Class 12 Alternating Current Guide 2027 — RLC Circuits, Resonance, Phasor Diagrams, Transformers

In short: Alternating Current (Chapter 7) carries 5-6 marks in CBSE Class 12 Physics 2027 and is the most formula-intensive chapter in the Electromagnetic Induction and AC unit. The key topics are AC through R, L, C, and LCR circuits with their phasor diagrams, the resonance condition (ω₀ = 1/√LC), power factor (cos φ), and transformers (step-up, step-down, efficiency). Board exams consistently test phasor diagrams, impedance calculations, resonance numericals, and transformer-based problems.

Alternating Current is Chapter 7 in the NCERT textbook and represents the practical application of electromagnetic induction. It carries 5-6 marks on the board paper, typically split between one 3-mark question on LCR circuits, phasor diagrams, or resonance, and one 2-3 mark question on transformers or AC values (rms, peak). This chapter requires comfort with formulas and diagrams — but once you understand the phasor approach, every circuit type follows the same logic.

For the complete marks distribution, see the Class 12 Physics chapter-wise weightage. For all derivations, see the Physics derivations list.

AC vs DC — Fundamentals

Direct Current (DC): Current flows in one direction only. Magnitude may be constant or varying. Example: battery.

Alternating Current (AC): Current periodically reverses direction. Follows a sinusoidal waveform.

AC voltage: V = V₀ sin ωt AC current: I = I₀ sin(ωt + φ)

Where:

• V₀, I₀ = peak (maximum) values
• ω = angular frequency = 2πf = 2π/T
• φ = phase difference between voltage and current
• f = frequency (50 Hz in India, 60 Hz in USA)
• T = time period = 1/f = 1/50 = 0.02 s

Important AC Values

RMS (Root Mean Square) is the most important value because:

• Household voltage (220 V in India) is the RMS value
• V₀ (peak) = 220√2 = 311 V
• Power formulas use RMS values: P = V_rms × I_rms × cos φ

Memory trick — "RMS = peak divided by √2. For Indian supply: 220 V rms = 311 V peak. Your appliances see the rms value."

AC Through Pure Resistor (R)

Circuit: AC source connected to a pure resistor R.

Voltage: V = V₀ sin ωt Current: I = V₀ sin ωt / R = I₀ sin ωt

Key points:

• Voltage and current are in phase (φ = 0)
• No phase difference — both reach maximum and zero at the same time
• Phasor diagram: V and I vectors point in the same direction

AC Through Pure Inductor (L)

Circuit: AC source connected to a pure inductor L.

Voltage: V = V₀ sin ωt Current: I = I₀ sin(ωt - π/2) = -I₀ cos ωt

Key points:

• Current lags voltage by π/2 (90°) — "ELI" (E leads I in L)
• Inductive reactance: X_L = ωL = 2πfL (in ohms)
• I₀ = V₀/X_L
• X_L increases with frequency (inductor opposes high-frequency AC more)
• For DC (f = 0): X_L = 0 (inductor is a short circuit for DC in steady state)

Memory trick — "ELI the ICE man. In L (inductor), E leads I. In C (capacitor), I leads E."

AC Through Pure Capacitor (C)

Circuit: AC source connected to a pure capacitor C.

Voltage: V = V₀ sin ωt Current: I = I₀ sin(ωt + π/2) = I₀ cos ωt

Key points:

• Current leads voltage by π/2 (90°) — "ICE" (I leads E in C)
• Capacitive reactance: X_C = 1/(ωC) = 1/(2πfC) (in ohms)
• I₀ = V₀/X_C
• X_C decreases with frequency (capacitor allows high-frequency AC more easily)
• For DC (f = 0): X_C = ∞ (capacitor blocks DC completely)

AC Through Series LCR Circuit — The Core Topic

This is the most important circuit in the chapter and the most tested topic on the board exam.

Setup

A resistor R, inductor L, and capacitor C are connected in series with an AC source.

Impedance

Z = √[R² + (X_L - X_C)²]

Where:

• Z = impedance (total opposition to AC, measured in ohms)
• R = resistance
• X_L = ωL (inductive reactance)
• X_C = 1/(ωC) (capacitive reactance)

Phase Angle

tan φ = (X_L - X_C)/R

Where φ is the angle by which voltage leads (or lags) the current.

