Alternating Current Class 12 Notes | CBSE Physics Chapter 7 (Free PDF)

Chapter summary

Alternating Current covers how current and voltage that reverse direction periodically behave in circuits, starting from peak, rms and average values and how resistors, inductors and capacitors each oppose AC through resistance and reactance. It then builds to the series LCR circuit, impedance, resonance and quality factor, AC power and power factor, and applications like transformers and LC oscillations. It is a high-yield, mostly numerical chapter in the NEET Physics syllabus, so mastering rms, reactance, impedance and resonance formulas reliably earns marks.

Chapter notes

Key Concepts

1. Alternating Current Basics

An alternating current changes direction periodically. It is described as:

I = I₀ sin ωt and V = V₀ sin ωt

where I₀, V₀ are peak values and ω = 2πf (angular frequency).

RMS (Root Mean Square) Values

The effective or DC-equivalent values used for AC calculations:

Irms = I₀/√2 ≈ 0.707 I₀

Vrms = V₀/√2 ≈ 0.707 V₀

Household AC: Vrms = 220 V → V₀ = 220√2 ≈ 311 V


2. AC Through Pure Components

ComponentImpedancePhase Relation
Resistor (R)RV and I are in phase
Inductor (L)XL = ωL = 2πfLV leads I by 90° (π/2)
Capacitor (C)XC = 1/(ωC) = 1/(2πfC)I leads V by 90° (π/2)

XL = inductive reactance; XC = capacitive reactance (both in Ω)


3. Series LCR Circuit

Impedance: Z = √[R² + (XL − XC)²]

Current: I = V/Z

Phase angle: tan φ = (XL − XC)/R

  • If XL > XC: circuit is inductive (V leads I)
  • If XC > XL: circuit is capacitive (I leads V)
  • If XL = XC: resonance (purely resistive)

Resonance

At resonance: XL = XC → ωL = 1/(ωC)

Resonant frequency: f₀ = 1/(2π√LC)

At resonance: Z = R (minimum impedance), I = V/R (maximum current)

Quality Factor (Q)

Q = ωL/R = 1/(ωCR) = (1/R)√(L/C)

Higher Q → sharper resonance peak → more selective tuning


4. Power in AC Circuits

P = VrmsIrms cos φ

where cos φ is the power factor.

CircuitPower FactorPower
Pure Rcos φ = 1P = VI (maximum)
Pure L or Ccos φ = 0P = 0 (wattless current)
LCR at resonancecos φ = 1P = V²/R (maximum)

5. Transformer

A device that changes AC voltage from one level to another using mutual induction.

Vs/Vp = Ns/Np = Ip/Is (for ideal transformer)

  • Step-up: Ns > Np → voltage increases, current decreases
  • Step-down: Ns < Np → voltage decreases, current increases
  • Efficiency: η = (output power)/(input power) × 100%. Ideal transformer: η = 100%

Energy Losses in Transformer

  • Copper loss: I²R heating in coils → use thick copper wires
  • Iron/core loss: Eddy currents in core → use laminated core
  • Hysteresis loss: Energy to magnetise/demagnetise core each cycle → use soft iron
  • Flux leakage: Not all flux links both coils → use close winding

Solved Examples

Example 1

A series LCR circuit has R = 100 Ω, L = 0.5 H, C = 10 μF connected to 200 V, 50 Hz. Find impedance and current.

Answer: XL = 2π × 50 × 0.5 = 157 Ω. XC = 1/(2π × 50 × 10⁻⁵) = 318 Ω.

Z = √[100² + (157 − 318)²] = √[10000 + 25921] = √35921 = 189.5 Ω

I = V/Z = 200/189.5 = 1.06 A

Example 2

Find the resonant frequency for L = 0.1 H and C = 100 μF.

Answer: f₀ = 1/(2π√LC) = 1/(2π√(0.1 × 10⁻⁴)) = 1/(2π × 3.16 × 10⁻³) = 50.3 Hz

Example 3

A transformer with 500 primary turns and 50 secondary turns is connected to 220 V AC. Find the output voltage.

