Alternating Current Class 12 Physics Notes 2026-27 | RMS, LCR & Resonance

Alternating Current Class 12 Physics Notes 2026–27

Chapter 7 — complete CBSE-focused notes on alternating current, peak and RMS values, reactance, impedance, series LCR circuits, resonance, AC power, power factor, wattless current, AC generator and transformer.

Class 12 PhysicsChapter 7CBSE 2026–27Concepts + NumericalsBoard Exam Revision

Quick chapter guide: These Alternating Current Class 12 Physics notes cover the high-intent CBSE Chapter 7 topics students commonly search for: AC waveform, peak and RMS value, average value, pure R/L/C circuits, inductive and capacitive reactance, impedance, phase difference, series LCR circuit, phasors, resonance, AC power, power factor, wattless current, AC generator and transformer. Build the chapter in this order: AC basics → RMS → R/L/C → reactance → series LCR → impedance/phase → resonance → power factor → generator → transformer.

Quick answers for common CBSE questions:

What is the RMS value of a sinusoidal AC? For peak current I₀, I_rms = I₀/√2; similarly V_rms = V₀/√2.

What is inductive reactance? X_L = ωL = 2πfL; it increases with frequency.

What is capacitive reactance? X_C = 1/(ωC) = 1/(2πfC); it decreases with frequency.

What is the impedance of a series LCR circuit? Z = √[R² + (X_L − X_C)²].

What is the condition for resonance? X_L = X_C, so f₀ = 1/(2π√LC), Z = R and current is maximum for a fixed source voltage.

What is power factor? cosφ; for a series LCR circuit, cosφ = R/Z.

2026–27 syllabus boundary: CBSE places Alternating Current in Chapter 7 of Unit IV: Electromagnetic Induction and Alternating Currents. The official syllabus includes alternating currents, peak and RMS values, reactance and impedance, the LCR series circuit (phasors only), resonance, power in AC circuits, power factor, wattless current, AC generator and transformer. Do not mix Chapter 6 electromagnetic-induction topics into this chapter, and do not add LC oscillations as Chapter 7 core content.

Official CBSE Physics Curriculum 2026–27 · Official Class XII 2026–27 SQP & Marking Scheme

Important syllabus note: Some online resources place AC generator at the end of the electromagnetic-induction discussion because of textbook chapter sequencing. For the current CBSE 2026–27 mapping, AC generator and transformer are explicitly listed under Chapter 7: Alternating Current. This page follows the official syllabus boundary.

Exam-focused scope: For CBSE 2026–27, prioritise the relationships and applications that connect RMS value, reactance, impedance, phase angle, resonance, power factor, AC generator and transformer. The LCR series-circuit treatment here is deliberately kept at the official phasors-only level. Topics found in broader online notes are not automatically treated as current CBSE requirements.

1. What You Will Learn in Chapter 7

Alternating current
Understand sinusoidally varying voltage/current, amplitude, time period, frequency and angular frequency.

Peak and RMS values
Connect peak values with effective values used in AC circuits and power calculations.

R, L and C in AC
Learn phase relationships and the opposition offered by resistance, inductance and capacitance.

Series LCR circuit
Use phasor relationships to obtain impedance, phase angle and current.

Resonance
Understand X_L = X_C, resonant frequency and maximum current in a series LCR circuit.

Power in AC
Understand average power, power factor and wattless current.

AC generator
Understand the principle, construction, working and induced EMF of an AC generator.

Transformer
Understand mutual induction, turns ratio, step-up/step-down operation, ideal power relation and losses.

2. The Big Idea: Why Alternating Current Is Different

An alternating current (AC) changes its magnitude and reverses its direction periodically. In the standard sinusoidal model:

i = I₀ sin(ωt + φ)

Similarly, an alternating voltage may be written as:

v = V₀ sin(ωt + φ_v)

Here I₀ and V₀ are peak values, ω is angular frequency, and the phase term describes timing relative to a chosen reference.

Time period
T is the time for one complete cycle.

Frequency
f = 1/T gives cycles per second in hertz.

Angular frequency
ω = 2πf = 2π/T.

Peak value
The maximum magnitude reached by the sinusoidal quantity.