• If X_L > X_C → φ is positive → voltage leads current (inductive behaviour)
• If X_C > X_L → φ is negative → current leads voltage (capacitive behaviour)
• If X_L = X_C → φ = 0 → voltage and current in phase (resonance)

Phasor Diagram for Series LCR

The phasor diagram is the single most important diagram in this chapter.

1. Take current I as the reference phasor (along x-axis, since current is common in series)
2. V_R is in phase with I (along I)
3. V_L leads I by 90° (drawn upward)
4. V_C lags I by 90° (drawn downward)
5. Net voltage V = vector sum of V_R and (V_L - V_C)

V = √[V_R² + (V_L - V_C)²] = IZ

Drawing tip: Draw I along the positive x-axis. V_R along I. V_L straight up. V_C straight down. The resultant V makes angle φ with I.

Current in LCR Circuit

I₀ = V₀/Z = V₀/√[R² + (X_L - X_C)²]

Current is maximum when Z is minimum — this happens at resonance.

Resonance — When LCR Circuit Peaks

Resonance Condition

At resonance, the inductive reactance equals the capacitive reactance:

X_L = X_C ωL = 1/(ωC) ω₀² = 1/(LC)

Resonant angular frequency: ω₀ = 1/√(LC) Resonant frequency: f₀ = 1/(2π√(LC))

Properties at Resonance

• Z = R (minimum impedance — only resistance remains)
• Current is maximum: I₀ = V₀/R
• Phase angle φ = 0 (voltage and current in phase)
• The circuit behaves as a pure resistor
• V_L = V_C (they cancel each other out, but each can be much larger than V_source)

Quality Factor (Q-factor)

Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C)

The Q-factor measures the sharpness of the resonance peak:

• Higher Q = sharper peak = more selective (used in radio tuning)
• Lower Q = broader peak = less selective

At resonance: V_L = V_C = QV_source (voltage across L or C can be Q times the source voltage)

Memory trick — "Resonance = React-CANCEL. X_L and X_C cancel each other, leaving only R. Current peaks, impedance dips."

Power in AC Circuits

Average Power

P = V_rms × I_rms × cos φ

Where cos φ is the power factor.

Power factor: cos φ = R/Z

Special Cases

Memory trick — "Pure L and C consume zero average power — they store and return energy every half cycle. Only R dissipates energy as heat."

Wattless Current

The component of current that does NOT contribute to power dissipation:

I_wattless = I sin φ

The power-contributing component: I_active = I cos φ

Transformers

A transformer converts AC voltage from one level to another using electromagnetic induction.

Principle

Two coils (primary and secondary) wound on a common soft iron core. Changing current in the primary creates a changing magnetic flux that induces an EMF in the secondary.

Transformation Ratio

V_s/V_p = N_s/N_p = k (turns ratio)

Where:

• V_p, V_s = primary and secondary voltages
• N_p, N_s = number of turns in primary and secondary
• k = transformation ratio

Types

Step-up transformer: N_s > N_p → V_s > V_p (voltage increases, current decreases) Step-down transformer: N_s < N_p → V_s < V_p (voltage decreases, current increases)

For an Ideal Transformer

Power in = Power out: V_p I_p = V_s I_s

Therefore: I_s/I_p = N_p/N_s = V_p/V_s

When voltage steps up, current steps down (and vice versa) — power is conserved.

Energy Losses in Real Transformers

Efficiency: η = (Output power/Input power) × 100 = (V_s I_s)/(V_p I_p) × 100

Large power transformers can achieve efficiency > 95%.

Why AC is used for long-distance power transmission: Power loss in transmission = I²R. By stepping up voltage (and stepping down current), I²R losses are dramatically reduced. At the receiving end, the voltage is stepped back down for safe domestic use.

Board Exam Question Types

Type 1: LCR Circuit and Phasor Diagram (3-5 marks)

"Derive an expression for the impedance of a series LCR circuit. Draw the phasor diagram."

Approach: Draw the phasor diagram first (I as reference, V_R along I, V_L up, V_C down). Show that V = √[V_R² + (V_L - V_C)²]. Divide by I to get Z = √[R² + (X_L - X_C)²]. Define tan φ = (X_L - X_C)/R. Label everything clearly.

