Ace Your Board Exams: Complete RBSE Class 12 Physics Chapter 3 Solutions (Current Electricity)

Master RBSE Class 12 Physics Chapter 3, Current Electricity! Find detailed, step-by-step NCERT solutions and essential practice questions for Rajasthan Board (RBSE) students to score high.

Introduction: Powering Up Your Physics Preparation!

Welcome, Rajasthan Board (RBSE) Class 12 students! Are you finding Chapter 3: Current Electricity of Physics a challenge? This crucial chapter forms the foundation for many advanced concepts and is a high-scoring unit in your board exams.

The good news is that the RBSE syllabus for Class 12 Physics aligns with the NCERT curriculum. This means you can leverage the comprehensive NCERT Solutions for Class 12 Physics Chapter 3 to excel in your RBSE exams!

This complete guide provides SEO-friendly, highly rankable solutions and essential tips, focusing specifically on the needs of RBSE students.

Why Focus on Chapter 3: Current Electricity?

Current Electricity (Chapter 3) is a vital unit for several reasons:

  1. High Weightage: Questions from Ohm’s Law, Kirchhoff’s Laws, Potentiometer, and Meter Bridge are frequently asked in the RBSE board exams, making it a high-scoring chapter.
  2. Conceptual Clarity: A strong grasp of concepts like drift velocity, resistivity, and internal resistance is essential for understanding subsequent chapters and for competitive exams like JEE and NEET.
  3. Numerical Practice: This chapter is rich in numerical problems, which are often direct application questions. Mastering these solutions guarantees marks.

Detailed NCERT Solutions for Class 12 Physics Chapter 3 (Current Electricity)

Ace Your Board Exams: Complete RBSE Class 12 Physics Chapter 3 Solutions (Current Electricity)

The core of your RBSE preparation lies in mastering the NCERT textbook questions. Here are solutions to the most important topics and exercise questions:

1. Concepts and Formulae You Must Know

Before diving into the solutions, ensure you are familiar with these key formulae:

ConceptFormulaDescription
Ohm’s LawV=IRRelates voltage, current, and resistance.
Drift VelocityI=neAvd​Relates current (I) to drift velocity (vd​), number density (n), charge (e), and area (A).
Resistivityρ=ne2τm​Resistivity (ρ) depends on mass (m), number density (n), charge (e), and relaxation time (τ).
Kirchhoff’s First Rule (Junction Rule)∑Iin​=∑Iout​Sum of currents entering a junction equals the sum of currents leaving it.
Kirchhoff’s Second Rule (Loop Rule)∑ΔV=0Algebraic sum of potential differences in a closed loop is zero.
Meter BridgeSR​=100−ll​Used to find an unknown resistance (S) using a known resistance (R) and balance point (l).
Potentiometer (EMF Comparison)E2​E1​​=l2​l1​​Compares EMFs of two cells.

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2. Solved Exercise Questions (RBSE Important)

Here is a glimpse of the detailed, step-by-step solutions to common numerical problems that frequently appear in the Rajasthan Board exams.

Question 3.1: Maximum Current from a Battery

Problem: The storage battery of a car has an e.m.f. of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

Solution: For maximum current, the external resistance (R) must be zero.

  • Emf (E) = 12 V
  • Internal Resistance (r) = 0.4 Ω
  • Maximum Current (Imax​) = ?

Using Ohm’s law for the circuit: I=R+rE​ For Imax​, set R=0.

Imax​=rE​=0.4 Ω12 V​=30 A

Answer: The maximum current that can be drawn from the battery is 30 A.

Question 3.5: Temperature Dependence of Resistance

Problem: At room temperature (27.0 °C), the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material is α=1.70×10−4 °C−1?

Solution: The resistance change with temperature is given by:

Rt​=R0​[1+α(t−t0​)]

  • R0​ (Resistance at t0​=27.0 °C) = 100 Ω
  • Rt​ (Resistance at unknown temperature t) = 117 Ω
  • α=1.70×10−4 °C−1

Substituting the values:

117=100[1+1.70×10−4(t−27)]100117​=1+1.70×10−4(t−27)1.17−1=1.70×10−4(t−27)0.17=1.70×10−4(t−27)t−27=1.70×10−40.17​=1.7×10−417×10−2​=10×102=1000

t=1000+27=1027 °C

Answer: The temperature of the element is 1027 °C.


RBSE Class 12 Physics Chapter 3: Important Topics for High Ranks

To ensure a high rank in the RBSE board exam, make sure you cover these frequently asked topics beyond the main exercises:

1. Drift Velocity and Mobility

  • Derivation: Be prepared to derive the expression for drift velocity: vd​=meEτ​.
  • Relation to Current: Understand the relation I=neAvd​.

2. Kirchhoff’s Laws

  • Statement: Clearly state both the Junction Rule (charge conservation) and the Loop Rule (energy conservation).
  • Numerical Application: Practice solving complex circuit problems using these laws.

3. Meter Bridge and Potentiometer

  • Principle: Understand the underlying principle for both instruments (Potentiometer: V∝l).
  • Application Diagrams: Memorize the circuit diagrams for comparing EMFs and measuring internal resistance of a cell using a potentiometer.
  • Advantage of Potentiometer: Know why a potentiometer is preferred over a voltmeter (it draws no current from the cell).

How to Use These Solutions for Maximum Score in RBSE

  1. Self-Study First: Try to solve the NCERT Chapter 3 exercises yourself before looking at the solutions.
  2. Verify Steps: Use these solutions to verify your final answers and, more importantly, your intermediate steps and conceptual understanding.
  3. Focus on Derivations: Write out the derivations for Meter Bridge and Potentiometer multiple times until you can reproduce them perfectly.
  4. RBSE Previous Year Questions: Search for RBSE Class 12 Physics Chapter 3 PYQ (Previous Year Questions) and practice them rigorously. You will notice a high repetition rate of concept-based numericals.
Chapter 3(Open)

Mastering these solutions and key concepts is your assured path to achieving a high score in the RBSE Class 12 Physics examination. Good luck!

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