Answer: Vs = Vp × Ns/Np = 220 × 50/500 = 22 V (step-down)


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Important Questions for Board Exams

1-Mark

  1. What is the power factor of a pure inductor?
  2. What is resonance in an LCR circuit?

3-Mark

  1. Derive the expression for impedance of a series LCR circuit.
  2. What is a transformer? Explain its working with a diagram.
  3. What is resonance? Derive the condition and resonant frequency for a series LCR circuit.

5-Mark

  1. Explain the working of a series LCR circuit. Derive expressions for impedance and resonant frequency. What is the Q factor?

Quick Revision Points

  • AC: I = I₀ sin ωt; Irms = I₀/√2; Vrms = V₀/√2
  • XL = ωL (V leads I by 90°); XC = 1/(ωC) (I leads V by 90°)
  • LCR: Z = √[R² + (XL − XC)²]; tan φ = (XL − XC)/R
  • Resonance: XL = XC; f₀ = 1/(2π√LC); Z = R (min); I = max
  • Power: P = VIcos φ; pure L or C → P = 0 (wattless)
  • Transformer: Vs/Vp = Ns/Np; losses: copper, iron, hysteresis, flux leakage

Previous: Ch 6 - Electromagnetic Induction
Next: Ch 8 - Electromagnetic Waves

🃏 Flash Cards: Alternating Current

Class 12 Physics · Chapter 7 – swipe through all 9 cards to understand the whole chapter.

🌊Start here1/9

AC, Peak & RMS Values

Alternating current reverses direction periodically, so we describe it by its peak and rms values.

i = i0 sin(ωt), ω = 2πf, I_rms = i0/√2 ≈ 0.707 i0

AC meters read rms. Indian mains: V_rms = 220 V, V0 ≈ 311 V, f = 50 Hz.

  • Average over a full cycle = 0; rms is the DC that gives the same I2R heating
  • Half-cycle mean = 2i0/π ≈ 0.637 i0 (don’t mix with rms)
  • Chain: peak → divide by √2 → rms; start power problems from rms
Core element2/9

Resistor in AC

A resistor obeys Ohm’s law instant by instant, so it never falls out of step.

V and I in phase (φ = 0), opposition = R

R is the same at every frequency; the only element that dissipates power.

  • Voltage and current peak at the same instant
  • Opposition stays R for all frequencies
  • Power is dissipated only here, never in pure L or C
🧲Core element3/9

Inductor & Capacitor (Reactance)

L and C react to the change in AC, so their opposition depends on frequency.

X_L = ωL = 2πf L, X_C = 1/(ωC) = 1/(2πf C)

CIVIL: in C, I leads V; in L, V leads I. Both lag/lead by 90°.

  • Inductor: current lags V by 90°; X_L rises with f (X_L = 0 at DC)
  • Capacitor: current leads V by 90°; X_C falls with f (blocks DC, X_C → ∞)
  • Pure L and pure C consume zero average power
📐Key tool4/9

Series LCR & Impedance

Same current flows through R, L, C but their voltages differ in phase, so add them as phasors.

Z = √(R2 + (X_L − X_C)2), V_rms = I_rms Z

V_L and V_C are 180° apart, so their phasors subtract — never add rms voltages arithmetically.

  • V = √(V_R2 + (V_L − V_C)2); combine as vectors, not numbers
  • Impedance Z is the AC version of resistance (in ohms)
  • V_L or V_C can exceed the source voltage — that is normal
🧭Phase angle5/9

Phase in a Series LCR Circuit

The phase angle φ tells you whether the circuit behaves inductively or capacitively.

tan φ = (X_L − X_C)/R

Keep the sign of (X_L − X_C) to judge inductive vs capacitive.

  • X_L > X_C → inductive, voltage leads current
  • X_C > X_L → capacitive, current leads voltage
  • X_L = X_C → purely resistive (φ = 0), this is resonance
🎯Resonance6/9

Resonance in Series LCR

When X_L = X_C the reactances cancel and the circuit draws maximum current.

ω0 = 1/√(LC), f0 = 1/(2π√(LC))

At resonance Z = R (minimum, not zero); ω0 depends only on L and C, not R.

  • Impedance minimum (Z = R), current maximum (I = V/R)
  • Power factor = 1; circuit acts purely resistive
  • Called an acceptor circuit — used to tune a radio
🔍Sharpness7/9

Quality Factor (Q)

Q measures how sharp and selective the resonance peak is.

