Chapter 3: Current Electricity - Comprehensive NEET Physics Notes

1. Electric Current and Ohm’s Law

1.1 Electric Current

Formula: $I=\frac{q}{t}$

Explanation:

• $I$: Electric current (A)
• $q$: Net charge flowing through the conductor (C)
• $t$: Time interval during which charge flows (s)

Electric current is defined as the rate of flow of charge through a conductor. The SI unit of current is the ampere (A).

1.2 Ohm’s Law

Formula: $V=IR$

Explanation:

• $V$: Voltage across the conductor (V)
• $I$: Current through the conductor (A)
• $R$: Resistance of the conductor (Ω)

Ohm's Law states that the current flowing through a conductor is directly proportional to the voltage across it, provided the temperature remains constant. The resistance $R$ is a property of the material and its dimensions.

1.3 Resistivity and Conductivity

Formula: $R=\rho \frac{l}{A}$

$\sigma =\frac{1}{\rho }$

Explanation:

• ρ: Resistivity of the material (Ω·m)
• σ: Conductivity of the material (S/m)
• $I$: Length of the conductor (m)
• $A$: Cross-sectional area of the conductor (m²)

Resistivity is a material property that quantifies how strongly a given material opposes the flow of electric current. The reciprocal of resistivity is conductivity, which indicates how well a material conducts electricity.

Example Application:

Problem: A copper wire has a length of 2 m and a cross-sectional area of 1 mm². If the resistivity of copper is $1.68\times 10^{-8}\Omega \cdot \text{m}$, calculate its resistance.

Solution: Given:

• $l=2\text{m}$
• $A=1{\text{mm}}^2=1\times 10^{-6}{\text{m}}^2$
• $\rho =1.68\times 10^{-8}\Omega \cdot \text{m}$

Using the formula: $R=\rho \frac{l}{A}=1.68\times 10^{-8}\times \frac{2}{1\times 10^{-6}}=3.36\times 10^{-2}\Omega$

NEET Tip: Always remember to convert the area into square meters (m²) when using the resistivity formula. Incorrect units can lead to significant errors in the final answer.

Common Mistake:

• Misconception: Many students confuse resistivity ($\rho$) with resistance ($R$). Remember, resistivity is a material property, while resistance depends on both the material and the dimensions of the conductor.

2. Electric Power and Energy

2.1 Power in Electrical Circuits

Formula: $P=VI$

$P=I^{2}R$

$P=\frac{V^2}{R}$

Explanation:

• $P$: Power dissipated in the circuit (W)
• $V$: Voltage across the component (V)
• $I$: Current through the component (A)
• $R$: Resistance of the component (Ω)

Power is the rate at which energy is consumed or generated in a circuit. The above formulas represent power in terms of voltage, current, and resistance.

Example Application:

Problem: A 60 W light bulb operates at 240 V. Calculate the current passing through the bulb and its resistance.

Solution: Given:

• $P=60\text{W}$
• $V=240\text{V}$

Using the formula: $I=\frac{P}{V}=\frac{60}{240}=0.25\text{A}$

$R=\frac{V^2}{P}=\frac{240^2}{60}=960\Omega$

Real-life Application: Electric bulbs and other resistive loads in households convert electrical energy into light and heat. Understanding power and energy calculations helps in determining the efficiency and safety of electrical devices.

Common Mistake:

• Misconception: Confusing power (W) with energy (Joules). Power is the rate of energy usage, not the total energy consumed.

Quick Recap

• Current:$I=\frac{q}{t}$
• Ohm's Law: $V=IR$
• Resistance: $R=\rho \frac{l}{A}$
• Power: $P=VI$, $P=I^{2}R$, $P=\frac{V^2}{R}$

Practice Questions

1. A conductor of length 3 m and cross-sectional area 2 mm² has a resistance of 0.15 Ω. Calculate the resistivity of the material.
2. What current flows through a 100 Ω resistor when connected to a 220 V power supply?
3. If a heater dissipates 1.5 kW of power when connected to a 250 V supply, find the resistance of the heater.

This structure covers the key formulae, their explanations, example applications, and common mistakes to avoid, tailored specifically for NEET UG Physics preparation.