What is the difference between AC and DC current?

AC (Alternating Current) periodically reverses direction and changes magnitude with time, typically following a sinusoidal pattern. DC (Direct Current) flows in one direction only with constant magnitude. AC is easier to transform to different voltages and is used for power transmission, while DC is used in batteries and electronic circuits.

What is the power factor in AC circuits?

Power factor is the cosine of the phase difference (φ) between voltage and current in an AC circuit: Power Factor = cos φ = R/Z. It ranges from 0 to 1. A power factor of 1 (purely resistive circuit) means maximum power transfer, while 0 means no real power is consumed (purely reactive circuit).

What is resonance in RLC circuits?

Resonance in an RLC circuit occurs when the inductive reactance (X_L) equals the capacitive reactance (X_C), making the circuit behave purely resistively. At resonance frequency ω₀ = 1/√(LC), impedance is minimum (Z = R), current is maximum, and power factor is unity. This is crucial for tuning circuits in radio and TV receivers.

What is the formula for impedance in series RLC circuit?

The impedance in a series RLC circuit is Z = √[R² + (X_L - X_C)²], where X_L = ωL is inductive reactance and X_C = 1/(ωC) is capacitive reactance. The phase angle φ = tan⁻¹[(X_L - X_C)/R]. At resonance, X_L = X_C, so Z = R (minimum).

What is the working principle of a transformer?

A transformer works on the principle of mutual induction. When alternating current flows through the primary coil, it creates a changing magnetic flux in the core. This changing flux induces an EMF in the secondary coil. The voltage ratio is V_s/V_p = N_s/N_p, where N is the number of turns. For an ideal transformer, V_p I_p = V_s I_s (power is conserved).

What is the quality factor (Q-factor) of an RLC circuit?

Quality factor (Q-factor) measures the sharpness of resonance in an RLC circuit. Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C), where ω₀ is the resonance frequency. Higher Q-factor means sharper resonance peak, lower energy loss, and better selectivity. For a good tuning circuit, Q should be high (typically > 100).

Alternating Current for JEE 2025-26 | Complete Notes with Formulas, Phasor Diagrams & Solved Examples

Alternating Current JEE notes, Formulas, PYQs, RLC Circuits, Phasor Diagrams — Alternating Current JEE Notes - Complete Guide with Formulas & PYQs

AC Voltage and Current

Alternating Current (AC) is electric current that periodically reverses direction and changes its magnitude continuously with time. Unlike DC which flows in one direction, AC is the standard form of electricity supplied to homes and industries worldwide.

1.1 AC vs DC: Fundamental Differences

Aspect	Alternating Current (AC)	Direct Current (DC)
Direction	Reverses periodically	Flows in one direction only
Magnitude	Varies sinusoidally with time	Constant magnitude
Frequency	50 Hz (India) or 60 Hz (USA)	0 Hz (no oscillation)
Transmission	Efficient over long distances	Significant power loss
Voltage Transformation	Easy using transformers	Complex and expensive
Applications	Household power, motors, lighting	Batteries, electronics, electroplating
Safety	More dangerous (can cause fibrillation)	Relatively safer

1.2 Mathematical Representation of AC

Instantaneous Values

AC Voltage

V = V_0 \sin(\omega t + \phi_V)

V₀ = Peak voltage (amplitude)
ω = Angular frequency (2πf)
φᵥ = Initial phase of voltage

AC Current

I = I_0 \sin(\omega t + \phi_I)

I₀ = Peak current (amplitude)
ω = Angular frequency (2πf)
φᵢ = Initial phase of current

Phase Difference (φ):

\phi = \phi_V - \phi_I

Phase difference determines whether voltage leads or lags current

1.3 Peak, Average, and RMS Values

Peak Value

V_0, \quad I_0

Maximum value of voltage or current in one cycle

For household AC in India:

V₀ = 311 V (from 220 V RMS)

Average Value

V_{avg} = \frac{2V_0}{\pi}

I_{avg} = \frac{2I_0}{\pi}

Average over half cycle (full cycle average = 0)

Ratio:

V_avg/V₀ = 2/π ≈ 0.637

RMS Value

V_{rms} = \frac{V_0}{\sqrt{2}}

I_{rms} = \frac{I_0}{\sqrt{2}}

Root Mean Square - effective value for power calculation

Ratio:

V_rms/V₀ = 1/√2 ≈ 0.707

💡 Why RMS Value is Important?

Power Equivalence: AC with RMS value produces same heating effect as DC with same magnitude
Standard Measurement: All AC voltmeters and ammeters show RMS values
Power Calculation: P = V_rms × I_rms (for purely resistive circuits)
Household Supply: 220V AC means 220V RMS (peak is 311V)

1.4 Important Relationships & Formulas

Relationship	Formula	Value
Peak to RMS	V₀ = √2 × V_rms	V₀ = 1.414 V_rms
Peak to Average	V₀ = (π/2) × V_avg	V₀ = 1.571 V_avg
RMS to Average	V_rms = (π/2√2) × V_avg	V_rms = 1.11 V_avg
Frequency	f = ω/(2π)	In India: 50 Hz
Time Period	T = 1/f = 2π/ω	T = 0.02 s (for 50 Hz)

📝 Solved Example 1

Question: An AC voltage is given by V = 311 sin(100πt) volts. Find: (a) Peak voltage (b) RMS voltage (c) Frequency (d) Time period

Solution:

Comparing with V = V₀ sin(ωt):

(a) Peak Voltage:

V_0 = 311 \text{ V}

(b) RMS Voltage:

V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{311}{1.414} = 220 \text{ V}

(c) Frequency:

ω = 100π rad/s

f = \frac{\omega}{2\pi} = \frac{100\pi}{2\pi} = 50 \text{ Hz}

(d) Time Period:

T = \frac{1}{f} = \frac{1}{50} = 0.02 \text{ s} = 20 \text{ ms}

Note: This represents standard household AC supply in India (220V, 50Hz)

⚠️ Common JEE Mistakes

Using peak instead of RMS: Always use RMS values unless specifically asked for peak
Wrong power formula: P = V₀I₀ is WRONG. Use P = V_rms × I_rms × cos φ
Forgetting √2 factor: V_rms = V₀/√2, not V₀/2
Phase angle confusion: φ = φ_V - φ_I (voltage phase minus current phase)
Average over full cycle: Average of sin wave over complete cycle is ZERO

Phasor Diagrams

Phasor diagrams are rotating vector representations of sinusoidally varying quantities (voltage and current). They provide a visual and simplified method to analyze AC circuits by converting differential equations into algebraic operations.

