State and Verify Ohm's Law | Draw V-I Characteristics - Param Himalaya
Statement of Ohm's Law :
According to Ohm's law, the current ($I$) flowing through a conductor is directly proportional to the potential difference ($V$) across the ends of the conductor, provided the physical conditions (like temperature, pressure, strain, etc.) of the conductor remain unchanged.
Mathematically, this can be expressed as:
$V \propto I$
Introducing a constant of proportionality, $R$, known as the electric resistance or simply resistance of the conductor, we get:
$V = RI$
This can also be written as:
$R = \frac{V}{I}$
The value of $R$ depends on the nature of the material of the conductor, its dimensions, and temperature. It does not depend on the values of $V$ and $I$.
Verification of Ohm's Law :
Ohm's law can be verified using the voltmeter-ammeter method. A circuit is set up with a battery connected to a conductor XY through a rheostat, an ammeter in series, and a key. A voltmeter is connected in parallel across the conductor XY to measure the potential difference across it.
When the key is closed, current passes through the conductor. Readings of the voltmeter and ammeter are noted for different positions of the rheostat. It is observed that the ratio $\frac{V}{I}$ remains constant. Thus, Ohm's law is verified.
V-I Characteristics :
The variation of electric current ($I$) through a conductor with the variation of potential difference ($V$) across the ends of the conductor at constant temperature is known as the V-I characteristic of the conductor.
According to Ohm's law, $V \propto I$, therefore, the V-I characteristic of a conductor is a straight line passing through the origin.
Determination of Resistance using V-I Characteristics :
Using V-I Graph
The slope of the V-I graph is given by $\frac{\Delta V}{\Delta I}$, which is equal to the resistance $R$.
$\text{Slope of V-I graph} = \frac{V}{I} = R $
Thus, the resistance of a conductor is equal to the slope of the V-I graph.
A higher slope indicates a higher resistance, and vice-versa.
Vector form of ohm's law. or Show that $\vec{J} = \sigma \vec{E}$
Consider a conductor AB of length $l$. If V is the potential difference across the ends of the conductor, then the uniform electric field in the conductor is given by
$E = \frac{V}{l} \quad \text{or} \quad V = El. \quad \ldots (1)$
According to ohm's law,
$ V = IR = \frac{I \rho l}{A} \quad \left( \because R = \frac{\rho l}{A} \right)$
Since
$\frac{I}{A} = J, \quad \text{therefore,}$
$V = J \rho l \quad \ldots (2)$
From eqns. (1) and (2), we get
$El = J \rho l \quad \text{or} \quad E = J \rho \quad \ldots (3)$
In vector form, eqn. (3) is written as
$ \vec{E} = \rho \vec{J} \quad \ldots (4)$
Since, $\rho = \frac{1}{\sigma}$, where $\sigma$ is conductivity of the conductor, therefore, eqn. (4) can be written as
$ \vec{E} = \frac{1}{\sigma} \vec{J}$
$ \vec{J} = \sigma \vec{E} \quad \ldots (5)$