Exam trap: Do not confuse instantaneous value, peak value, average value and RMS value. They answer different questions.

3. Average Value and RMS Value

Average value of a sinusoidal AC

Over a complete cycle, a pure sinusoidal current has equal positive and negative contributions, so its algebraic average is zero. For a sine wave, the commonly used average magnitude over a half cycle is:

I_{avg, half} = 2I₀/π ≈ 0.637 I₀

Similarly:

V_{avg, half} = 2V₀/π

RMS value

The root-mean-square (RMS) value is the effective value of AC: the value of steady DC that would produce the same heating effect in a resistor under the same conditions.

I_rms = I₀/√2     V_rms = V₀/√2

Quick example: If the peak current is 10 A, then I_rms = 10/√2 ≈ 7.07 A.

Common confusion: For a pure sine wave, average over a full cycle = 0, while RMS is not zero. RMS is based on the square of the instantaneous value and therefore measures effective heating capability.

4. AC Circuit Containing a Pure Resistor

For a pure resistance R connected to an AC source:

v = V₀ sinωt

i = I₀ sinωt,   I₀ = V₀/R

Voltage and current are in phase, so the phase difference is φ = 0.

Phase
V and I are in phase.

Impedance
Z = R.

Power factor
cosφ = 1.

Average power
P = V_rmsI_rms = I_rms²R.

5. AC Circuit Containing a Pure Inductor

For an ideal inductor of inductance L:

X_L = ωL

X_L is called inductive reactance. It measures the opposition offered by the inductor to AC and increases with frequency.

I₀ = V₀/X_L

In a pure inductive circuit, current lags voltage by 90° (π/2).

Energy viewpoint: An ideal inductor stores energy temporarily in its magnetic field and returns it to the circuit. Therefore, its average power consumption over a complete cycle is zero.

P_avg = 0    (pure ideal inductor)

Memory cue: In an ideal inductor, V leads I by 90°, or equivalently, I lags V by 90°.

6. AC Circuit Containing a Pure Capacitor

For a capacitor of capacitance C:

X_C = 1/(ωC)

X_C is the capacitive reactance. It decreases as frequency increases.

I₀ = V₀/X_C

In a pure capacitive circuit, current leads voltage by 90° (π/2).

Energy viewpoint: An ideal capacitor stores electrical energy and returns it to the source during another part of the cycle. Hence its average power consumption over a complete cycle is zero.

P_avg = 0    (pure ideal capacitor)

Memory cue: In a capacitor, I leads V by 90°.

7. Reactance — The AC Opposition of L and C

Frequency-change reasoning

Change	What happens	Reason
Frequency f increases	XL increases	XL = 2πfL
Frequency f increases	XC decreases	XC = 1/(2πfC)
Frequency f decreases	XL decreases	Inductive reactance is directly proportional to f.
Frequency f decreases	XC increases	Capacitive reactance is inversely proportional to f.

Fast exam rule: Higher frequency means an ideal inductor opposes AC more strongly, while an ideal capacitor opposes it less strongly.

Element	Reactance / opposition	Frequency dependence	Phase relation
Resistor R	R	Ideal R is frequency-independent	V and I in phase
Inductor L	XL = ωL	Increases with f	I lags V by 90°
Capacitor C	XC = 1/(ωC)	Decreases with f	I leads V by 90°

High-yield comparison: As frequency rises, an ideal inductor offers more reactance, while an ideal capacitor offers less reactance.

8. Series LCR Circuit — Phasor Method

A series LCR circuit contains resistance R, inductance L and capacitance C connected in series to an AC source. The same current flows through all three elements, but the voltage across each element has a different phase relationship with the current.

CBSE boundary: The 2026–27 syllabus specifies “LCR series circuit (phasors only)”. The purpose here is to understand the phasor relationship, resultant voltage, impedance and phase angle. Do not expand this into unnecessary advanced phasor mathematics.

Voltage relationships

V_R = IR    V_L = IX_L    V_C = IX_C

Taking current as the reference phasor, V_R is in phase with I, V_L leads I by 90°, and V_C lags I by 90°. Therefore the net reactive voltage is proportional to X_L − X_C.

