## Concepts of Current Electricity JEE Main Physics

Current Electricity is a crucial chapter in the JEE Main syllabus, and it's all about understanding how electric currents work and how they interact with materials. In simple terms, it's like learning how electricity flows in wires and devices. This chapter is fundamental as it forms the basis for a multitude of advanced electrical and electronic concepts.

In this article, explore various aspects of Current Electricity, including key concepts like Electric Current (the flow of electricity), Ohm's Law (which explains the relationship between voltage and current), Electrical Resistance (how materials slow down electric flow), and Drift Velocity (how fast electrons move). We will also discuss different materials' resistances, Ohmic and non-Ohmic conductors, and the basics of Electrical Energy and Power. Understanding electrical resistivity, the color code for resistors, and how resistors can be combined in series and parallel is also on our list. Plus, we will delve into the temperature effects on resistance.

## JEE Main Physics Chapters 2024

### Important Topics of Current Electricity Chapter

Electric current density and drift velocity

Ohm’s law and temperature depend on resistivity.

Grouping of Resistors

Colour code of carbon resistor

Grouping of cells

Kirchhoff’s Law

Wheatstone bridge

Metre bridge

Potentiometer

### Current Electricity Important Concept for JEE Main

## Ohm’s Law Formula

### Types of Electric Current

1. Alternating current (AC) — In its systems, the movement of electric charge reverses direction periodically. AC is the form of electric power that is most commonly delivered to businesses and residences. Audio and radio signals that are carried on electrical wires are some examples of AC.

2. Direct current (DC) — It is the unidirectional flow of electric charge and a system in which electric charge moves in one direction only. Sources of direct current are batteries, thermocouples, solar cells, and commutator-type electric machines of the dynamo type. DC flows in a conductor like wire, but can also flow through semiconductors, insulators, or a vacuum as in electron or ion beams.

### Relation Between Resistance and Resistivity

Resistance: To obstruct the flow of electric current is called the resistance.

Let’s say we have a conductor, and we apply a potential across its ends.

Then the resistance of this material is,

R = V/I..(1)

So on applying the potential difference, ΔV across the ends of the conductor, if the current I flow through it, then the ratio of V/I is the resistance.

This relationship isn’t for Ohm’s law because this law states that V ∝ I or V = IR

Where R is constant, but resistance can be variable.

So for the devices having a variable resistance, we can consider the potential difference at an instant and use the above equation (1):

Therefore, Ohm’s law isn’t valid everywhere.

We know that:

R ∝ 1/L

R ∝ A

On removing the constant of proportionality sign, we get:

R = ρ L/A

The proportionality constant, ρ is called the specific resistance or the electrical resistivity of a material which is measured in Ohm-m or Ω m.

### What is Conventions?

In a conductive object, the moving charged particles that constitute the electric current are known as charge carriers. In metals, the positively charged atomic nuclei of the atoms are held in a fixed position whereas the negatively charged electrons that are charge carriers that are free to move in the metal. In semiconductors, the charge carriers can be positive or negative, depending on the dopant used. A flow of positive charge gives the electric current and has an effect in a circuit the same as an equal flow of negative charges in the opposite direction.

### Measurement of Current

Current can be measured using an ammeter. It’s also measured by detecting the magnetic field associated with the current.

Different Techniques to Measure Current are:

Shunt resistors

Hall effect current sensor transducers

Transformers

Magnetoresistive field sensors

Rogowski coils

Current clamps

### Resistive Heating

Joule heating or resistive heating is a process of power dissipation through which the passage of an electric current through a conductor increases the internal energy of the conductor, converting thermodynamic work into heat. James Prescott Joule, the inventor of this phenomenon, immersed a length of wire in a fixed mass of water and measured the temperature rise due to a known current through the wire for 30 minutes. By varying the current and length of the wire, he deduced that the heat produced was proportional to the square of the current multiplied by the electrical resistance of the wire.

P ∝ I²R

This relationship is known as Joule's Law. The SI unit of energy was named joule and given the symbol J. The SI unit of power, watt (W), is equivalent to one joule per second.

### List of Important Formulas for Current Electricity Chapter

### JEE Main Current Electricity Solved Examples

1. The internal resistance of a cell of emf 2V is 0.1 Ω. It is connected to a resistance of 3.9 Ω. Find the voltage across the cell.

Sol:

Given:

Emf of the cell, E = 2V

Internal resistance of the cell, r = 0.1 Ω

Resistance connected to the cell, R = 3.9 Ω

Let V be the terminal voltage across the cell.

We know that the voltage drop across the resistor,

$V=IR$....(1)

Applying Kirchhoff's voltage law,

$E-V-Ir=0$

$E-IR-Ir=0$

$2V-3.9I-0.1I=0$

$I=\dfrac{2}{4}=0.5~A$

Substitute the value of current I in equation (1) to obtain the voltage across the cell V.

$V=0.5\times 3.9=1.95~V$

Therefore, the voltage across the cell is 1.95 V

Key point: Since resistor R and the cell are in parallel connection, the voltage across them are equal.

2. The resistance of a wire is R ohm. If it is melted and stretched to n times its original length, its new resistance will be_________.

