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Alternating Current and Electromagnetic Waves – JEE Main Physics Formula Sheet | AC Circuits, LCR Series, Resonance, Power Factor, EM Waves, Speed of Light & All Formulas
This is the complete JEE Main Physics Formula Sheet for Alternating Current and Electromagnetic Waves from the Aakash Rapid Revision & Formula Bank. This chapter covers: AC Fundamentals — V=V₀sinωt, I=I₀sinωt, V_rms=V₀/√2, I_rms=I₀/√2, average value 2V₀/π; AC Circuit Elements — pure R (in phase), pure L (V leads I by 90°, X_L=ωL), pure C (I leads V by 90°, X_C=1/ωC); Series LCR Circuit — Z=√(R²+(X_L–X_C)²), tanφ=(X_L–X_C)/R, I_rms=V_rms/Z; Resonance — ω₀=1/√(LC), f₀=1/2π√(LC), Z_min=R, I_max=V/R, V_L=V_C=QV; Quality Factor — Q=ω₀L/R=1/ω₀CR=(1/R)√(L/C), bandwidth Δω=R/L; Power in AC — P=V_rms I_rms cosφ, cosφ=R/Z, wattless current; Transformer — V₁/V₂=N₁/N₂=I₂/I₁; Maxwell's Equations — Gauss (electric and magnetic), Faraday, Ampere-Maxwell; Displacement Current — I_d=ε₀dΦ_E/dt; EM Waves — c=1/√(μ₀ε₀)=3×10⁸ m/s, c=Eₘ/Bₘ, v=c/n, E and B are perpendicular and in phase, intensity I=½cε₀E₀²=cε₀E_rms²; Radiation Pressure — P=I/c (absorbed), P=2I/c (reflected); EM Spectrum — wavelength ranges and uses of all seven types. Alternating Current and Electromagnetic Waves contributes 3–5 questions in every JEE Main session. Download the Free PDF for all formulas in one JEE Main exam-ready reference.
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Scroll to explore all AC and Electromagnetic Waves formulas — JEE Main Physics Formula Sheet
Topics Covered in This Alternating Current and Electromagnetic Waves Formula Sheet
Introduction: Why AC and Electromagnetic Waves Is a High-Priority JEE Main Physics Chapter
Alternating Current and Electromagnetic Waves represents the culmination of two chapters of classical electromagnetism. AC circuits bring together resistors, capacitors, and inductors into analysable networks where the concepts of reactance, impedance, phase angle, and resonance give a complete picture of how electrical energy is transmitted, stored, and dissipated. Electromagnetic waves are the physical consequence of Maxwell's equations — the insight that a changing electric field creates a magnetic field (displacement current) and a changing magnetic field creates an electric field (Faraday's law), enabling a self-sustaining wave to propagate through free space at the speed c=1/√(μ₀ε₀)=3×10⁸ m/s.
For JEE Main physics, AC and Electromagnetic Waves contributes 3–5 questions per session. AC questions test: LCR impedance Z=√(R²+(X_L–X_C)²), resonance ω₀=1/√(LC), quality factor Q=ω₀L/R, power factor cosφ=R/Z, and transformer ratio V₁/V₂=N₁/N₂. EM wave questions test: speed c=1/√(μ₀ε₀), E₀/B₀=c, intensity I=½cε₀E₀², radiation pressure, and EM spectrum classification.
Download the Free PDF for AC and Electromagnetic Waves to access all AC circuit formulas, resonance, quality factor, power, transformer, displacement current, Maxwell's equations, EM wave properties, intensity, radiation pressure, and EM spectrum in one structured JEE Main physics revision reference.
Key Concepts and Formulas in Alternating Current and Electromagnetic Waves
AC Fundamentals — Sinusoidal Voltage, RMS Values, Average Values
Why V_rms=V₀/√2 and Average Value 2V₀/π Are the Essential AC Fundamentals for JEE Main
AC Voltage and Current (from Aakash PDF — AC and Electromagnetic Waves JEE Main):
AC voltage: V(t) = V₀ sin(ωt) or V₀ cos(ωt)
AC current: I(t) = I₀ sin(ωt + φ) where φ = phase of I with respect to V
Angular frequency: ω = 2πf = 2π/T. Frequency f in Hz. Period T in seconds.
V₀ = peak (maximum) voltage; I₀ = peak current.
RMS (Root Mean Square) Values (from Aakash PDF — AC and EM Waves JEE Main):
RMS value = value of DC that would produce the same heating effect.