2.1 What is a Phasor?

Mathematical Definition

A phasor is a complex number that represents the amplitude and phase of a sinusoidal function.

V(t) = V_0 \sin(\omega t + \phi)

↓ Represented as ↓

\vec{V} = V_0 \angle \phi

V₀ = Magnitude (length of phasor)
φ = Phase angle (from reference)
Rotates at angular frequency ω

Key Properties

Rotation: All phasors rotate counterclockwise at ω
Projection: Vertical projection gives instantaneous value
Reference: Usually current or voltage is taken as reference (phase = 0°)
Addition: Phasors add vectorially (parallelogram law)
Phase difference: Angle between two phasors
Snapshot: Diagram shows state at t = 0

2.2 Phasor Addition and Subtraction

Vector Addition Rules

If V₁ = V₀₁∠φ₁ and V₂ = V₀₂∠φ₂

Resultant magnitude:

V = \sqrt{V_{01}^2 + V_{02}^2 + 2V_{01}V_{02}\cos(\phi_2 - \phi_1)}

Resultant phase:

\tan \phi = \frac{V_{01}\sin\phi_1 + V_{02}\sin\phi_2}{V_{01}\cos\phi_1 + V_{02}\cos\phi_2}

Special Cases

When φ₂ - φ₁ = 0° (in phase):

V = V_{01} + V_{02}

When φ₂ - φ₁ = 180° (opposite phase):

V = |V_{01} - V_{02}|

When φ₂ - φ₁ = 90° (perpendicular):

V = \sqrt{V_{01}^2 + V_{02}^2}

🎯 Phasor Diagram Shortcuts for JEE

Reference choice: Always take current as reference (I along x-axis) for easier calculation
Leading/Lagging: Counterclockwise from reference = leading, Clockwise = lagging
Capacitor: V_C lags I by 90° (I leads V_C by 90°)
Inductor: V_L leads I by 90° (I lags V_L by 90°)
Resistor: V_R is in phase with I (0° difference)

📝 Solved Example 2

Question: Two AC voltages V₁ = 100 sin(ωt) and V₂ = 80 sin(ωt + π/3) are applied across a circuit. Find the resultant voltage.

Solution:

Given: V₀₁ = 100 V, φ₁ = 0°

V₀₂ = 80 V, φ₂ = 60° (π/3 rad)

Step 1: Find resultant magnitude

V_0 = \sqrt{V_{01}^2 + V_{02}^2 + 2V_{01}V_{02}\cos(\phi_2 - \phi_1)}

V_0 = \sqrt{100^2 + 80^2 + 2(100)(80)\cos 60°}

V_0 = \sqrt{10000 + 6400 + 16000 \times 0.5}

V_0 = \sqrt{10000 + 6400 + 8000} = \sqrt{24400} = 156.2 \text{ V}

Step 2: Find resultant phase

\tan \phi = \frac{V_{01}\sin\phi_1 + V_{02}\sin\phi_2}{V_{01}\cos\phi_1 + V_{02}\cos\phi_2}

\tan \phi = \frac{100 \times 0 + 80 \times \sin 60°}{100 \times 1 + 80 \times \cos 60°}

\tan \phi = \frac{80 \times 0.866}{100 + 80 \times 0.5} = \frac{69.28}{140} = 0.495

\phi = \tan^{-1}(0.495) = 26.3° = 0.459 \text{ rad}

\boxed{V = 156.2 \sin(\omega t + 26.3°) \text{ volts}}

AC Through Pure R, L, and C

Understanding how AC behaves through individual circuit elements (Resistor, Inductor, Capacitor) is fundamental before analyzing complex RLC combinations. Each element creates a unique phase relationship between voltage and current.

3.1 AC Through Pure Resistance (R)

Key Relationships

Applied Voltage:

V = V_0 \sin \omega t

Current (Ohm's Law):

I = \frac{V}{R} = \frac{V_0}{R} \sin \omega t = I_0 \sin \omega t

Peak Current:

I_0 = \frac{V_0}{R}

RMS Relation:

I_{rms} = \frac{V_{rms}}{R}

Important Properties

Phase Difference: φ = 0° (V and I in phase)
Power Factor: cos φ = cos 0° = 1 (maximum)
Average Power: P = V_rms I_rms = I²_rms R
Impedance: Z = R (purely resistive)
Phasor: V and I both along same direction
Energy: Completely dissipated as heat

Phasor Diagram

V and I are in phase (φ = 0°)

3.2 AC Through Pure Inductance (L)

Key Relationships

Applied Voltage:

V = V_0 \sin \omega t

Current (lags by 90°):

I = I_0 \sin(\omega t - \frac{\pi}{2}) = -I_0 \cos \omega t

Inductive Reactance:

X_L = \omega L = 2\pi f L

Peak Current:

I_0 = \frac{V_0}{X_L} = \frac{V_0}{\omega L}

RMS Relation:

I_{rms} = \frac{V_{rms}}{X_L}

Important Properties

Phase Difference: φ = +90° (I lags V by 90°)
Power Factor: cos φ = cos 90° = 0
Average Power: P = 0 (no power consumed)
Impedance: Z = X_L = ωL
Frequency dependence: X_L ∝ f (increases with frequency)
DC behavior: X_L = 0 (acts as short circuit)
Energy: Stored in magnetic field, returned to source

Phasor Diagram

V leads I by 90° (or I lags V)

3.3 AC Through Pure Capacitance (C)

Key Relationships

Applied Voltage:

V = V_0 \sin \omega t

Current (leads by 90°):

I = I_0 \sin(\omega t + \frac{\pi}{2}) = I_0 \cos \omega t

Capacitive Reactance:

X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}

Peak Current:

I_0 = \frac{V_0}{X_C} = V_0 \omega C

RMS Relation:

I_{rms} = \frac{V_{rms}}{X_C}

Important Properties

Phase Difference: φ = -90° (I leads V by 90°)
Power Factor: cos φ = cos 90° = 0
Average Power: P = 0 (no power consumed)
Impedance: Z = X_C = 1/(ωC)
Frequency dependence: X_C ∝ 1/f (decreases with frequency)
DC behavior: X_C = ∞ (acts as open circuit)
Energy: Stored in electric field, returned to source

Phasor Diagram

I leads V by 90° (or V lags I)