V = √[V_R² + (V_L − V_C)²]

Impedance

Z = √[R² + (X_L − X_C)²]

Current

I_rms = V_rms/Z

Phase angle

tanφ = (X_L − X_C)/R

X_L > X_C
Circuit behaves net inductively; current lags the source voltage.

X_C > X_L
Circuit behaves net capacitively; current leads the source voltage.

X_L = X_C
Reactive parts cancel; the circuit is at series resonance.

At resonance
Z = R and the current is maximum for a fixed source RMS voltage.

9. Impedance Triangle and Power Factor

The impedance triangle provides a compact visual relationship between resistance, net reactance and impedance:

Z² = R² + (X_L − X_C)²

cosφ = R/Z

The power factor is the cosine of the phase angle between voltage and current. For a series LCR circuit:

cosφ = R/Z

Exam trap: Impedance Z is not simply R + X_L + X_C. Inductive and capacitive reactances have opposite phase effects and combine through the phasor relationship.

10. Resonance in a Series LCR Circuit

Series resonance occurs when the inductive and capacitive reactances are equal:

X_L = X_C

ω₀ = 1/√(LC)

f₀ = 1/(2π√(LC))

Impedance
Z = R, the minimum value for the series circuit.

Current
For a fixed source voltage, current is maximum: I_rms = V_rms/R.

Phase
φ = 0 and voltage/current are in phase.

Power factor
cosφ = 1.

Quick example: If L = 0.20 H and C = 50 μF, then ω₀ = 1/√(0.20 × 50×10⁻⁶) ≈ 316 rad s⁻¹. The corresponding f₀ is about 50.3 Hz.

Graph cue: X_L increases with frequency while X_C decreases. Their intersection corresponds to resonance.

11. Power in AC Circuits

For a general AC circuit with RMS voltage V_rms, RMS current I_rms and phase difference φ:

P_avg = V_rmsI_rmscosφ

For a series LCR circuit, since cosφ = R/Z:

P_avg = I_rms²R

Pure resistor
φ = 0 → cosφ = 1 → real power is non-zero.

Pure inductor
φ = 90° → cosφ = 0 → average power = 0.

Pure capacitor
|φ| = 90° → cosφ = 0 → average power = 0.

Series LCR at resonance
φ = 0 → power factor = 1.

12. Wattless Current

In a purely reactive circuit, current flows but the average power transferred over a complete cycle is zero. Such current is commonly called wattless current.

I_w = I_rms sinφ

Important: “Zero average power” does not mean “zero current.” Energy can be stored and returned during different parts of the AC cycle.

13. AC Generator — Principle

Flux and EMF check: For a coil of N turns, the flux through one turn can be written as Φ = BA cos(ωt) for the standard orientation, so the flux linkage is NΦ. Faraday's law then gives the sinusoidal generator EMF, e = E₀ sinωt, with E₀ = NBAω.

An AC generator converts mechanical energy into electrical energy using electromagnetic induction. A rotating coil changes the magnetic flux through it, producing an alternating induced EMF.

Chapter connection: Chapter 6 supplied the Faraday-law foundation. Chapter 7 applies that principle to a rotating coil to produce an alternating voltage.

Construction

Armature coil
A coil of N turns and area A rotates in a magnetic field.

Field magnet
Provides the magnetic field B through the rotating coil.

Slip rings
Each ring remains connected to the same end of the rotating coil, allowing the external output to alternate.

Brushes
Stationary contacts collect current from the rotating slip rings.

Working

As the coil rotates, the magnetic flux through it changes periodically. Faraday’s law therefore produces an induced EMF whose direction reverses periodically.

e = E₀ sinωt

E₀ = NBAω

Slip-ring vs split-ring: An AC generator uses slip rings so the external output reverses direction periodically. A split-ring commutator is associated with obtaining a unidirectional output in a DC generator.

14. Transformer — Principle and Construction

A transformer transfers AC electrical energy from one circuit to another through mutual induction. It can change voltage and current levels without changing frequency in the ideal transformer model.

Primary coil
Connected to the AC input source.

Secondary coil
Supplies transformed output to the load.