Q = (1/R)√(L/C) = ω0L/R = 1/(ω0CR) = f0/Δf

Bandwidth Δf = f0/Q; high Q means a narrow band.

  • High Q → tall, narrow peak → good selectivity
  • Small R raises Q (R is in the denominator)
  • High Q and high bandwidth are inverses, not the same
🔋Power8/9

Power & Power Factor

AC power depends on the phase between voltage and current, captured by the power factor cos φ.

P_avg = V_rms I_rms cos φ, cos φ = R/Z

Use rms values, never peak. Power is dissipated only in R.

  • Apparent power = V_rms I_rms (VA); true power = V_rms I_rms cos φ (W)
  • Pure R: cos φ = 1 (max); pure L/C: cos φ = 0 → wattless current
  • At resonance cos φ = 1, so power is maximum
🔌Applications9/9

Transformer & LC Oscillations

Both run on changing magnetic flux: a transformer reshapes AC voltage, an LC circuit oscillates on its own.

V_s/V_p = N_s/N_p = I_p/I_s ; f = 1/(2π√(LC))

Transformer works only for AC (needs changing flux); ideal: V_p I_p = V_s I_s.

  • Step-up: N_s > N_p (V up, I down); step-down: N_s < N_p (V down, I up)
  • Real losses: copper (I2R), eddy currents (laminated core), hysteresis, leakage
  • LC: energy swaps between capacitor’s electric field and inductor’s magnetic field (like SHM)
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📝 Practice Alternating Current — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2021
In a series LCR circuit the potential differences across the inductor, the capacitor and the resistor are V_L=40 V, V_C=10 V and V_R=40 V. The rms voltage of the AC source is:
Correct answer: B. V_L and V_C are 180° out of phase and subtract: V=√(V_R²+(V_L-V_C)²)=√(40²+(40-10)²)=√(1600+900)=√(2500)=50 V.
🔎 See the full step-by-step solution in the app →
Q2NEET 2021
A series LCR circuit contains a 5.0 H inductor, an 80 μF capacitor and a 40 Ω resistor, connected to a 230 V variable-frequency AC source. The angular frequencies of the source at which the power transferred is half the power at the resonant angular frequency are likely to be:
Correct answer: C. Resonant angular frequency ω₀=(1)/(√(LC))=(1)/(√(5×80×10⁻⁶))=(1)/(√(4×10⁻⁴))=50 rad/s. The half-power points are at ω₀±Δω with Δω=(R)/(2L)=(40)/(2(5))=4 rad/s, giving 46 rad/s and 54 rad/s.
🔎 See the full step-by-step solution in the app →
Q3NEET 2021
A step-down transformer connected to a 220 V AC mains supply is used to operate a lamp rated 11 V, 44 W. Ignoring power losses, the current in the primary circuit is:
Correct answer: A. For an ideal transformer, primary power = secondary power = lamp power = 44 W. Iₚ=(P)/(Vₚ)=(44)/(220)=0.2 A.
🔎 See the full step-by-step solution in the app →
Q4NEET 2020
A 40 μF capacitor is connected to a 200 V, 50 Hz AC supply. The rms value of the current in the circuit is nearly:
Correct answer: B. X_C=(1)/(2π f C)=(1)/(2π(50)(40×10⁻⁶))≈ 79.6 Ω. Then Iᵣₘₛ=(Vᵣₘₛ)/(X_C)=(200)/(79.6)≈ 2.5 A.
🔎 See the full step-by-step solution in the app →
Q5NEET 2020
A series LCR circuit is connected to an AC voltage source. When L is removed from the circuit, the phase difference between current and voltage is (π)/(3). If instead C is removed, the phase difference is again (π)/(3). The power factor of the (full) circuit is:
Correct answer: B. With L removed: tan(π)/(3)=(X_C)/(R)=sqrt3. With C removed: tan(π)/(3)=(X_L)/(R)=sqrt3. Hence X_L=X_C, so the full circuit is at resonance: φ=0 and power factor cosφ=1.0.
🔎 See the full step-by-step solution in the app →
Q6NEET 2019
An AC voltage source of 12 V drives a current of 0.2 A through a certain circuit. The same circuit driven by a 12 V DC source carries 0.4 A. The circuit is most likely a:
Correct answer: A. On DC the circuit carries 0.4 A, so it has no series capacitor (a capacitor blocks DC). DC resistance R=12/0.4=30 Ω. On AC the impedance Z=12/0.2=60 Ω>R, so there must be reactance in series with R. Since DC still flows, the reactance is inductive: a series L–R circuit.
🔎 See the full step-by-step solution in the app →
Q7NEET 2016
An inductor 20 mH, a capacitor 50 μF and a resistor 40 Ω are connected in series across a source of emf V=10sin(340 t) V. The power loss in the AC circuit is:
Correct answer: D. ω=340 rad/s. X_L=ω L=340(0.02)=6.8 Ω; X_C=(1)/(ω C)=(1)/(340×50×10⁻⁶)≈58.8 Ω. Z=√(40²+(6.8-58.8)²)=√(1600+2704)≈65.6 Ω. Vᵣₘₛ=(10)/(sqrt2)=7.07 V; Iᵣₘₛ=(7.07)/(65.6)≈0.108 A. P=Iᵣₘₛ² R=(0.108)²(40)≈0.51 W.
🔎 See the full step-by-step solution in the app →
Q8NEET 2016
In a series LCR circuit, the rms voltages across the resistance, capacitance and inductance are 80 V, 40 V and 100 V respectively. The power factor of the circuit is:
Correct answer: C. Source voltage V=√(V_R²+(V_L-V_C)²)=√(80²+(100-40)²)=√(6400+3600)=100 V. Power factor cosφ=(V_R)/(V)=(80)/(100)=0.8.
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Q9NEET 2015
A resistance R draws power P when connected to an AC source. If an inductor is now placed in series with the resistance so that the impedance of the circuit becomes Z, the power drawn will be:
Correct answer: A. With only R: P=(Vᵣₘₛ²)/(R). With the inductor in series the current is I=(Vᵣₘₛ)/(Z) and real power is dissipated only in R: P’=I²R=(Vᵣₘₛ²)/(Z²)R=(Vᵣₘₛ²)/(R)·(R²)/(Z²)=P((R)/(Z))².
🔎 See the full step-by-step solution in the app →
Q10NEET 1994
In an AC circuit, the rms value iᵣₘₛ of an alternating current is related to its peak value i₀ by:
Correct answer: D. The rms value is the square root of the mean of over a full cycle. For i=i₀sinω t, the mean of sin²ω t over a cycle is 1/2, so iᵣₘₛ=i₀/√(2)≈ 0.707 i₀.
🔎 See the full step-by-step solution in the app →
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Frequently Asked Questions