3.4 Comparison Table: R vs L vs C

Property	Resistor (R)	Inductor (L)	Capacitor (C)
Reactance/Resistance	R (constant)	X_L = ωL	X_C = 1/(ωC)
Phase Difference	0°	+90° (I lags V)	-90° (I leads V)
Power Factor	1 (maximum)	0	0
Average Power	V_rms I_rms	0	0
Frequency Dependence	Independent	Increases with f	Decreases with f
DC Response (f = 0)	R	0 (short)	∞ (open)
High Frequency	R	∞ (open)	0 (short)
Energy Storage	None (dissipated)	Magnetic field	Electric field

💡 Memory Trick: "CIVIL"

C - In Capacitor

I - Current (I)

V - leads Voltage

I - In Inductor

L - Voltage (Leads current)

Or remember: "ELI the ICE man"
ELI: In L (inductor), E (voltage) leads I (current)
ICE: In C (capacitor), I (current) leads E (voltage)

📝 Solved Example 3 (JEE Main 2022 Pattern)

Question: A 200 mH inductor is connected to a 220 V, 50 Hz AC source. Calculate: (a) Inductive reactance (b) RMS current (c) Peak current

Solution:

Given: L = 200 mH = 0.2 H, V_rms = 220 V, f = 50 Hz

(a) Inductive Reactance:

X_L = 2\pi f L = 2 \times 3.14 \times 50 \times 0.2

X_L = 62.8 \text{ Ω}

(b) RMS Current:

I_{rms} = \frac{V_{rms}}{X_L} = \frac{220}{62.8} = 3.5 \text{ A}

(c) Peak Current:

I_0 = \sqrt{2} \times I_{rms} = 1.414 \times 3.5

\boxed{I_0 = 4.95 \text{ A}}

Additional Insights:

Current lags voltage by 90° in pure inductor
No power is consumed (P = 0)
Energy oscillates between source and magnetic field of inductor

Series RLC Circuit

The series RLC circuit combines all three elements - Resistor, Inductor, and Capacitor - in series. This is one of the most important circuits in AC theory and forms the basis for understanding resonance, filters, and tuning circuits.

4.1 Impedance of Series RLC Circuit

Impedance Derivation

In series RLC circuit:

Same current I flows through all elements
Total voltage V = V_R + V_L + V_C (phasor sum)
V_L and V_C are 180° out of phase
Net reactive voltage = V_L - V_C

Impedance Formula:

Z = \sqrt{R^2 + (X_L - X_C)^2}

Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}

Phase Angle

\tan \phi = \frac{X_L - X_C}{R}

\phi = \tan^{-1}\left(\frac{\omega L - \frac{1}{\omega C}}{R}\right)

Three Cases:

1. X_L > X_C: φ > 0° (inductive)

Current lags voltage

2. X_L < X_C: φ < 0° (capacitive)

Current leads voltage

3. X_L = X_C: φ = 0° (resonance)

Current in phase with voltage

Current in Series RLC:

I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + (X_L - X_C)^2}}

Current is same throughout the circuit (series connection)

4.2 Voltage Relationships in Series RLC

Individual Voltages

Across Resistor:

\[V_R = I R\]

(In phase with current)

Across Inductor:

V_L = I X_L = I \omega L

(Leads current by 90°)

Across Capacitor:

V_C = I X_C = \frac{I}{\omega C}

(Lags current by 90°)

Total Voltage

Phasor Addition:

V = \sqrt{V_R^2 + (V_L - V_C)^2}

⚠️ Important Note:

V ≠ V_R + V_L + V_C

Because they are phasors, not scalar quantities. Must use vector addition!

Verification:

V = IZ = I\sqrt{R^2 + (X_L - X_C)^2}

4.3 Phasor Diagram for Series RLC

Complete Phasor Diagram (X_L > X_C case)

Construction Steps:

Take current I as reference (horizontal)
Draw V_R along I (in phase)
Draw V_L perpendicular upward (leads by 90°)
Draw V_C perpendicular downward (lags by 90°)
Net reactive voltage = V_L - V_C
Resultant V using Pythagoras theorem

Key Observations:

When X_L > X_C: Circuit is inductive (φ positive)
When X_L < X_C: Circuit is capacitive (φ negative)
V_L and V_C can individually exceed total voltage V
φ varies from -90° to +90° depending on frequency

4.4 Important Formulas Summary

Quantity	Formula	Unit
Impedance (Z)	√[R² + (X_L - X_C)²]	Ω (ohm)
Current (I)	V/Z	A (ampere)
Phase angle (φ)	tan⁻¹[(X_L - X_C)/R]	degrees or radians
Power factor	cos φ = R/Z	dimensionless
Total voltage	√[V_R² + (V_L - V_C)²]	V (volt)
Average Power	V_rms I_rms cos φ = I²R	W (watt)

📝 Solved Example 4 (JEE Advanced Pattern)

Question: A series RLC circuit has R = 30 Ω, L = 0.5 H, and C = 20 μF. It is connected to 220 V, 50 Hz AC source. Calculate: (a) Impedance (b) Current (c) Phase angle (d) Power factor (e) Voltages across R, L, C

Solution:

Given: R = 30 Ω, L = 0.5 H, C = 20 μF = 20 × 10⁻⁶ F

V = 220 V, f = 50 Hz, ω = 2πf = 314 rad/s

Step 1: Calculate Reactances

X_L = \omega L = 314 \times 0.5 = 157 \text{ Ω}

X_C = \frac{1}{\omega C} = \frac{1}{314 \times 20 \times 10^{-6}} = \frac{1}{0.00628} = 159.2 \text{ Ω}

(a) Impedance:

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + (157 - 159.2)^2}

Z = \sqrt{900 + (-2.2)^2} = \sqrt{900 + 4.84} = \sqrt{904.84}

\boxed{Z = 30.08 \text{ Ω} \approx 30 \text{ Ω}}

(b) Current:

I = \frac{V}{Z} = \frac{220}{30.08} = 7.31 \text{ A}

(c) Phase Angle:

\tan \phi = \frac{X_L - X_C}{R} = \frac{157 - 159.2}{30} = \frac{-2.2}{30} = -0.073

\phi = \tan^{-1}(-0.073) = -4.2°

Negative angle → Circuit is slightly capacitive (current leads voltage)

(d) Power Factor:

\cos \phi = \frac{R}{Z} = \frac{30}{30.08} = 0.997 \approx 1

Nearly unity power factor (almost resonance condition)

(e) Individual Voltages:

V_R = IR = 7.31 \times 30 = 219.3 \text{ V}

V_L = IX_L = 7.31 \times 157 = 1147.7 \text{ V}

V_C = IX_C = 7.31 \times 159.2 = 1163.8 \text{ V}

⚡ Important Observations:

V_L and V_C are both much larger than supply voltage (1147 V vs 220 V)
This is possible because V_L and V_C are 180° out of phase and nearly cancel
Net reactive voltage = |V_L - V_C| = |1147.7 - 1163.8| = 16.1 V (very small)
Circuit is near resonance (X_L ≈ X_C)

Verification:

V = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{219.3^2 + 16.1^2} = 220 \text{ V} \checkmark

⚠️ Common Mistakes in RLC Problems

Adding voltages algebraically: V ≠ V_R + V_L + V_C (must use phasor addition)
Forgetting ω in reactance: X_L = ωL, not just L; X_C = 1/(ωC), not just 1/C
Wrong phase angle sign: X_L > X_C gives positive φ (inductive), not negative
Using peak instead of RMS: Always use RMS unless specified
Power calculation: P = I²R only (not P = I²Z or P = V²/Z)
Units confusion: Convert μF to F, mH to H before calculation

Resonance in AC Circuits

Resonance is the most important phenomenon in AC circuits, occurring when inductive and capacitive reactances are equal. It's the principle behind radio tuning, filters, and many communication devices. This topic carries 15-20% weightage in JEE Advanced.

5.1 Condition for Resonance

Resonance Occurs When

Mathematical Condition

\[X_L = X_C\]

\omega_0 L = \frac{1}{\omega_0 C}

\omega_0^2 = \frac{1}{LC}

Resonant Frequency:

\omega_0 = \frac{1}{\sqrt{LC}}

f_0 = \frac{1}{2\pi\sqrt{LC}}

Physical Meaning

Energy oscillates between L and C
Magnetic energy in L = Electric energy in C
No net reactive power
Circuit behaves purely resistively
Maximum power transfer occurs
Used in radio/TV tuning circuits

"At resonance, the circuit 'sings' at its natural frequency, just like a tuning fork or a swing resonates at its natural frequency when pushed periodically."

5.2 Properties at Resonance

Impedance

Z = \sqrt{R^2 + (X_L - X_C)^2}

At resonance (X_L = X_C):

\boxed{Z_{\min} = R}

Minimum impedance!

Current

I = \frac{V}{Z}

At resonance (Z = R):

\boxed{I_{\max} = \frac{V}{R}}

Maximum current!

Phase Angle

\tan \phi = \frac{X_L - X_C}{R}

At resonance:

\boxed{\phi = 0°}

In phase!

Property	Below Resonance (f < f₀)	At Resonance (f = f₀)	Above Resonance (f > f₀)
Reactance	X_C > X_L (capacitive)	X_L = X_C	X_L > X_C (inductive)
Impedance	Z > R	Z = R (minimum)	Z > R
Current	I < V/R	I = V/R (maximum)	I < V/R
Phase angle	φ < 0° (I leads V)	φ = 0° (in phase)	φ > 0° (I lags V)
Power factor	cos φ < 1	cos φ = 1 (unity)	cos φ < 1
Power	P < P_max	P = V²/R (maximum)	P < P_max

5.3 Quality Factor (Q-factor)

Definition & Formulas

Quality Factor (Q): Measure of sharpness of resonance

Formula 1:

Q = \frac{\omega_0 L}{R}

Formula 2:

Q = \frac{1}{\omega_0 CR}

Formula 3:

Q = \frac{1}{R}\sqrt{\frac{L}{C}}

Physical Significance

High Q (>100): Sharp resonance, narrow bandwidth, high selectivity
Low Q (<10): Broad resonance, wide bandwidth, low selectivity
Q = 2π × (Energy stored / Energy lost per cycle)
Q = ω₀ / Δω (ratio of resonance frequency to bandwidth)

Voltage Magnification

At resonance, voltage across L or C can be Q times the applied voltage:

V_L = V_C = Q \times V

Example: If Q = 50 and V = 10 V, then V_L = V_C = 500 V!

This is why capacitors in resonant circuits need high voltage ratings.

5.4 Bandwidth and Sharpness of Resonance

Bandwidth (Δω or Δf)

Range of frequencies over which current is ≥ 0.707 I_max (or power ≥ 0.5 P_max)

\Delta \omega = \frac{R}{L} = \frac{\omega_0}{Q}

\Delta f = \frac{f_0}{Q}

Relationship:

Q = \frac{f_0}{\Delta f}

Higher Q → Smaller bandwidth → Sharper resonance

Resonance Curve

Red (sharp): High Q, Narrow bandwidth
Blue (broad): Low Q, Wide bandwidth

🎯 Applications of Resonance

High Q Circuits (Narrow Bandwidth):

Radio/TV station selection
Frequency filters
Signal processing
Oscillators

Low Q Circuits (Wide Bandwidth):

Audio amplifiers
Broadband filters
Power factor correction
Impedance matching

📝 Solved Example 5 (JEE Advanced 2021 Type)

Question: An RLC circuit has L = 10 mH, C = 100 μF, R = 5 Ω. Calculate: (a) Resonance frequency (b) Q-factor (c) Bandwidth (d) If applied voltage is 10 V at resonance, find voltage across L and C

Solution:

Given:

L = 10 mH = 10 × 10⁻³ H

C = 100 μF = 100 × 10⁻⁶ F

R = 5 Ω, V = 10 V

(a) Resonance Frequency:

\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10 \times 10^{-3} \times 100 \times 10^{-6}}}

\omega_0 = \frac{1}{\sqrt{10^{-6}}} = \frac{1}{10^{-3}} = 1000 \text{ rad/s}

f_0 = \frac{\omega_0}{2\pi} = \frac{1000}{2 \times 3.14} = 159.2 \text{ Hz}

(b) Quality Factor:

Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{5}\sqrt{\frac{10 \times 10^{-3}}{100 \times 10^{-6}}}

Q = \frac{1}{5}\sqrt{\frac{10^{-2}}{10^{-4}}} = \frac{1}{5}\sqrt{100} = \frac{10}{5}

\boxed{Q = 2}

(Low Q → Broad resonance)

(c) Bandwidth:

\Delta f = \frac{f_0}{Q} = \frac{159.2}{2} = 79.6 \text{ Hz}

Alternatively:

\Delta \omega = \frac{R}{L} = \frac{5}{10 \times 10^{-3}} = 500 \text{ rad/s}

\Delta f = \frac{500}{2\pi} = 79.6 \text{ Hz} \checkmark

(d) Voltages across L and C at Resonance:

At resonance: I = V/R = 10/5 = 2 A

X_L = \omega_0 L = 1000 \times 10 \times 10^{-3} = 10 \text{ Ω}

V_L = I \times X_L = 2 \times 10 = 20 \text{ V}

V_C = I \times X_C = 2 \times 10 = 20 \text{ V}

Verification using Q-factor:

V_L = V_C = Q \times V = 2 \times 10 = 20 \text{ V} \checkmark

Key Insight: Even though applied voltage is only 10 V, voltage across L and C is 20 V each (double). This is voltage magnification due to resonance. With higher Q-factor circuits, this magnification can be 50-100 times!

