Ohm's Law: Class 12 Physics Notes, Definition, Working Principle, Formula & Real-Life Applications

Electrical resistance is the property of a substance that opposes the flow of electric current through the substance. The electrical resistance depends on the size, geometry, temperature and internal structure of the conductor. As we know that:
$\mathrm{i}=\frac{\mathrm{n}\mathrm{A}{\mathrm{e}}^2\tau }{\mathrm{m}\mathcal{l}}\mathrm{V}$
Here $i\propto V$ which is known as Ohm's law

$\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}\Rightarrow \mathrm{R}=\frac{\mathrm{m}\mathcal{l}}{\mathrm{n}\mathrm{A}\mathrm{e}{\mathrm{e}}^2\tau }\mathrm{}\Rightarrow \mathrm{}\mathrm{V}=\mathrm{I}\mathrm{R}$

hence $R=\frac{m\mathcal{l}}{nA{e}^2\tau }\cdot \frac{\mathcal{l}}{A}\mathrm{}$ so $Here\mathrm{}R=\frac{\rho \mathcal{l}}{A}\Rightarrow V=I\times \frac{\rho \mathcal{l}}{A}$
$\Rightarrow \frac{V}{\mathcal{l}}=\frac{I}{A}\rho \Rightarrow E=J\rho \Rightarrow J=\frac{I}{A}=$ current density  $\rho$ is called resistivity (or specific resistance), and

$\rho =\frac{m}{n{e}^2\tau }=\frac{1}{\sigma },\sigma$ is called conductivity.

Therefore, the current in conductors is proportional to the potential difference that is applied across its ends. This is known as the Ohm's Law. Units in ohm's law are $\mathrm{R}\to \mathrm{o}\mathrm{h}\mathrm{m}\left(\mathrm{\Omega }\right),\rho \to \mathrm{o}\mathrm{h}\mathrm{m}-\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}(\mathrm{\Omega }-\mathrm{m})$  also called siemens, $\to {\mathrm{\Omega }}^{-1}{\mathrm{m}}^{-1}$ . 

Please note that current electricity is an important chapter when it comes to different entrance examinations in the country. It is therefore, important to go through NCERT solutions of Current Electricity of What is Ohm's law class 12th. 

Ohm's law works in the following way:

• Ohm's Law explain the relationship between voltage, current, and resistance in electrical circuits. Current measures how much electricity is actually flowing through a circuit, whereas voltage gives push to electricity through the circuit.
• Resistance limits how easily electricity moves through materials. The formula states that current equals voltage divided by resistance (I = V/R).
• More voltage creates more current in a circuit. More resistance reduces the current in a circuit. Whenever the voltage is doubled, the current doubles as well. If resistance is doubled, current is halved.

Engineers use Ohm's law for designing safe electrical devices, and electricians apply Ohm's Law to solve practical electrical problems every day. This simple relationship forms the foundation of electrical engineering and helps us understand how electricity behaves in all electronic systems.

As per NCERT, Ohm's Law statement is as follows:

"Imagine a conductor through which a current I is flowing and let V be the potential difference between the ends of the conductor.
Then Ohm’s law states that:
 V ∝ I (ohm's law formula)
or, V = R I 
where the constant of proportionality R is called the resistance of the conductor. The SI units of resistance is ohm, and is denoted by the symbol W. The resistance R not only depends on the material of the conductor but also on the dimensions of the conductor."

Here is the Ohm's Law diagram:

$R=\rho \frac{\mathcal{l}}{A}=\frac{m}{n{e}^2\tau }\cdot \frac{\mathcal{l}}{A}$  Therefore R depends as
(1) $\propto \mathcal{l}$
(2) $\propto \frac{1}{A}$
(3) $\propto \frac{1}{n}\propto \frac{1}{\tau }$
(4) and in metals $\tau$  decreases as $T$  increases $\Rightarrow R$  also increases.