Type 2: Resonance Numerical (3 marks)

"An LCR circuit has L = 5 H, C = 80 μF, R = 40 Ω. Find the resonant frequency and current at resonance if V = 230 V."

Approach: f₀ = 1/(2π√LC). Convert C to Farads. At resonance, Z = R, so I = V/R = 230/40 = 5.75 A.

Type 3: Transformer Problem (2-3 marks)

"A transformer has 500 turns in the primary and 10,000 turns in the secondary. If the primary voltage is 220 V, find the secondary voltage. If the primary current is 5 A, find the secondary current."

Approach: V_s = V_p × (N_s/N_p) = 220 × (10000/500) = 4400 V. For ideal transformer: I_s = I_p × (N_p/N_s) = 5 × (500/10000) = 0.25 A.

Type 4: AC Values and Power (2 marks)

"The rms voltage of an AC supply is 220 V. Find the peak voltage and the average power consumed by a 100 Ω resistor."

Approach: V₀ = V_rms × √2 = 220 × 1.414 = 311 V. P = V²_rms/R = 220²/100 = 484 W.

Type 5: Comparison Questions (2 marks)

"Distinguish between step-up and step-down transformers." or "Compare X_L and X_C variation with frequency."

Approach: Make a clear comparison table with at least 3-4 points. For reactance vs frequency: X_L = ωL (increases linearly with f), X_C = 1/(ωC) (decreases with f). They cross at resonance frequency.

Common Mistakes That Cost Marks

1. Confusing phase relationships in L and C. In an inductor, voltage LEADS current by 90° (ELI). In a capacitor, current LEADS voltage by 90° (ICE). Students regularly reverse these. The mnemonic "ELI the ICE man" is essential.

2. Using peak values instead of RMS in power formulas. Power formulas use RMS values: P = V_rms I_rms cos φ. If you use peak values, your answer will be off by a factor of 2. Remember: household 220 V is RMS.

3. Drawing wrong phasor diagrams. The most common error is drawing V_L and V_C in the wrong directions. V_L is ALWAYS 90° ahead of I (upward in the standard diagram), and V_C is ALWAYS 90° behind I (downward). The resultant voltage V makes angle φ with V_R (not with I directly).

4. Forgetting that impedance at resonance equals R, not zero. At resonance, X_L and X_C cancel, but R remains. Z = R (not Z = 0). Current is maximum because Z is minimum, but the circuit still has resistance.

5. Confusing turns ratio direction in transformers. V_s/V_p = N_s/N_p but I_s/I_p = N_p/N_s. Voltage and turns ratio go in the same direction. Current and turns ratio go in the OPPOSITE direction. Students often apply the voltage ratio to current, getting the wrong answer.

6. Writing that transformers work with DC. Transformers work ONLY with AC. DC creates a constant magnetic flux, so no EMF is induced in the secondary (Faraday's law requires CHANGING flux). If DC is applied, only a momentary EMF appears when the circuit is switched on or off.

4-Week Practice Plan

Week 1: AC Fundamentals and Pure R, L, C Circuits

• Read NCERT Chapter 7, Sections 7.1-7.4
• Derive relationships between peak, rms, and average values
• Study AC through pure R, L, C separately with phasor diagrams
• Memorise X_L = ωL and X_C = 1/(ωC) with their frequency dependence
• Reference: NCERT Exercise Q1-Q5

Week 2: Series LCR Circuit and Resonance

• Derive impedance formula for series LCR circuit
• Draw the phasor diagram from memory 5 times
• Derive resonance condition (ω₀ = 1/√LC) and properties at resonance
• Solve 10 numerical problems on impedance and resonance frequency
• Study Q-factor and its significance
• Reference: NCERT Exercise Q6-Q10

Week 3: Power, Transformers, and Applications

• Study average power formula P = V_rms I_rms cos φ
• Understand wattless current and power factor
• Learn transformer principle, types, turns ratio, and energy losses
• Solve transformer-based numerical problems
• Study why AC is preferred for long-distance power transmission
• Reference: NCERT Exercise Q11-Q16

Week 4: Board Paper Practice

• Solve AC questions from the last 5 years of CBSE board papers
• Time yourself: LCR phasor diagram + derivation in 8-10 minutes
• Focus on numerical problems involving resonance and transformers
• Final drill: draw all phasor diagrams from memory and write all formulas