AC Power

Power in AC circuits is more complex than in DC circuits due to phase differences between voltage and current. Understanding true power, reactive power, apparent power, and power factor is crucial for both JEE and practical electrical engineering.

6.1 Types of Power in AC Circuits

True Power (P)

P = V_{rms} I_{rms} \cos \phi

P = I_{rms}^2 R

Unit: Watt (W)
Also called: Active power, Real power
Represents: Power actually consumed/dissipated
Always: P ≥ 0

Reactive Power (Q)

Q = V_{rms} I_{rms} \sin \phi

Q = I_{rms}^2 (X_L - X_C)

Unit: VAR (Volt-Ampere Reactive)
Represents: Power oscillating between source and L/C
Not consumed: Returns to source
Can be: Positive or negative

Apparent Power (S)

S = V_{rms} I_{rms}

S = \sqrt{P^2 + Q^2}

Unit: VA (Volt-Ampere)
Represents: Total power supplied
Rating: Transformers rated in kVA
Always: S ≥ P

Power Triangle

Relationships

\[S^2 = P^2 + Q^2\]

P = S \cos \phi

Q = S \sin \phi

\tan \phi = \frac{Q}{P}

6.2 Power Factor

Definition and Significance

Mathematical Definition

\text{Power Factor (pf)} = \cos \phi

\cos \phi = \frac{R}{Z} = \frac{P}{S}

\cos \phi = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}}

Range:

0 ≤ Power Factor ≤ 1

Often expressed as percentage (0% to 100%)

Physical Meaning

cos φ = 1: Unity pf (best case)
- Pure resistive circuit
- 100% power utilization
- φ = 0°, V and I in phase
cos φ = 0: Zero pf (worst case)
- Pure reactive circuit (L or C)
- No power consumption
- φ = ±90°
0 < cos φ < 1: Practical circuits
- Mix of R and X
- Partial power utilization
- Most real circuits

⚡ Why Power Factor Matters?

Efficiency: Low pf means more current needed for same power → higher losses
Cost: Industries penalized for pf < 0.85 in electricity bills
Equipment size: Low pf requires larger transformers and cables
Voltage drop: Low pf causes excessive voltage drop in transmission

6.3 Leading vs Lagging Power Factor

Aspect	Lagging Power Factor	Leading Power Factor
Phase Relation	Current lags voltage	Current leads voltage
Phase Angle	φ > 0° (positive)	φ < 0° (negative)
Circuit Nature	Inductive (X_L > X_C)	Capacitive (X_C > X_L)
Common In	Motors, transformers, inductors	Capacitor banks, long transmission lines
Reactive Power	Q > 0 (consumed)	Q < 0 (supplied)
Notation	pf = 0.8 lagging	pf = 0.8 leading
Practical Impact	Requires reactive power compensation	Less common, can improve grid pf

6.4 Power Factor Improvement

Why and How to Improve Power Factor?

Method: Capacitor Bank

Connect capacitors in parallel with inductive load to cancel reactive power.

Required Capacitance:

C = \frac{P(\tan \phi_1 - \tan \phi_2)}{\omega V^2}

φ₁ = initial angle, φ₂ = desired angle

In terms of power:

Q_C = P(\tan \phi_1 - \tan \phi_2)

Q_C = Reactive power of capacitor needed

Benefits of Improvement

Reduced current: For same power, I decreases
Lower losses: I²R losses reduced in cables
Better voltage: Less voltage drop
Cost savings: Avoid penalty charges
Increased capacity: Existing equipment can handle more load
Improved efficiency: Overall system efficiency increases

Industry Standard: Most industries maintain pf > 0.95 to avoid penalties and maximize efficiency.

📝 Solved Example 6 (JEE Main 2023 Pattern)

Question: A 1000 W load operates at 220 V, 50 Hz with power factor 0.6 lagging. Find: (a) Current drawn (b) Impedance (c) Resistance (d) Reactive power (e) Capacitance needed to improve pf to 0.9

Solution:

Given: P = 1000 W, V = 220 V, f = 50 Hz

pf = cos φ₁ = 0.6 lagging, ω = 314 rad/s

(a) Current Drawn:

P = V I \cos \phi

I = \frac{P}{V \cos \phi} = \frac{1000}{220 \times 0.6} = \frac{1000}{132} = 7.58 \text{ A}

(b) Impedance:

Z = \frac{V}{I} = \frac{220}{7.58} = 29.02 \text{ Ω}

(c) Resistance:

cos φ = R/Z

R = Z \cos \phi = 29.02 \times 0.6 = 17.41 \text{ Ω}

(d) Reactive Power:

First find φ₁: cos φ₁ = 0.6 → φ₁ = 53.13° → sin φ₁ = 0.8

Q = V I \sin \phi = 220 \times 7.58 \times 0.8 = 1333.3 \text{ VAR}

Or: Q = P tan φ = 1000 × (0.8/0.6) = 1333.3 VAR

(e) Capacitance for pf improvement to 0.9:

cos φ₂ = 0.9 → φ₂ = 25.84° → tan φ₂ = 0.484

tan φ₁ = 0.8/0.6 = 1.333

Q_C = P(\tan \phi_1 - \tan \phi_2) = 1000(1.333 - 0.484)

Q_C = 1000 \times 0.849 = 849 \text{ VAR}

For capacitor: Q_C = V²ωC

C = \frac{Q_C}{V^2 \omega} = \frac{849}{220^2 \times 314} = \frac{849}{15209600}

\boxed{C = 55.8 \text{ μF}}

Result Summary:

Before: I = 7.58 A, pf = 0.6
After: I = 1000/(220×0.9) = 5.05 A, pf = 0.9
Current reduced by 33% → Significant savings!
Required: 55.8 μF capacitor connected in parallel

💡 Quick Formula Sheet - AC Power

Power Formulas:

P = V I cos φ
P = I²R
P = (V²/Z²)R
Q = V I sin φ
S = V I
S² = P² + Q²

Power Factor:

pf = cos φ = R/Z
pf = P/S
cos φ = R/√(R²+(X_L-X_C)²)
tan φ = (X_L-X_C)/R
sin φ = (X_L-X_C)/Z

Transformers

A transformer is a static electrical device that transfers electrical energy between two circuits through electromagnetic induction. It's one of the most important devices in AC power transmission and is a heavily tested topic in JEE (3-4% weightage in Physics).