Soft magnetic core
Improves magnetic coupling between the coils.

Alternating flux
Changing core flux links the coils and induces EMF in the secondary.

Transformer turns ratio

V_s/V_p = N_s/N_p

For an ideal transformer:

V_pI_p = V_sI_s

I_s/I_p = N_p/N_s

Step-up transformer
N_s > N_p → V_s > V_p; ideal secondary current is lower.

Step-down transformer
N_s < N_p → V_s < V_p; ideal secondary current is higher.

Why transformers require AC

A transformer relies on changing magnetic flux. A steady DC supply cannot maintain the required changing flux after the initial transient.

Transformer losses

Loss	Cause	Reduction method
Copper loss	Resistance of windings	Use suitable low-resistance conductors.
Eddy-current loss	Induced currents in the core	Use a laminated core.
Hysteresis loss	Repeated magnetisation	Use suitable magnetic materials.
Flux leakage	Not all primary flux links secondary	Improve magnetic coupling and core design.

η = (P_out/P_in) × 100%

Ideal vs real: V_pI_p = V_sI_s is an ideal-transformer relation. A real transformer has losses.

15. AC Generator vs Transformer

Feature	AC Generator	Transformer
Main function	Converts mechanical energy into electrical energy.	Transfers AC electrical energy between circuits while changing voltage/current levels.
Main principle	Electromagnetic induction through rotation.	Mutual induction through changing magnetic flux.
Input	Mechanical energy + magnetic field	AC electrical input
Key component	Rotating coil + slip rings	Primary/secondary coils + magnetic core

16. High-Yield Formula and Relationship Map

Concept	Relationship	Use / condition
AC current	i = I0sin(ωt + φ)	Sinusoidal AC
Angular frequency	ω = 2πf = 2π/T	AC waveform
RMS current	Irms = I0/√2	Sinusoidal AC
RMS voltage	Vrms = V0/√2	Sinusoidal AC
Inductive reactance	XL = ωL	Ideal inductor
Capacitive reactance	XC = 1/(ωC)	Ideal capacitor
LCR impedance	Z = √[R² + (XL − XC)²]	Series LCR
Phase angle	tanφ = (XL − XC)/R	Series LCR
Resonant angular frequency	ω0 = 1/√(LC)	Series resonance
Resonant frequency	f0 = 1/(2π√LC)	Series resonance
AC power	Pavg = VrmsIrmscosφ	AC circuit
Power factor	cosφ = R/Z	Series LCR
Wattless current	Iw = Irmssinφ	Reactive component
Generator EMF	e = E0sinωt	Ideal AC generator
Generator peak EMF	E0 = NBAω	N-turn rotating coil
Transformer turns ratio	Vs/Vp = Ns/Np	Ideal transformer
Ideal transformer power	VpIp = VsIs	Neglect losses

17. Formula Selection: What to Use When

If the question gives...	Start with...	Then check...
Peak value of a sine wave	Irms = I0/√2 or Vrms = V0/√2	Whether the question asks RMS or peak.
Frequency and L	XL = 2πfL	Units of f and L.
Frequency and C	XC = 1/(2πfC)	Convert μF/nF to farads.
R, XL and XC	Z = √[R² + (XL − XC)²]	Whether the circuit is net inductive or capacitive.
Series LCR resonance	XL = XC, f0 = 1/(2π√LC)	At resonance, Z = R and power factor = 1.
AC voltage, current and phase	Pavg = VrmsIrmscosφ	Use RMS values in the power formula.
Transformer turns	Vs/Vp = Ns/Np	Step-up vs step-down direction.

18. High-Yield Problem-Solving Method

Step 1: Identify waveform, pure R/L/C, LCR, resonance, power, generator or transformer.
Step 2: Convert peak ↔ RMS only for a sinusoidal waveform.
Step 3: For L/C, calculate reactance and phase relation.
Step 4: For series LCR, find X_L, X_C, then Z and φ.
Step 5: Check resonance early: X_L = X_C.
Step 6: For power, identify cosφ before substituting.
Step 7: For generator/transformer, write the principle and relevant relation first.
Step 8: Check units and whether the question asks for RMS, peak, average, phase or direction.

