Important Derivations for Class 12 Physics Chapter 3 Current Electricity

Important Derivations for Class 12 Physics Chapter 3 Current Electricity

Deduction of Ohm’s law.
When a potential difference ‘V’ is applied across a conductor of length ‘l’, the drift velocity in terms of ‘V’ is given by $$ v_d=\frac{e E \tau}{m}=\frac{e V \tau}{m l} $$ If the area of cross-section of the conductor is ‘A’ and the number of electrons per unit volume or the electron density of the conductor is ‘n’, then the current through the conductor will be $$ I=e n A v_d=e n A \cdot \frac{e V \tau}{m l} $$ or $$ \frac{V}{I}=\frac{m l}{n e^2 \tau A} . $$ At a fixed temperature, the quantities and A, all have constant values for a given conductor. Therefore, $$ \frac{V}{I}=\text { a constant, } R $$ This proves Ohm’s law for a conductor and here $$ R=\frac{m l}{n e^2 \tau A} $$ is the resistance of the conductor.

---

Relation between and .
For an electron, $$ \begin{array}{ll}q & =-e \\ \text { and } \quad \overrightarrow{v_d} & =-\frac{e \vec{E} \tau}{m} \\ \quad \vec{j} & =n q \overrightarrow{v_d}=n(-e)\left(-\frac{e \vec{E} \tau}{m}\right)=\frac{n e^2 \tau}{m} \vec{E}\end{array} $$ But $$\frac{n e^2 \tau}{m}=\frac{1}{\rho}=\sigma$$, conductivity of the conductor $$ \quad \vec{j}=\sigma \vec{E}$$ or $$\quad \vec{E}=\rho \vec{j}$$ This is Ohm’s law in terms of vector quantities like current density and electric field .

---

Relation between Electric Current and Drift Velocity

Number of electrons in length ‘l’ of the conductor volume of the conductor
Total charge contained in length ‘l’ of the conductor is
All the electrons which enter the conductor at the right end will pass through the conductor at the left end in time, $$ t=\frac{\text { distance }}{\text { velocity }}=\frac{l}{v_d} $$ therefore, Current, $$I=\frac{q}{t}=\frac{e n A l}{l / v_d}$$ or $$I=e n A v_d$$ This equation relates the current ‘I’ with the drift velocity . The current density ‘j’ is given by $$ j=\frac{I}{A}=e n v_d $$ In vector form $$\quad \vec{j}=e n \vec{v}_d$$ The above equation is valid for both positive and negative values of ‘q’.

---

Expression for Internal Resistance of a Cell

To derive the expression for the internal resistance (\( r \)) of a cell in terms of electromotive force (e.m.f) (\( \mathcal{E} \)) and terminal potential difference (\( V \)) of a cell, we can use Ohm’s law and the concept of potential difference across the internal resistance.
Ohm’s Law states that the current (\( I \)) flowing through a circuit is directly proportional to the voltage (\( V \)) applied across it and inversely proportional to the total resistance (\( R \)) of the circuit: \[ V = I \cdot R \] In the case of a cell, the terminal potential difference (\( V \)) is the voltage across the cell’s terminals when a current flows through it. Let’s assume that the internal resistance of the cell is \( r \).
The e.m.f (\( \mathcal{E} \)) of the cell represents the maximum potential difference that the cell can provide when no current is flowing through it. In other words, when there is no current (\( I = 0 \)), the voltage across the cell terminals is equal to the e.m.f (\( \mathcal{E} \)).
Now, when a current (\( I \)) flows through the cell, there will be a voltage drop across the internal resistance (\( r \)), which we can represent as \( I \cdot r \). The remaining voltage (\( V – I \cdot r \)) will be the terminal potential difference (\( V \)).
Therefore, we can write the equation for terminal potential difference as: \[ V = \mathcal{E} – I \cdot r \] Now, we can rearrange this equation to solve for the internal resistance (\( r \)): \[ r = \frac{\mathcal{E} – V}{I} \] This expression gives the internal resistance (\( r \)) of a cell in terms of the e.m.f (\( \mathcal{E} \)), terminal potential difference (\( V \)), and current (\( I \)) flowing through the cell.

---

Expression for Resistance and Resistivity

To derive the expression for the resistivity (\( \rho \)) of a conductor in terms of the number density of free electrons (\( n \)) and the relaxation time (\( \tau \)), we consider the concept of electron drift in a conductor.