Quick Revision Card for Exam Day

• V = V₀ sin ωt | V_rms = V₀/√2 | Indian supply: 220 V rms = 311 V peak
• X_L = ωL (increases with f) | X_C = 1/(ωC) (decreases with f)
• Phase: ELI the ICE man — L: V leads I by 90°, C: I leads V by 90°
• LCR impedance: Z = √[R² + (X_L - X_C)²] | tan φ = (X_L - X_C)/R
• Resonance: ω₀ = 1/√(LC), Z = R (minimum), I = V/R (maximum), φ = 0
• Q-factor: ω₀L/R = (1/R)√(L/C) — sharpness of resonance
• Power: P = V_rms I_rms cos φ | cos φ = R/Z (power factor)
• Pure L or C: cos φ = 0 → zero average power (wattless)
• Transformer: V_s/V_p = N_s/N_p | I_s/I_p = N_p/N_s | Works only with AC
• Losses: Copper (I²R), Iron (eddy), Hysteresis, Flux leakage

For guided practice on AC circuits, phasor diagrams, and transformer problems, try Padhaao's AI tutor.

Frequently Asked Questions

How many marks does Alternating Current carry in CBSE Class 12 Physics 2027?

Alternating Current carries 5-6 marks in CBSE Class 12 Physics 2027. The series LCR circuit with its phasor diagram is the most frequently tested topic, usually as a 3-5 mark derivation or numerical question. Transformer problems and resonance frequency calculations appear as 2-3 mark questions. The chapter is formula-heavy but predictable — the same types of questions repeat across years.

What is the resonance condition in a series LCR circuit?

Resonance occurs when the inductive reactance equals the capacitive reactance: X_L = X_C, which gives ωL = 1/(ωC). Solving for the resonant angular frequency: ω₀ = 1/√(LC), and the resonant frequency: f₀ = 1/(2π√(LC)). At resonance, impedance is minimum (Z = R), current is maximum (I = V/R), and the phase angle is zero (voltage and current are in phase).

Why does a transformer not work with DC?

A transformer relies on electromagnetic induction, which requires a CHANGING magnetic flux through the coils. AC produces a continuously changing current, which creates a continuously changing magnetic flux in the core, inducing an EMF in the secondary coil. DC produces a constant current and constant magnetic flux — since the flux is not changing, no EMF is induced in the secondary (by Faraday's law, EMF = -dΦ/dt, and dΦ/dt = 0 for constant flux). A brief transient EMF appears only when DC is switched on or off.

What is the difference between inductive reactance and capacitive reactance?

Inductive reactance (X_L = ωL = 2πfL) is the opposition to AC by an inductor. It increases linearly with frequency — inductors oppose high-frequency AC more strongly. Capacitive reactance (X_C = 1/(ωC) = 1/(2πfC)) is the opposition by a capacitor. It decreases with frequency — capacitors allow high-frequency AC to pass more easily. At the resonant frequency, X_L = X_C, and they cancel each other in a series LCR circuit.

Why is the power consumed in a pure inductor or capacitor zero?

In a pure inductor, current lags voltage by exactly 90°, so the power factor cos φ = cos 90° = 0. The average power P = V_rms I_rms cos φ = 0. Physically, the inductor stores energy in its magnetic field during one quarter-cycle and returns it to the circuit during the next quarter-cycle. The same applies to a pure capacitor (current leads voltage by 90°). Energy oscillates between the source and the reactive element but is not dissipated. This current is called wattless current.

How does power transmission at high voltage reduce losses?

Power transmitted = V × I. For a given power P, if voltage is increased by a factor of n (using a step-up transformer), current decreases by a factor of n (I = P/V). Power loss in transmission lines = I²R. Since I decreases by factor n, the power loss decreases by n². For example, stepping up from 220 V to 22,000 V (100 times) reduces transmission losses by 100² = 10,000 times. At the receiving end, a step-down transformer reduces the voltage back to safe levels for domestic use.

Alternating Current is where physics meets electrical engineering. The phasor diagram is your Swiss army knife — it solves every AC circuit by converting a time-dependent problem into a simple vector addition. Master the LCR phasor diagram, understand resonance as the point where reactive effects cancel, and learn how transformers enable efficient power distribution — those 5-6 marks reward both conceptual clarity and formula fluency.