V_rms = V₀/√2 ≈ 0.707V₀
I_rms = I₀/√2 ≈ 0.707I₀
Derivation: V²_rms = ⟨V²⟩ = ⟨V₀²sin²ωt⟩ = V₀²/2 → V_rms = V₀/√2.
Household supply (India): V_rms = 220V → V₀ = 220√2 ≈ 311V.
Average Values (from Aakash PDF — AC and EM Waves JEE Main):
Average over complete cycle: ⟨sinωt⟩ = 0 → ⟨V⟩_full = 0
Average over half cycle (0 to π): ⟨V⟩_half = 2V₀/π ≈ 0.637V₀
Derivation: ⟨V⟩_half = (1/π)∫₀^π V₀ sinωt d(ωt) = V₀(2/π) = 2V₀/π
Peak factor (crest factor): V₀/V_rms = √2 ≈ 1.414
Form factor: V_rms/V_avg(half) = (V₀/√2)/(2V₀/π) = π/(2√2) ≈ 1.11
Download the Free PDF for AC and Electromagnetic Waves for all AC fundamentals examples for JEE Main.
AC Circuit Elements — Pure R, Pure L, Pure C and Phase Relations
Why X_L=ωL, X_C=1/ωC and CIVIL Mnemonic Are the Core AC Circuit JEE Main Formulas
Pure Resistor in AC (from Aakash PDF — AC and EM Waves JEE Main):
V = V₀sinωt → I = (V₀/R)sinωt = I₀ sinωt
V and I are in phase (φ = 0). I₀ = V₀/R; I_rms = V_rms/R.
Power: P = V_rms I_rms = I²_rms R = V²_rms/R. Resistor dissipates real power.
Pure Inductor in AC (from Aakash PDF — AC and EM Waves JEE Main):
V = V₀sinωt; Back EMF from inductor: V = L dI/dt → I = –(V₀/ωL)cosωt = I₀sin(ωt–π/2)
Voltage leads current by 90° (current lags voltage by 90°).
Inductive reactance: X_L = ωL = 2πfL (unit: Ω)
I₀ = V₀/X_L; I_rms = V_rms/X_L
X_L ∝ f: at high frequency, X_L is large (inductor blocks high frequency). At DC (f=0): X_L=0 (inductor is short circuit).
Power: P = V_rms I_rms cos90° = 0. Inductor stores and returns energy — no net dissipation.
Pure Capacitor in AC (from Aakash PDF — AC and EM Waves JEE Main):
V = V₀sinωt; Q = CV = CV₀sinωt; I = dQ/dt = CV₀ω cosωt = I₀sin(ωt+π/2)
Current leads voltage by 90° (voltage lags current by 90°).
Capacitive reactance: X_C = 1/(ωC) = 1/(2πfC) (unit: Ω)
I₀ = V₀/X_C; I_rms = V_rms/X_C
X_C ∝ 1/f: at low frequency (DC, f=0), X_C→∞ (capacitor blocks DC). At high frequency, X_C→0 (capacitor passes high frequency).
Power: P = V_rms I_rms cos90° = 0. Capacitor stores and returns energy — no net dissipation.
CIVIL Mnemonic (from Aakash PDF — AC and EM Waves JEE Main):
C-I-V-I-L: In a Capacitor, I leads V. In an Inductor (L), V leads I.
Memory: "ELI the ICE man" — E leads I in L (inductor); I leads E in C (capacitor).
Impedance analogy: X_L and X_C behave like resistances but cause phase shifts. R, X_L, X_C all in Ohms.
Download the Free PDF for AC and Electromagnetic Waves for all AC element examples for JEE Main.
Series LCR Circuit — Impedance, Phase, Resonance, Quality Factor, Power
Why LCR Impedance Z, Resonance ω₀=1/√(LC), Q Factor and cosφ=R/Z Are the Highest-Tested AC JEE Main Formulas
Series LCR Impedance (from Aakash PDF — AC and EM Waves JEE Main):
R, L, C in series with AC V=V₀sinωt. Common current I flows through all. Phasor voltages: V_R along I; V_L 90° ahead; V_C 90° behind.
Net voltage phasor: V = √(V_R² + (V_L – V_C)²)
Z = V/I = √(R² + (X_L – X_C)²) = √(R² + (ωL – 1/ωC)²)
tanφ = (X_L – X_C)/R
I_rms = V_rms/Z; I₀ = V₀/Z
φ > 0 (X_L > X_C): inductive → V leads I. φ < 0 (X_C > X_L): capacitive → I leads V. φ = 0: resistive (resonance).