7.1 Working Principle

Mutual Induction

A transformer works on the principle of mutual induction:

AC flows through primary coil
Creates changing magnetic flux in iron core
Flux links both primary and secondary coils
Changing flux induces EMF in secondary coil
EMF drives current through external circuit

Key Requirements:

Works ONLY with AC (not DC)
Both coils wound on same core
High permeability core (iron/ferrite)
No electrical connection between coils

Basic Transformer Diagram

N_p turns on primary, N_s turns on secondary

7.2 Transformer Equations

Fundamental Relationships

EMF Equation

Induced EMF in a coil:

\mathcal{E} = -N\frac{d\Phi}{dt}

For sinusoidal flux Φ = Φ₀ sin ωt:

\mathcal{E}_{rms} = \frac{2\pi}{\sqrt{2}} f N \Phi_0 = 4.44 f N \Phi_0

Where:

f = frequency (Hz)
N = number of turns
Φ₀ = maximum flux (Wb)

Turns Ratio

For an ideal transformer:

\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{\mathcal{E}_s}{\mathcal{E}_p}

Power conservation (ideal):

\[V_p I_p = V_s I_s\]

\frac{I_s}{I_p} = \frac{N_p}{N_s} = \frac{V_p}{V_s}

⚡ Important: Turns Ratio (k)

k = \frac{N_s}{N_p} = \frac{V_s}{V_p} = \frac{I_p}{I_s}

k > 1: Step-up Transformer

V_s > V_p (voltage increases)

I_s < I_p (current decreases)

k < 1: Step-down Transformer

V_s < V_p (voltage decreases)

I_s > I_p (current increases)

7.3 Types of Transformers

Type	Step-up Transformer	Step-down Transformer
Turns Ratio	N_s > N_p (k > 1)	N_s < N_p (k < 1)
Voltage Change	V_s > V_p (increased)	V_s < V_p (decreased)
Current Change	I_s < I_p (decreased)	I_s > I_p (increased)
Wire Thickness	Secondary: thin, more turns	Secondary: thick, fewer turns
Application	Power transmission (11 kV → 440 kV)	Home supply (11 kV → 220 V)
Other Examples	CRT TV, ignition coils	Phone chargers, laptop adapters

7.4 Efficiency and Energy Losses

Transformer Efficiency

Efficiency Formula

\eta = \frac{\text{Output Power}}{\text{Input Power}} \times 100\%

\eta = \frac{V_s I_s \cos \phi_s}{V_p I_p \cos \phi_p} \times 100\%

For ideal transformer:

η = 100% (no losses)

V_p I_p = V_s I_s

Practical transformers: η = 90-99%

Energy Losses (Important for JEE)

1. Copper Losses (I²R)

Due to resistance of windings

Reduction: Use thick wires (low R)
2. Iron/Core Losses

a) Eddy current losses in core

b) Hysteresis losses (magnetization cycle)

Reduction: Use laminated core, soft iron
3. Flux Leakage

Not all flux links secondary

Reduction: Proper core design
4. Magnetostriction

Humming sound, energy as sound

⚠️ Why Transformer Doesn't Work with DC?

DC produces constant flux (no change with time)
EMF = -N(dΦ/dt) = 0 when Φ is constant
No induced EMF in secondary coil
Also: DC would create very high current in primary (only R limits it, no X_L after steady state)
This high current would burn the primary coil!

7.5 Long Distance Power Transmission

Why High Voltage Transmission?

Power transmitted: P = VI

Power loss in transmission line: P_loss = I²R

For same power P transmitted:

I = \frac{P}{V}

P_{loss} = I^2 R = \frac{P^2 R}{V^2}

\boxed{P_{loss} \propto \frac{1}{V^2}}

Higher voltage → Lower current → Much lower losses!

Power Station

11 kV

Generated

Transmission Line

132-765 kV

Step-up transformer

Home/Industry

220V / 11kV

Step-down transformer

Numerical Example:

If voltage is increased 10 times (11 kV → 110 kV):

Current decreases 10 times
Power loss decreases 100 times (10² = 100)
This is why we use very high voltage for transmission!

📝 Solved Example 7 (JEE Main Pattern)

Question: A transformer has 500 turns in primary and 2000 turns in secondary. If primary voltage is 220 V and primary current is 4 A, find: (a) Secondary voltage (b) Secondary current (c) Type of transformer (d) If efficiency is 90%, find power delivered to load

Solution:

Given: N_p = 500, N_s = 2000, V_p = 220 V, I_p = 4 A, η = 90%

(a) Secondary Voltage:

\frac{V_s}{V_p} = \frac{N_s}{N_p}

V_s = V_p \times \frac{N_s}{N_p} = 220 \times \frac{2000}{500} = 220 \times 4

\boxed{V_s = 880 \text{ V}}

(b) Secondary Current (for ideal transformer):

\frac{I_s}{I_p} = \frac{N_p}{N_s}

I_s = I_p \times \frac{N_p}{N_s} = 4 \times \frac{500}{2000} = 4 \times 0.25

\boxed{I_s = 1 \text{ A (ideal)}}

(c) Type of Transformer:

Turns ratio k = N_s/N_p = 2000/500 = 4 > 1

Also, V_s (880 V) > V_p (220 V)

Step-up Transformer

(d) Power delivered with 90% efficiency:

Input power = V_p × I_p = 220 × 4 = 880 W

\eta = \frac{P_{output}}{P_{input}} \times 100

P_{output} = \frac{\eta \times P_{input}}{100} = \frac{90 \times 880}{100}

\boxed{P_{output} = 792 \text{ W}}

Actual Secondary Current:

P_output = V_s × I_s (actual)

I_s (actual) = 792/880 = 0.9 A

Note: Less than ideal (1 A) due to losses

LC Oscillations

LC oscillations occur when a charged capacitor is connected to an inductor. Energy oscillates between the electric field of capacitor and magnetic field of inductor, similar to mechanical oscillations (spring-mass system). This is a high-weightage topic in JEE Advanced.

8.1 LC Circuit Analysis

Energy Exchange in LC Circuit

Initial Condition

Capacitor charged to voltage V₀ (or charge q₀), then connected to inductor:

Energy in Capacitor:

U_E = \frac{1}{2}CV^2 = \frac{q^2}{2C}

Energy in Inductor:

U_B = \frac{1}{2}LI^2

Total Energy (Constant):

U = U_E + U_B = \frac{q_0^2}{2C} = \frac{1}{2}LI_0^2

Energy Exchange Cycle

Energy oscillates between C and L, total remains constant

8.2 Differential Equation and Solution

Mathematical Analysis

Differential Equation:

Applying Kirchhoff's voltage law around the LC loop:

\frac{q}{C} + L\frac{dI}{dt} = 0

Since I = dq/dt:

\frac{q}{C} + L\frac{d^2q}{dt^2} = 0

\boxed{\frac{d^2q}{dt^2} + \frac{1}{LC}q = 0}

This is SHM equation with ω² = 1/LC

Solution: Charge Variation

q(t) = q_0 \cos(\omega t + \phi)

q(t) = q_0 \sin(\omega t + \phi)

φ depends on initial conditions

Current Variation

I = \frac{dq}{dt} = -\omega q_0 \sin(\omega t + \phi)

I = I_0 \cos(\omega t + \phi + \frac{\pi}{2})

Where I₀ = ωq₀ = q₀/√(LC)

Key Parameters:

Angular Frequency:

\omega = \frac{1}{\sqrt{LC}}

Time Period:

T = 2\pi\sqrt{LC}

Frequency:

f = \frac{1}{2\pi\sqrt{LC}}

Maximum Current:

I_0 = \frac{q_0}{\sqrt{LC}} = q_0\omega

8.3 Analogy with Mechanical SHM

Mechanical (Spring-Mass)	Electrical (LC Circuit)
Mass (m)	Inductance (L)
Spring constant (k)	1/C (reciprocal of capacitance)
Displacement (x)	Charge (q)
Velocity (v = dx/dt)	Current (I = dq/dt)
Momentum (mv)	Magnetic flux linkage (LI)
KE = ½mv²	Magnetic Energy = ½LI²
PE = ½kx²	Electric Energy = ½q²/C
Damping (friction)	Resistance (R)
ω = √(k/m)	ω = 1/√(LC)
T = 2π√(m/k)	T = 2π√(LC)

8.4 Damped LC Oscillations (LCR Circuit)

Effect of Resistance

With Resistance (Damped)

Modified differential equation:

\frac{d^2q}{dt^2} + \frac{R}{L}\frac{dq}{dt} + \frac{1}{LC}q = 0

Solution (underdamped case):

q(t) = q_0 e^{-\gamma t}\cos(\omega' t + \phi)

Damping coefficient: γ = R/(2L)

Damped frequency: ω' = √(ω₀² - γ²)

where ω₀ = 1/√(LC)

Types of Damping

Underdamped (R² < 4L/C)

Oscillations with decreasing amplitude

Most common case, used in radio tuning

Critically Damped (R² = 4L/C)

Fastest return to equilibrium without oscillation

R_critical = 2√(L/C)

Overdamped (R² > 4L/C)

Slow return, no oscillation

Energy dissipated quickly

📝 Solved Example 8 (JEE Advanced Pattern)

Question: An LC circuit has L = 20 mH and C = 50 μF. The capacitor is initially charged to 100 μC. Find: (a) Angular frequency (b) Frequency (c) Time period (d) Maximum current (e) Maximum energy stored

Solution:

Given:

L = 20 mH = 20 × 10⁻³ H

C = 50 μF = 50 × 10⁻⁶ F

q₀ = 100 μC = 100 × 10⁻⁶ C

(a) Angular Frequency:

\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{20 \times 10^{-3} \times 50 \times 10^{-6}}}

\omega = \frac{1}{\sqrt{10^{-6}}} = \frac{1}{10^{-3}} = 1000 \text{ rad/s}

(b) Frequency:

f = \frac{\omega}{2\pi} = \frac{1000}{2 \times 3.14} = 159.2 \text{ Hz}

(c) Time Period:

T = \frac{1}{f} = \frac{1}{159.2} = 6.28 \text{ ms}

Or: T = 2π√(LC) = 2π × 10⁻³ = 6.28 ms

(d) Maximum Current:

I_0 = \omega q_0 = 1000 \times 100 \times 10^{-6}

\boxed{I_0 = 0.1 \text{ A} = 100 \text{ mA}}

(e) Maximum Energy Stored:

U = \frac{q_0^2}{2C} = \frac{(100 \times 10^{-6})^2}{2 \times 50 \times 10^{-6}}

U = \frac{10^{-8}}{10^{-4}} = 10^{-4} \text{ J}

\boxed{U = 0.1 \text{ mJ} = 100 \text{ μJ}}

Verification:

U = ½LI₀² = ½ × 20 × 10⁻³ × (0.1)² = ½ × 20 × 10⁻³ × 0.01

U = 10⁻⁴ J = 0.1 mJ ✓

💡 Quick Formulas - LC Oscillations

Frequency & Time Period:

ω = 1/√(LC)
f = 1/[2π√(LC)]
T = 2π√(LC)

Energy Relations:

U_total = q₀²/(2C) = ½LI₀²
I₀ = ωq₀ = q₀/√(LC)
q₀ = CV₀ (initial charge)

Instantaneous Values:

q = q₀ cos(ωt)
I = -ωq₀ sin(ωt) = I₀ sin(ωt + π)
U_E = q²/(2C)
U_B = ½LI²

At Special Times:

t = 0: q = q₀, I = 0, All energy in C
t = T/4: q = 0, I = I₀, All energy in L
t = T/2: q = -q₀, I = 0, All energy in C
t = 3T/4: q = 0, I = -I₀, All energy in L

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ RLC Series Circuit: 30%
✓ AC Power & Power Factor: 25%
✓ Transformers: 20%
✓ Resonance: 15%
✓ LC Oscillations: 10%

JEE Advanced (Last 5 Years)

✓ Resonance & Q-factor: 35%
✓ LC Oscillations: 25%
✓ Complex RLC Problems: 20%
✓ Phasor Analysis: 15%
✓ Transformer Efficiency: 5%

Weightage Analysis

JEE Main: 8-12 marks (2-3 questions)

JEE Advanced: 12-18 marks (3-4 questions)

Difficulty Level: Medium to High

Time Required: 4-5 hours study

Recent JEE Questions (2023-2024)

JEE Main 2024 (Jan):

In a series LCR circuit, R = 100Ω, L = 0.5H, C = 10μF. If an AC source of 200V is connected, find the current at resonance.

Answer: I = V/R = 200/100 = 2A (at resonance Z = R)

JEE Main 2024 (April):

A transformer has primary turns 1000 and secondary turns 2000. Primary voltage is 200V. Find secondary voltage and type of transformer.

Answer: V_s = 400V, Step-up transformer

JEE Advanced 2023:

In an LC circuit, L = 1mH and C = 1μF. If initial charge on capacitor is 10μC, find the frequency of oscillation and maximum current.

