)$ and $(ii)$ we get, $$\begin{equation} \begin{aligned} \Rightarrow \frac{{5 - {R_0}}}{{6 - {R_0}}} = \frac{{50\alpha {R_0}}}{{100\alpha {R_0}}} \\ \Rightarrow \frac{{5 - {R_0}}}{{6 - {R_0}}} = \frac{1}{2} \\ \Rightarrow 10 - 2{R_0} = 6 - {R_0} \\ \Rightarrow {R_0} = 4\Omega \\\end{aligned} \end{equation} $$