Results

(a) On stretching a wire (volume constant)

If length of wire is taken into account then $\frac{R_1}{R_2}=\frac{{\mathcal{l}}_1^2}{{\mathcal{l}}_2^2}$
If radius of cross section is taken into account then, where $R_1$  and $R_2$  are initial and final resistances and ${\mathcal{l}}_1,{\mathcal{l}}_2$  are initial and final lengths and $r_1$  and $r_2$  initial and final radii respectively. (If elasticity of the material is taken into consideration, the variation of area of cross-section is calculated with the help of Young's modulus and Poison's ratio)
(b) Effect of percentage change in length of wire
$\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}=\frac{{\mathcal{l}}^2{\left[1+\frac{\mathrm{x}}{100}\right]}^2}{{\mathcal{l}}^2}$  where $\mathcal{l}$  - original length and $\mathrm{x}-\mathrm{\%}$  increment
if $x$  is quite small (say $<5\mathrm{\%}$  ) then % change in $R$  is

$\frac{R_2-R_1}{R_1}\times 100=\left(\frac{{\left(1+\frac{x}{100}\right)}^2-1}{1}\right)\times 100\cong 2x\mathrm{\%}$

Note :

Above method is applicable when % change is very small.

The resistivity of a metallic conductor nearly increases with increasing temperature. This is because, with the increase in temperature the ions of the conductor vibrate with greater amplitude, and the collision between electrons and ions become more frequent. Over a small temperature range (upto ${100}^{\circ }\mathrm{C}$ ), the resistivity of a metal can be represented approximately by the equation where, ${\rho }_0$  is the resistivity at a reference temperature $T_0$  (often taken as $0^{\circ }\mathrm{C}$  or ${20}^{\circ }\mathrm{C}$  ) and $\rho \left(\mathrm{T}\right)$  is the resistivity at temperature T , which may be higher or lower than ${\mathrm{T}}_0$ . The factor $\alpha$  is called the temperature coefficient of resistivity.
The resistance of a given conductor depends on its length and area of cross-section besides the resistivity. As temperature changes, the length and area also change. But these changes are quite small and the factor $\mathcal{l}/A$  may be treated as constant.

Then, $R\propto \rho$ and hence,. In this equation $R\left(T\right)$  is the resistance at temperature $T$  and $R_0$  is the resistance at temperature $T_0$ , often taken to be $0^{\circ }\mathrm{C}$  or ${20}^{\circ }\mathrm{C}$ . The temperature coefficient of resistance $\alpha$  is the same constant that appears.

Note :                  

• The $\rho$ -T equation written above can be derived from the relation,
  $\alpha =$ fractional change in resistivity per unit change in temperature
  $=\frac{d\rho }{\rho dT}=\alpha \mathrm{}$ or, $\frac{d\rho }{dT}=\alpha \rho$
  $\therefore \mathrm{}\frac{d\rho }{\rho }=\alpha dT\mathrm{}$  ( $\alpha$  can be assumed constant for small temperature variation )

$\begin{array}{c}\therefore {\int }_{{\rho }_0}^{\rho } \frac{d\rho }{\rho }=\alpha {\int }_{T_0}^T dT\#\left(iii\right)\end{array}$

$e^{\alpha \left(T-T_0\right)}$ can approximately be written as Hence,
In the above discussion we have assumed  to be constant. If it is a function of temperature it will come inside the integration in Eq. (iii).

In a resistor current flows from high potential to low potential High potential is represented by positive (+) sign and low potential is represented by negative ( - ) sign.

$V_A-V_B=iR$

 If $V_1>V_2$  then current will flow from  to $A$ and $B$

and $i=\frac{V_1-V_2}{R}$

If $V_1<V_2$  then current will go from  to $B$ and $A$ and $i=\frac{V_2-V_1}{R}$
Example 8. Calculate current (i) flowing in part of the circuit shown in figure?

Solution: