Current Electricity is a crucial chapter in the JEE Main syllabus, forming the foundation for understanding electrical circuits and their applications. This chapter equips students with the knowledge to analyze various electrical phenomena and solve complex circuit problems.

## Key Concepts:

## 1. Electric Current

Electric current is the flow of electric charge through a conductor. It is measured in amperes (A) and represented by the symbol I.

I = Q/t

Where:

Q = charge in coulombs (C)

t = time in seconds (s)

**Types of Electric Current:**a) Alternating Current (AC): The direction of charge flow periodically reverses.

b) Direct Current (DC): Charge flows in one direction only.

## 2. Current Density and Drift Velocity

Current density (J) is the current per unit cross-sectional area of a conductor.

J = I/A

Where:

I = current

A = cross-sectional area

Drift velocity (v_d) is the average velocity of charge carriers in a conductor under an electric field.

v_d = I / (nAq)

Where:

n = number density of charge carriers

q = charge of each carrier

## 3. Ohm’s Law

Ohm’s law states that the current through a conductor is directly proportional to the potential difference across it, provided physical conditions remain constant.

V = IR

Where:

V = potential difference

I = current

R = resistance

## 4. Resistance and Resistivity

Resistance (R) is the opposition to current flow in a conductor. It is measured in ohms (Ω).

R = V/I

Resistivity (ρ) is the intrinsic property of a material that determines its resistance.

R = ρL/A

Where:

ρ = resistivity

L = length of the conductor

A = cross-sectional area

## 5. Temperature Dependence of Resistivity

The resistivity of most materials changes with temperature:

ρ = ρ_0 [1 + α(T – T_0)]

Where:

ρ_0 = resistivity at reference temperature T_0

α = temperature coefficient of resistivity

T = current temperature

## 6. Combination of Resistors

a) Series Combination:

R_eq = R_1 + R_2 + R_3 + …

b) Parallel Combination:

1/R_eq = 1/R_1 + 1/R_2 + 1/R_3 + …

## 7. Electrical Energy and Power

Energy consumed:

E = VIt = I^2Rt = V^2t/R

**Power:**P = VI = I^2R = V^2/R

## 8. Kirchhoff’s Laws

a) Kirchhoff’s Current Law (KCL): The algebraic sum of currents meeting at a junction is zero.

∑I = 0

b) Kirchhoff’s Voltage Law (KVL): The algebraic sum of potential differences in a closed loop is zero.

∑V = 0

## 9. Wheatstone Bridge

A Wheatstone bridge is used to measure unknown resistance. It is balanced when:

R_1/R_2 = R_3/R_4

## 10. Potentiometer

A potentiometer is used to measure EMF and compare potential differences. It works on the principle of null deflection.

## 10. Cells in Series and Parallel

**a) Series Combination:**E_eq = E_1 + E_2 + E_3 + …

R_eq = r_1 + r_2 + r_3 + …

**b) Parallel Combination:**1/E_eq = 1/E_1 + 1/E_2 + 1/E_3 + …

1/R_eq = 1/r_1 + 1/r_2 + 1/r_3 + …

**Table of Important Formulas and Definitions:**

Concept | Formula/Definition |
---|---|

Electric Current | I = Q/t |

Current Density | J = I/A |

Drift Velocity | v_d = I / (nAq) |

Ohm’s Law | V = IR |

Resistance | R = V/I |

Resistivity | R = ρL/A |

Temperature Dependence of Resistivity | ρ = ρ_0 [1 + α(T – T_0)] |

Series Resistors | R_eq = R_1 + R_2 + R_3 + … |

Parallel Resistors | 1/R_eq = 1/R_1 + 1/R_2 + 1/R_3 + … |

Electrical Energy | E = VIt = I^2Rt = V^2t/R |

Electrical Power | P = VI = I^2R = V^2/R |

Kirchhoff’s Current Law | ∑I = 0 |

Kirchhoff’s Voltage Law | ∑V = 0 |

Wheatstone Bridge Balance | R_1/R_2 = R_3/R_4 |

Cells in Series (EMF) | E_eq = E_1 + E_2 + E_3 + … |

Cells in Parallel (EMF) | 1/E_eq = 1/E_1 + 1/E_2 + 1/E_3 + … |

**Important Topics for JEE Main:**

**1. Electric Current and Current Density**Understanding the concept of electric current and its relation to charge flow is crucial. Students should be able to calculate current density and drift velocity in various scenarios.

**2. Ohm’s Law and Its Applications**Ohm’s law forms the basis for analyzing simple circuits. Students should be comfortable applying it to solve problems involving voltage, current, and resistance.

**3. Resistance and Resistivity**The relationship between resistance and resistivity is important. Students should be able to calculate resistance for different materials and geometries.

**4. Temperature Effects on Resistance**Understanding how temperature affects resistance is essential for solving problems involving heating effects in conductors.

**5. Combination of Resistors**Students should be proficient in calculating equivalent resistance for series and parallel combinations, as well as more complex networks.

**6. Electrical Energy and Power**Calculating energy consumption and power in electrical circuits is a common problem type in JEE Main.

**7. Kirchhoff’s Laws**These laws are fundamental for analyzing complex circuits. Students should be able to apply both KCL and KVL to solve multi-loop circuit problems.

**8. Wheatstone Bridge and Meter Bridge**Understanding the working principle of Wheatstone bridge and meter bridge is important. Students should be able to analyze balanced and unbalanced bridge circuits.

**9. Potentiometer**The principle and applications of potentiometer, including measurement of EMF and comparison of potential differences, are important topics.

**10. Cells and Batteries**Students should understand how cells combine in series and parallel, and be able to calculate equivalent EMF and internal resistance.

## Solved Examples:

**1. Resistor Network Problem:**Question: Five equal resistances of 2Ω each are connected in a network as shown. Find the equivalent resistance between points A and B.

**Solution:**Step 1: Recognize that this is a balanced Wheatstone bridge.

Step 2: In a balanced bridge, no current flows through the central resistor.

Step 3: The circuit reduces to two parallel branches of 2Ω each.

Step 4: Calculate the equivalent resistance:

R_eq = (2Ω × 2Ω) / (2Ω + 2Ω) = 1Ω

Therefore, the equivalent resistance between A and B is 1Ω.

**2. Potentiometer Problem:**Question: A potentiometer wire of length 1000 cm and resistance 10Ω is connected to a 2V battery. A cell of EMF 1.5V and internal resistance 0.5Ω is connected to the potentiometer. The balance point is found at 600 cm. Calculate the internal resistance of the 2V battery.

**Solution:**Step 1: Calculate the current in the potentiometer wire:

I = V/R = 2V / 10Ω = 0.2A

Step 2: Calculate the potential gradient of the wire:

Potential gradient = V/L = 2V / 1000cm = 0.002 V/cm

Step 3: At balance point, EMF of cell = Potential drop across 600 cm

1.5V = 0.002 V/cm × 600 cm

Step 4: Calculate the current drawn from the 2V battery:

I_2 = (2V – 1.5V) / (Internal resistance of 2V battery)

0.2A = 0.5V / R_internal

Step 5: Solve for R_internal:

R_internal = 0.5V / 0.2A = 2.5Ω

Therefore, the internal resistance of the 2V battery is 2.5Ω.

### Conclusion:

The Current Electricity chapter is fundamental to understanding electrical circuits and forms the basis for more advanced topics in electromagnetism. Mastering the concepts, formulas, and problem-solving techniques in this chapter is crucial for success in **JEE Main**. Regular practice with a variety of problems, especially those involving complex circuits and applications of Kirchhoff’s laws, will help students build confidence and proficiency in this important area of physics.