Ohm's Law : Definition
When a potential difference V is applied across a conductor, it causes a current I to flow through it.
According to Ohm’s Law
“Under given physical conditions the potential diļ¬erence (V) applied across a conductor produces a proportionate amount of current (I) in the conductor”
Mathematical Formula of Ohm Law
\[V \propto I\]
\[V= IR\]
\( V \) is the voltage (in volts),
\( I \) is the current (in amperes),
\( R \) is the resistance (in ohms $\Omega$).
Resistance of conductor
The resistance of a conductor is defined as the amount of potential difference required to drive a unit current through it. It is the ratio of the voltage applied across the conductor to the current it conducts.The resistance of a conductor is said to be one ohm, if one ampere of current flows through it, when a potential difference of one volt is applied across it.
$\displaystyle R=\frac{V}{I}$ S.I unit is $\Omega (ohm)$
Cause of Resistance:
Resistance in a conductor refers to the opposition it offers to the flow of electric charge. When a potential difference is applied across the conductor, an electric field is created, causing free electrons to accelerate. As these electrons move, they frequently collide with the atoms and ions within the material. These collisions slow down the movement of electrons, creating opposition to the current flow. This opposition is what we call the resistance of the conductor.
Definition of Resistivity
Resistivity or specific resistance of the material of a conductor is the resistance offered by a wire of this material of unit length and unit area of crosssection.
The resistance of a conductor depends on:
1. Length (\(L\)) : Resistance is directly proportional to the length of the conductor.
\[R \propto L\]
2. Area of Cross-section (\(A\): Resistance is inversely proportional to the area of cross-section.
\[ R \propto \frac{1}{A}\]
Combining the two relationships:
\[R \propto \frac{L}{A}\]
Introducing the constant of proportionality \( \rho \), called the resistivity or specific resistance:
\[R = \rho \frac{L}{A}\]
Where
\( R \) is the resistance in ohms (\( \Omega \)),
\( \rho \) is the resistivity in ohm-meters (\( \Omega \cdot \text{m} \)),
\( L \) is the length in meters (m),
\( A \) is the cross-sectional area in square meters (m\(^2\)).
Example 1: Calculate the resistivity of the material of a wire 1.0 m long, 0.4 mm in diameter and having a resistance of 2.0 ohm
To calculate the resistivity \( \rho \) of the material, we use the formula:
\[\displaystyle R = \rho \frac{L}{A}\]
Rearranging for \( \rho \):
\[\displaystyle \rho = R \cdot \frac{A}{L}\]
Given:
$\displaystyle L= 10 \ \text{m}$
$\displaystyle \text{Diameter}, \ d = 0.4 \ \text{mm} = 0.0004 \ \text{m} $
$\displaystyle \text{Radius}, \ r = \frac{d}{2} = 0.0002 \ \text{m}$
$\displaystyle R = 2 \ \Omega \\$
$\displaystyle A = \pi r^2 = \pi (0.0002)^2 = \pi \cdot 4 \times 10^{-8} = 1.2566 \times 10^{-7} \ \text{m}^2$
Substitute into the formula:
\[\rho = 2 \cdot \frac{1.2566 \times 10^{-7}}{10} = 2.5132 \times 10^{-8} \ \Omega \cdot \text{m}\]
Final Answer:
\[\boxed{\rho = 2.51 \times 10^{-8} \ \Omega \cdot \text{m}}\]
Example 2: A negligibly small current is passed through a wire of length 15 m and uniform cross-section $6.0 \times 10^{–7} m^2$, and its resistance is measured to be $5.0 \Omega$. What is the resistivity of the material at the temperature of the experiment?
Solution
We use the formula for resistance:
\[R = \rho \frac{L}{A}\]
Rearranging for resistivity \( \rho \):
\[\rho = R \cdot \frac{A}{L}\]
Given:
$\begin{align*}
R &= 5.0 \ \Omega \\
L &= 15 \ \text{m} \\
A &= 6.0 \times 10^{-7} \ \text{m}^2
\end{align*}$
Substitute the values:
\[\rho = 5.0 \cdot \frac{6.0 \times 10^{-7}}{15}= 5.0 \cdot 4.0 \times 10^{-8}= 2.0 \times 10^{-7} \ \Omega \cdot \text{m}\]
Final Answer:
\[\boxed{\rho = 2.0 \times 10^{-7} \ \Omega \cdot \text{m}}\]
Example 3: Calculate resistance of a conductor carrying a current of 2 A, when potential difference across it is 20 V.
Given:$\begin{align*}
I &= 2 \ \text{A} \\
V &= 20 \ \text{V}
\end{align*}$
Using Ohm's Law:
\[R = \frac{V}{I}\]
Substituting the values:
\[R = \frac{20}{2} = 10 \ \Omega\]
Final Answer
\[\boxed{R=10 \ \Omega}\]