19. Worked Example — RMS Value

Question: A sinusoidal voltage has a peak value of 311 V. Find its RMS value.
Solution: V_rms = V₀/√2 = 311/√2 ≈ 220 V.

20. Worked Example — Series LCR Impedance

Question: A series LCR circuit has R = 6 Ω, X_L = 10 Ω and X_C = 2 Ω. Find its impedance.
Solution: Net reactance = 8 Ω. Z = √(6² + 8²) = 10 Ω. Therefore cosφ = R/Z = 0.6, and the circuit is net inductive.

21. Worked Example — Transformer

Question: An ideal transformer has 1000 primary turns and 200 secondary turns. If the primary voltage is 220 V, find the secondary voltage.
Solution: V_s/V_p = N_s/N_p = 0.2. Therefore V_s = 44 V. It is a step-down transformer.

22. Common Exam Traps

Trap 1: RMS is not peak; for a sine wave divide peak by √2.

Trap 2: Average of a pure sine wave over a complete cycle is zero; do not confuse it with RMS.

Trap 3: In pure L, current lags voltage by 90°; in pure C, current leads voltage by 90°.

Trap 4: X_L increases with frequency; X_C decreases.

Trap 5: Series LCR impedance is not R + X_L + X_C.

Trap 6: At resonance, X_L = X_C, Z = R, current is maximum and power factor is unity.

Trap 7: Zero average power in an ideal reactive circuit does not mean zero current.

Trap 8: Transformer turns ratio and current ratio are inverse in the ideal model.

Trap 9: A transformer requires changing flux; steady DC cannot sustain normal transformer action.

Trap 10: Generator converts mechanical energy to electrical energy; transformer transfers AC electrical energy between circuits.

23. Chapter 7 at a Glance

Topic	What to learn	Exam skill
Alternating current	Sinusoidal waveform, T, f and ω	Concept + formula
Peak/RMS values	Irms = I0/√2 and Vrms = V0/√2	Numerical
Pure R	V and I in phase	Phase + power
Pure L	XL = ωL; current lags	Reactance + phase
Pure C	XC = 1/(ωC); current leads	Reactance + phase
Series LCR	Phasor relationship, Z and φ	Numerical + reasoning
Resonance	XL = XC; ω0 = 1/√LC	Graph + numerical
AC power	P = VrmsIrmscosφ	Application
Power factor	cosφ = R/Z	Reasoning + numerical
Wattless current	Reactive component and zero average power	Conceptual
AC generator	Principle, construction, working and e = E0sinωt	Diagram + explanation
Transformer	Principle, ratio, ideal relation and losses	Numerical + long answer

24. Frequently Asked Questions

What is Alternating Current in Class 12 Physics?

Alternating current is current whose magnitude and direction vary periodically. A standard sinusoidal representation is i = I₀sin(ωt + φ).

What is the RMS value of AC?

For a sinusoidal current, I_rms = I₀/√2. It represents the equivalent steady-current value for the same heating effect in a resistor.

What is inductive reactance?

Inductive reactance is X_L = ωL and increases with frequency.

What is capacitive reactance?

Capacitive reactance is X_C = 1/(ωC) and decreases as frequency increases.

What is impedance in a series LCR circuit?

Z = √[R² + (X_L − X_C)²].

What is resonance in an LCR circuit?

Series resonance occurs when X_L = X_C, giving ω₀ = 1/√(LC), minimum impedance Z = R and maximum current for a fixed source voltage.

What is power factor?

Power factor is cosφ, where φ is the phase difference between voltage and current. For a series LCR circuit, cosφ = R/Z.

What is wattless current?

It is the reactive component of AC current associated with zero average power transfer over a complete cycle in an ideal purely reactive circuit.

What is an AC generator?

An AC generator converts mechanical energy into electrical energy using electromagnetic induction and produces an alternating EMF in the ideal sinusoidal model.

What is a transformer?

A transformer transfers AC electrical energy between circuits by mutual induction and can step voltage up or down according to the turns ratio.

Why does a transformer not work normally on DC?

Normal transformer action requires continuously changing magnetic flux. A steady DC supply cannot maintain the required changing flux after the initial transient.