When an electric field (\( E \)) is applied across a conductor, the free electrons within the conductor experience a force in the direction opposite to the electric field. Due to this force, the electrons undergo random motion and also drift in the direction opposite to the electric field. The average velocity acquired by the electrons due to this drift is called the drift velocity (\( v_d \)).

The drift velocity (\( v_d \)) of the free electrons is related to the electric field (\( E \)) and the relaxation time (\( \tau \)) through the following equation: \[ v_d = \frac{e \cdot E \cdot \tau}{m} \] where: \( e \) = Charge of a single electron,
\( m \) = Mass of a single electron,
\( \tau \) = Relaxation time of electrons,
\( E \) = Magnitude of the applied electric field.

The drift velocity (\( v_d \)) can also be expressed in terms of the number density of free electrons (\( n \)) and the cross-sectional area of the conductor (\( A \)). The number density of free electrons (\( n \)) is the number of free electrons per unit volume. Hence, the total number of free electrons in the conductor is \( n \cdot A \cdot L \), where \( L \) is the length of the conductor along the direction of the electric field.

The total charge flowing through the conductor per unit time (\( I \), current) is given by: \[ I = n \cdot A \cdot e \cdot v_d \] Substitute the expression for \( v_d \) in terms of \( E \) and \( \tau \): \[ I = n \cdot A \cdot e \cdot \frac{e \cdot E \cdot \tau}{m} \] Now, the electric current (\( I \)) flowing through the conductor is related to the electric field (\( E \)) by Ohm’s law: \[ I = \frac{E}{R} \] where \( R \) is the resistance of the conductor. Substitute the expression for \( I \) in terms of \( E \) and \( R \): \[ \frac{E}{R} = n \cdot A \cdot e \cdot \frac{e \cdot E \cdot \tau}{m} \] Simplify and isolate \( R \): \[ R = \frac{m}{n \cdot e^2 \cdot \tau} \] Finally, the resistivity (\( \rho \)) of the conductor is defined as the reciprocal of conductivity (\( \sigma \)), which is the inverse of resistivity: \[ \sigma = \frac{1}{\rho} \] Thus, the expression for the resistivity (\( \rho \)) in terms of the number density of free electrons (\( n \)) and the relaxation time (\( \tau \)) is: \[ \rho = \frac{m}{n \cdot e^2 \cdot \tau} \]

---

Related Posts

You May Also Like:

Category Lists (All Posts)

Why students face difficulty in physics derivations?

There can be several reasons why students face difficulty in physics derivations, including:

1. Lack of foundation: A lack of understanding of the underlying concepts and principles can make it difficult to follow the logical steps in a derivation.
2. Mathematical background: Physics often involves complex mathematical calculations, and students who struggle with math may find it challenging to perform the necessary calculations.
3. Limited practice: Regular practice is key to developing the skills required for physics derivations, and students who have limited practice opportunities may struggle.
4. Poor problem-solving skills: Physics problems often require creative problem-solving skills, and students who struggle with this aspect of physics may find derivations particularly challenging.
5. Limited exposure to different problems: Physics derivations can vary widely in their level of complexity, and students who have limited exposure to different types of problems may struggle when faced with a new challenge.
6. Confusion with notation: Physics often uses a specialized notation, and students who are unfamiliar with this notation may struggle to follow the steps in a derivation.

Remembering physics derivations can be a challenge, but here are some tips that may help:

1. Practice, practice, practice: Regularly practicing physics derivations helps to build muscle memory and improve recall.
2. Visualize the process: Try to visualize the steps involved in a derivation, as well as the physical meanings behind each equation.
3. Make connections: Try to relate each step of the derivation to a concept or formula you already know, which can help to reinforce your understanding.
4. Write it down: Writing out a derivation helps to solidify your understanding and makes it easier to remember.
5. Understand the physical meaning: Try to understand the physical meaning behind each equation, which can help you to remember the derivation in a broader context.
6. Teach someone else: Teaching someone else the derivation can be a great way to reinforce your understanding and remember it more easily.
7. Use mnemonics: Creating mnemonics or acronyms can be a helpful way to remember a series of steps in a derivation.

Remember, it takes time and consistent effort to remember physics derivations. Keep practicing and seeking help when needed, and you will likely see improvement over time.

Download Books – Exam Special

CBSE Books	ICSE Books
OLYMPIAD Books	FOUNDATION Books
JEE Books	NEET Books

➡ Click below titles to expand