RBSE Class 12 Physics Chapter 7: Alternating Current (AC) – Complete Solutions & Concepts for Board Exam Success!

Are you a Rajasthan Board (RBSE) Class 12 student aiming for a top score in Physics? Chapter 7, Alternating Current (AC), is a high-scoring unit packed with fundamental concepts and essential numerical problems. Mastering this chapter is crucial for your board exams.

This comprehensive guide offers detailed explanations of all key concepts, step-by-step solutions to every major problem type (based on the NCERT curriculum adopted by RBSE), and pro tips to help you secure a top rank!

Chapter 7: Alternating Current (AC) – Essential Concepts

Alternating Current (AC) is the electrical current whose magnitude changes continuously with time and whose direction reverses periodically. Unlike Direct Current (DC), AC is widely used for power transmission because it can be easily stepped up or down using transformers, minimizing energy loss over long distances.

1. AC Voltage and Current (Phasor Representation)

• AC Voltage: v=vm​sin(ωt)
  • vm​: Peak Voltage
  • ω: Angular frequency (ω=2πf)
• AC Current: i=im​sin(ωt+ϕ)
  • im​: Peak Current
  • ϕ: Phase difference between voltage and current.
• Root Mean Square (RMS) Value: The RMS value is the equivalent DC current/voltage that would produce the same heat in a resistor.
  • Vrms​=2​vm​​≈0.707vm​
  • Irms​=2​im​​≈0.707im​
  • Note: Household supply (e.g., 220V) is the RMS value.
• Phasors: Rotating vectors used to represent sinusoidally varying quantities like AC voltage and current.

2. AC Circuits with Single Components

Circuit Element	Reactance (X)	Current-Voltage Phase Relation	Power Consumed (P)
Resistor (R)	R (Resistance)	In phase (ϕ=0∘)	P=Vrms​Irms​ (Maximum Power)
Inductor (L)	Inductive Reactance: XL​=ωL	Current lags voltage by 90∘(π/2)	P=0 (Wattless current)
Capacitor (C)	Capacitive Reactance: XC​=ωC1​	Current leads voltage by 90∘(π/2)	P=0 (Wattless current)

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3. Series LCR Circuit

This is the most critical circuit configuration.

• Impedance (Z): The total effective resistance offered by the LCR circuit. Z=R2+(XL​−XC​)2​
  • Current Amplitude: im​=Zvm​​
  • RMS Current: Irms​=ZVrms​​
• Phase Angle (ϕ): The phase difference between the resultant voltage and current. tanϕ=RXL​−XC​​
• Power in AC Circuit (Average Power): Pavg​=Vrms​Irms​cosϕ
  • Power Factor (cosϕ): cosϕ=ZR​. It represents the fraction of total apparent power that is actually consumed.

4. Resonance in LCR Circuit

The condition where the circuit allows maximum current to flow.

• Condition for Resonance: Inductive reactance equals capacitive reactance: XL​=XC​
• Resonant Frequency (ωr​): ωr​=LC​1​orfr​=2πLC​1​
• At Resonance:
  • Impedance Z is minimum and equal to R.
  • Current is maximum: Imax​=Vrms​/R.
  • Phase angle ϕ=0∘ and Power Factor cosϕ=1 (maximum power).

5. Quality Factor (Q-factor)

It defines the sharpness of the resonance curve. A high Q-factor means the circuit is more selective.

Q=Rωr​L​=ωr​CR1​=R1​CL​​

6. Transformer

A device used to change (step-up or step-down) an alternating voltage. It works on the principle of Mutual Induction.

• Transformer Ratio (k): k=Np​Ns​​=Vp​Vs​​≈Is​Ip​​
  • For an Ideal Transformer (Input Power = Output Power): Vp​Ip​=Vs​Is​
• Step-Up Transformer: Ns​>Np​ (k>1). Increases voltage.
• Step-Down Transformer: Ns​<Np​ (k<1). Decreases voltage.
• Energy Losses: Flux leakage, Resistance (Copper Loss), Eddy Currents (Iron Loss), Hysteresis Loss.

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RBSE Class 12 Physics Chapter 7 Solutions & Key Numericals

Most numerical problems in this chapter revolve around calculating RMS/peak values, reactances (XL​,XC​), Impedance (Z), Resonance frequency, and Power Factor.

Example Problem 1: RMS and Peak Values

Question: The peak voltage of an AC supply is 300V. What is the RMS voltage? Solution:

Vrms​=2​vm​​

Vrms​=2​300V​≈1.414300​≈212.2V

Example Problem 2: Inductive Reactance and Current

Question: A 44mH inductor is connected to a 220V, 50Hz AC supply. Determine the RMS value of the current. Solution:

1. Given: L=44×10−3H, Vrms​=220V, f=50Hz.
2. Angular Frequency: ω=2πf=2×3.14×50=314rad/s.
3. Inductive Reactance: XL​=ωL=314×44×10−3≈13.82Ω.
4. RMS Current: Irms​=XL​Vrms​​=13.82220​≈15.92A.
5. Power Absorbed: Since it’s a pure inductor, the power factor cosϕ=cos(90∘)=0. Hence, Pavg​=Vrms​Irms​cosϕ=0.

Example Problem 3: Series LCR Circuit Resonance

Question: An LCR circuit has L=2.0H, C=32μF, and R=10Ω. Calculate the resonant angular frequency and the Q-factor. Solution:

1. Given: L=2.0H, C=32×10−6F, R=10Ω.
2. Resonant Angular Frequency (ωr​): ωr​=LC​1​=2.0×32×10−6​1​=64×10−6​1​ωr​=8×10−31​=125rad/s
3. Quality Factor (Q): Q=R1​CL​​=101​32×10−62.0​​=101​16×10−61​​Q=101​×4×10−31​=401000​=25

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RBSE Board Exam Strategy: Tips for Chapter 7

1. Prioritize LCR Circuit: Questions on LCR circuits, impedance, resonance, and Q-factor are the most frequently asked. Master the formulas for Z, tanϕ, ωr​, and Q.
2. Phasor Diagrams: Practice drawing phasor diagrams for R, L, C, and series LCR circuits. They help visualize the phase relationships (ϕ=0,±π/2).
3. Power Factor & Wattless Current: Understand the concept of the power factor (cosϕ) and why power dissipation is zero in pure L or pure C circuits (Wattless Current). This is a common theory question.
4. Transformer Theory: Prepare a detailed answer (Principle, Construction, Working, Losses, Uses) on the Transformer. The turns ratio formula and efficiency problems are also important.

Chapter 7

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By thoroughly understanding these concepts and practicing the numerical problems, you will be well-prepared to ace the Alternating Current chapter in your RBSE Class 12 Physics exam!