Answer: f = 5033 Hz, I₀ = 0.316A

Download Complete AC PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

An AC voltage is V = 220√2 sin(100πt). Find peak voltage, RMS voltage, and frequency.
Calculate X_L for L = 50 mH at f = 50 Hz. If V = 110V, find current.
Calculate X_C for C = 100 μF at f = 50 Hz. If current is 2A, find voltage.
In a pure resistor circuit, if V = 100V and I = 2A, find power consumed.
A transformer has N_p = 200, N_s = 1000. If V_p = 220V, find V_s.
Find resonance frequency of LC circuit with L = 10 mH and C = 10 μF.
In series RLC circuit, R = 10Ω, X_L = 20Ω, X_C = 12Ω. Find Z and φ.
Power factor of a circuit is 0.8. If apparent power is 1000 VA, find true power.

Level 2: Intermediate (JEE Main/Advanced)

In series RLC circuit, R = 100Ω, L = 1H, C = 10μF connected to 200V, 50Hz supply. Find: (a) Impedance (b) Current (c) Voltage across each element (d) Phase angle (e) Power factor
A 100V AC source of variable frequency is connected to RLC series circuit. At resonance, current is 5A. If Q-factor is 10, find bandwidth.
A coil of inductance 0.1H and resistance 10Ω is connected to 220V, 50Hz supply. Find power consumed and wattless current.
In an LC circuit, L = 0.5H and C = 8μF. Initially C has charge of 4μC. Find frequency, max current, and energy stored.
A transformer has efficiency 90%. If input is 200V, 5A and output is 500V, find output current.
Power transmitted at 132 kV with 10% loss. If voltage is increased to 264 kV, find new loss percentage (same power transmitted).
An LCR circuit has L = 20mH, C = 50nF, R = 40Ω. Find Q-factor and bandwidth.
Two AC voltages V₁ = 100 sin ωt and V₂ = 100 cos ωt are applied in series. Find resultant voltage.

Level 3: Advanced (JEE Advanced/Olympiad)

In series RLC circuit, if frequency is changed from f₁ (below resonance) to f₂ (above resonance), show that resonance frequency f₀ = √(f₁f₂) if current is same at both frequencies.
A series LCR circuit with R = 10Ω has Q = 100 at resonance frequency 1000 rad/s. Find the voltages across L and C if source voltage is 10V.
An LC circuit with L = 1H and C = 1μF has initial charge 10μC on capacitor. At what time after connection is the magnetic energy equal to electric energy?
A load of power factor 0.6 lagging draws 10 kW at 230V. Find capacitance required to improve power factor to 0.9 at 50 Hz.
In a transformer, primary coil has self-inductance 4H and secondary has 1H. Coefficient of coupling is 0.9. Find mutual inductance and turns ratio.
An LCR circuit is connected to AC source. At angular frequency ω₁, impedance is Z₁ = √2 R. At ω₂ = 2ω₁, impedance is Z₂. Find Z₂ in terms of R.
Show that in a series RLC circuit, the maximum voltage across capacitor occurs at frequency ω_max = ω₀√(1 - 1/2Q²) where ω₀ is resonance frequency.
A damped LC circuit has L = 10mH, C = 1μF, R = 2Ω. Find: (a) Natural frequency (b) Damping coefficient (c) Time for amplitude to decay to 1/e (d) Is it underdamped, critically damped, or overdamped?

Get Solutions PDF with Step-by-Step Working

Year	JEE Main	JEE Advanced	Topic Focus
2024	3 Questions (12 marks)	4 Questions (14 marks)	RLC resonance, Transformer, LC oscillations
2023	2 Questions (8 marks)	3 Questions (11 marks)	Power factor, Impedance, Q-factor
2022	3 Questions (12 marks)	3 Questions (12 marks)	Series RLC, Transformer efficiency
2021	2 Questions (8 marks)	4 Questions (15 marks)	Resonance, LC energy, Phasor
2020	2 Questions (8 marks)	3 Questions (10 marks)	RLC circuit, Transmission

📑 Quick Navigation - Alternating Current

Alternating Current JEE Main & Advanced 2025-26

AC Voltage and Current

1.1 AC vs DC: Fundamental Differences

1.2 Mathematical Representation of AC

Instantaneous Values

AC Voltage

AC Current

1.3 Peak, Average, and RMS Values

Peak Value

Average Value

RMS Value

💡 Why RMS Value is Important?

1.4 Important Relationships & Formulas

📝 Solved Example 1

⚠️ Common JEE Mistakes

Phasor Diagrams

2.1 What is a Phasor?

Mathematical Definition

Key Properties

2.2 Phasor Addition and Subtraction

Vector Addition Rules

If V₁ = V₀₁∠φ₁ and V₂ = V₀₂∠φ₂

Special Cases

🎯 Phasor Diagram Shortcuts for JEE

📝 Solved Example 2

AC Through Pure R, L, and C

3.1 AC Through Pure Resistance (R)

Key Relationships

Important Properties

3.2 AC Through Pure Inductance (L)

Key Relationships

Important Properties

3.3 AC Through Pure Capacitance (C)

Key Relationships

Important Properties

3.4 Comparison Table: R vs L vs C

💡 Memory Trick: "CIVIL"

📝 Solved Example 3 (JEE Main 2022 Pattern)

Series RLC Circuit

4.1 Impedance of Series RLC Circuit

Impedance Derivation

Phase Angle

Current in Series RLC:

4.2 Voltage Relationships in Series RLC

Individual Voltages

Total Voltage

4.3 Phasor Diagram for Series RLC

Complete Phasor Diagram (X_L > X_C case)

4.4 Important Formulas Summary

📝 Solved Example 4 (JEE Advanced Pattern)

⚠️ Common Mistakes in RLC Problems

Resonance in AC Circuits

5.1 Condition for Resonance

Resonance Occurs When

Mathematical Condition

Physical Meaning

5.2 Properties at Resonance

Impedance

Current

Phase Angle

5.3 Quality Factor (Q-factor)

Definition & Formulas

Physical Significance

Voltage Magnification

5.4 Bandwidth and Sharpness of Resonance

Bandwidth (Δω or Δf)

Resonance Curve

🎯 Applications of Resonance

📝 Solved Example 5 (JEE Advanced 2021 Type)

AC Power

6.1 Types of Power in AC Circuits

True Power (P)

Reactive Power (Q)

Apparent Power (S)

Power Triangle

Relationships

6.2 Power Factor

Definition and Significance

Mathematical Definition