Sol:

Given:

The resistance of the wire is R.

Let the length and area of the cross-section of wire be l and A, respectively and the resistivity of the wire be 𝜌.

If the length of the wire is stretched to n times, then the area of the cross-section of the wire is reduced by n times.

Let the length and area of the cross-section of the stretched wire be l’ and A’, respectively.

The resistance of the stretched wire is given by

$R'=\rho\dfrac{l'}{A'}$

$R'=\rho\dfrac{(nl)}{\left(\dfrac{A}{n}\right)}$

$R'=\rho\dfrac{n^2l}{A}$

$R'=n^2\times\left(\rho\dfrac{l}{A}\right)$

$R'=n^2R$

Therefore, the resistance of the stretched wire is n2 times the initial resistance.

Key point: When a wire is stretched, its length increases and the area of cross-section decreases while keeping the volume constant.

### Previous Years’ Questions from JEE Paper

1. Five equal resistances are connected in a network as shown in the figure. The net resistance between points A and B is______. (Feb -2021)

a. 3R/2

b. R/2

c. R

d. 2R

Sol:

If you observe the diagram closely, you can see that the above circuit diagram is a balanced Wheatstone bridge since all the resistances are the same. Therefore, it satisfies the condition

$\dfrac{R_2}{R_{1}}=\dfrac{R_4}{R_3}$

Since it is balanced, the current does not pass through the resistor in the bridge and can be removed while calculating the effective resistance.

The two 2R resistors are in parallel, so the equivalent resistance between A and B is

$R_{eq}=\dfrac{2R\times 2R}{2R+2R}$

$R_{eq}=\dfrac{4R^2}{4R}$

$R_{eq}=R$

Therefore, the equivalent resistance between A and B is R and the correct answer is option C.

Key point to remember: When a difficult circuit diagram having resistors are given, always look for the possibility of a balanced Wheatstone bridge.

2. The length of a potentiometer wire is 1200 𝑐𝑚 and it carries a current of 60 𝑚𝐴. For a cell of emf 5 𝑉 and internal resistance of 20Ω, the null point on it is found to be at 1000 𝑐𝑚. The resistance of the whole wire is_____. (JEE 2020)

a. 80Ω

b. 100Ω

c. 120Ω

d. 60Ω

Sol:

Given:

The null point of the potentiometer = 1000 cm

The current through the potentiometer wire, I = 60 mA

Emf of the cell in the secondary circuit, E = 5 V

At the null point, no current passes through the galvanometer and hence no current is drawn from the cell. Therefore, the emf of the cell is equal to the potential difference across the section of the wire having a length of 1000 cm.

So, the resistance of the wire of 1000 cm (R1) can be calculated.

$R_1=\dfrac{E}{I}$

$R_1=\dfrac{5~V}{60~\text{mA}}$

$R_1=83.33~ \Omega $

Let the resistance of the wire per unit length be r.

$r=\dfrac{83.33~ \Omega}{1000~\text{cm} }$

$r=83.33\times 10^{-3}~ \Omega/cm$

Now, the resistance of the 1200 cm potentiometer wire (R) is obtained by

$R=83.33\times 10^{-3}~ \Omega/cm~\times 1200~\text{cm}$

$R=100~\Omega$

Therefore, the correct answer is option B.

Key point to remember: At the balancing point, no current is drawn from the cell and hence the terminal voltage and emf are equal.

### Practice Questions

1. A uniform wire of resistance 9 Ω is cut into three equal parts. They are connected in the form of an equilateral triangle ABC. A cell of emf 2V and negligible internal resistance is connected across B and C. Potential difference across AB is____. (Ans: 1 V)

2. A filament bulb (500 W, 100V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and consumes 500 W. The value of R is_______. (Ans: 26 Ω)

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### Conclusion

In this guide, we'll dive into the world of Current Electricity, a key topic in JEE Main physics. We'll demystify the core concepts, providing simple explanations and problem-solving strategies. You'll learn about electric circuits, Ohm's law, and how to calculate current and voltage. We've compiled everything you need in one place, including free downloadable PDFs with detailed explanations and practice questions. This resource is a valuable tool to help you ace your exams.

## FAQs on Current Electricity Chapter - Physics JEE Main

1. What is the weightage of the current electricity in JEE?

Around 2-3 questions are asked in JEE from various parts of the chapter. Most of the questions are based on electric current example problems and questions can also be asked by linking them to concepts from other chapters.

2. What section has to be given more importance in the chapter on current electricity?

More focus must be laid on solving electric current example problems and getting the answer without committing any mistakes. Therefore, students should be able to solve any problems related to this chapter by practising more and more problems.

3. How to score good marks in JEE from the chapter on Current Electricity?

First of all, learn and understand the concepts and current electricity formulae in each chapter in depth. This should be followed by solving all the problems given in NCERT textbooks. Then go through the question papers of the last 20 years. Try to solve them and observe the pattern to score good marks in JEE.

4. What is the importance of the Current Electricity chapter in Physics for JEE Main?

Understanding Current Electricity is crucial for JEE Main as it forms the foundation for many concepts and problem-solving skills in Physics.

5. How can Vedantu help in mastering the Current Electricity chapter for JEE Main?

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