What are the important topics in Alternating Current for CBSE 2026–27?

Focus on peak and RMS values, reactance, impedance, series LCR phasors, resonance, AC power, power factor, wattless current, AC generator and transformer, within the current CBSE Chapter 7 scope.

How many marks are assigned to Chapter 7?

CBSE assigns 18 marks to Unit IV: Electromagnetic Induction and Alternating Currents, covering Chapters 6 and 7 together. The curriculum does not prescribe a separate fixed mark allocation for Chapter 7 alone.

25. How to Study Alternating Current

1. First: master sinusoidal AC, peak value, time period, frequency and angular frequency.
2. Second: learn RMS value and the distinction between full-cycle average and half-cycle average.
3. Third: study pure R, L and C separately and memorise the phase relationships.
4. Fourth: combine them in the series LCR circuit using the phasor relationships.
5. Fifth: master resonance and its consequences for impedance, current and power factor.
6. Sixth: practise AC power, power factor and wattless-current questions.
7. Seventh: learn the construction, working and equation of the AC generator.
8. Finally: learn transformer ratios, ideal power relation, losses and efficiency, then practise the Chapter 7 MCQs, numericals, PYQs and chapter test.

26. Chapter 6 → Chapter 7 Connection

Chapter 6 established electromagnetic induction: changing magnetic flux produces induced EMF. Chapter 7 uses that foundation in two major applications.

AC generator
Mechanical rotation changes flux through a coil and produces an alternating EMF.

Transformer
Changing current in the primary produces changing magnetic flux and induces EMF in the secondary.

Chapter 6 — Electromagnetic Induction Notes

27. Continue the Class 12 Physics Sequence

Chapter 7: This page is the complete notes hub for Alternating Current.

Use the full Chapter 7 practice chain:

Important QuestionsBoard-style questions with answers and higher-order practice. MCQs with AnswersConcept, numerical and competency-focused MCQ practice. Solved NumericalsRMS, reactance, LCR, resonance, power and transformer problems. Case-Based QuestionsApplication and competency-style Chapter 7 practice. Assertion–ReasonStatement-based conceptual reasoning practice. PYQsPrevious-year question practice for recurring exam patterns. Formula SheetFast revision of formulas, relationships and exam traps. Chapter Test50-mark original test for final Chapter 7 practice. Chapter 6 NotesRevise electromagnetic induction before generator and transformer applications.

28. Final Exam-Readiness Check

✓ I can distinguish peak, average and RMS values.

✓ I know the phase relation for pure R, L and C circuits.

✓ I can calculate X_L and X_C.

✓ I can calculate impedance and phase angle in a series LCR circuit.

✓ I can identify resonance and resonant frequency.

✓ I can calculate average AC power and power factor.

✓ I understand wattless current.

✓ I can explain the principle and working of an AC generator.

✓ I can use transformer turns ratio and ideal power relation.

✓ I can distinguish ideal-transformer relations from real-transformer losses.

Continue your preparation: Use the already-published Chapter 6 resources to revise the electromagnetic-induction foundation needed for the AC generator and transformer sections.

Chapter 6 Notes
Electromagnetic Induction Class 12 Physics Notes

Chapter 6 Important Questions
Electromagnetic Induction Important Questions

Chapter 6 MCQs
Electromagnetic Induction MCQs with Answers

Chapter 6 Numericals
Electromagnetic Induction Numericals

Chapter 6 Case-Based Questions
Electromagnetic Induction Case-Based Questions

Chapter 6 Assertion–Reason
Electromagnetic Induction Assertion and Reason

Chapter 6 PYQs
Electromagnetic Induction PYQs

Chapter 6 Formula Sheet
Electromagnetic Induction Formula Sheet

Chapter 6 Chapter Test
Electromagnetic Induction Chapter Test

Source discipline: The official CBSE 2026–27 curriculum is the authority for Chapter 7 scope. Competitor resources were used only to identify search intent and common student needs. The content above follows the official syllabus boundary, including the “phasors only” limitation for the LCR series circuit.

CBSE Physics Curriculum 2026–27 · CBSE Class XII 2026–27 SQP & Marking Schemes