What is the rms value of an alternating current and why do we use it?

The rms (root-mean-square) value is the steady DC current that would produce the same heating (I squared R) effect as the AC. We use it because the average of a sinusoidal AC over a full cycle is zero, and for a sinusoid I_rms = i0 divided by root 2, about 0.707 times the peak value.

What is the formula for impedance in a series LCR circuit?

Impedance is the total opposition to AC and equals Z = square root of [R squared plus (X_L minus X_C) squared], where X_L = 2 pi f L and X_C = 1 divided by (2 pi f C). The current then follows V_rms = I_rms times Z.

What is resonance in a series LCR circuit?

Resonance occurs when the inductive and capacitive reactances are equal (X_L = X_C), so they cancel and impedance becomes minimum (Z = R) while current is maximum. The resonant frequency is f0 = 1 divided by (2 pi root LC), and the circuit is called an acceptor circuit because it accepts maximum current at this frequency.

Is Alternating Current important for NEET and how much weightage does it carry?

Yes, Alternating Current is part of the NEET Physics syllabus from Class 12 Electromagnetic Induction and AC, and it is largely numerical and high-yield. Questions commonly come from rms values, reactance, impedance, resonance and power factor, so it usually contributes a few marks each year.

What is the difference between reactance and impedance?

Reactance is the opposition of a single inductor (X_L = omega L) or capacitor (X_C = 1 divided by omega C) and depends on frequency, while impedance Z is the combined total opposition of R, L and C together in the circuit. Reactance involves only L or C, whereas impedance also includes resistance and is found by phasor (vector) addition.

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