Voltage across elements at resonance: V_R=IR (in phase); V_L=IX_L=IωL (leads I); V_C=IX_C=I/ωC (lags I).
Resonance (from Aakash PDF — AC and EM Waves JEE Main):
Condition: X_L = X_C → ωL = 1/ωC
ω₀ = 1/√(LC) (resonant angular frequency)
f₀ = 1/(2π√(LC)) (resonant frequency)
At resonance: Z = R (minimum); I = V_rms/R (maximum); φ = 0; cosφ = 1; P = I²_rms R (maximum).
V_L = V_C = I×X_L₀ = I×X_C₀ = QV (can exceed supply voltage by factor Q).
Quality Factor Q (from Aakash PDF — AC and EM Waves JEE Main):
Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C)
Voltage magnification: V_L = V_C = Q × V (at resonance). A high-Q circuit produces voltages Q times larger than input across L and C.
Bandwidth: Δω = R/L = ω₀/Q
Half-power frequencies: ω₁ = ω₀ – R/2L; ω₂ = ω₀ + R/2L. At these: I = I_max/√2; P = P_max/2; Z = R√2.
High Q → narrow bandwidth → selective circuit (radio tuner uses this for station selection).
Power in AC Circuit (from Aakash PDF — AC and EM Waves JEE Main):
P = V_rms I_rms cosφ (average power)
Power factor: cosφ = R/Z
For pure R: cosφ=1, P=V_rms I_rms. For pure L or C: cosφ=0, P=0 (wattless). For LCR at resonance: cosφ=1, maximum P.
Wattless current: component I_rms sinφ (in quadrature with V) produces zero power.
Apparent power: S=V_rms I_rms (VA). Real power: P=S cosφ (W). Reactive power: Q_r=S sinφ (VAR).
Download the Free PDF for AC and Electromagnetic Waves for all LCR circuit examples for JEE Main.
Transformer — Turn Ratio, Step-Up, Step-Down, Efficiency, Losses
Why Transformer Ratio V₁/V₂=N₁/N₂=I₂/I₁ Is the Most Direct AC JEE Main Formula
Transformer Principle (from Aakash PDF — AC and EM Waves JEE Main):
A transformer transfers AC electrical energy using mutual induction. Primary coil (N₁ turns) and secondary coil (N₂ turns) share a common soft iron core.
Same flux through both coils (ideal): V₁/N₁ = V₂/N₂ = dΦ/dt
Voltage transformation: V₁/V₂ = N₁/N₂
Current transformation (ideal, P₁=P₂): I₁/I₂ = N₂/N₁ = V₂/V₁
Turn ratio: K = N₂/N₁
V₂ = KV₁; I₂ = I₁/K (for ideal transformer).
Types (from Aakash PDF — AC and EM Waves JEE Main):
Step-Up transformer: N₂ > N₁ → K > 1 → V₂ > V₁; I₂ < I₁. Used in power transmission (high voltage, low current → less I²R loss).
Step-Down transformer: N₂ < N₁ → K < 1 → V₂ < V₁; I₂ > I₁. Used at households (bring high-voltage transmission down to 220V).
Ideal transformer: V₁I₁ = V₂I₂ (input power = output power, efficiency 100%).
Losses in Real Transformer (from Aakash PDF — AC and EM Waves JEE Main):
(1) Copper loss (winding resistance): I²R heat in primary and secondary coils. Reduced by thick wire.
(2) Eddy current loss: induced circular currents in iron core → heat. Reduced by laminating the core.
(3) Hysteresis loss: energy to magnetise and demagnetise the core each AC cycle. Reduced by using soft iron or silicon steel (narrow hysteresis loop).
(4) Flux leakage: not all magnetic flux from primary links secondary. Reduced by closed iron core geometry.
Efficiency: η = (V₂I₂/V₁I₁) × 100% (typically 95–99% for well-designed power transformers).
Why AC, not DC: DC produces constant flux → no dΦ/dt → no induced EMF in secondary → transformer does NOT work with DC.
Download the Free PDF for AC and Electromagnetic Waves for all transformer examples for JEE Main.
Displacement Current and Maxwell's Equations
Why Displacement Current I_d=ε₀dΦ_E/dt and Maxwell's Modified Ampere Law Complete the Electromagnetic Framework
Displacement Current (from Aakash PDF — AC and EM Waves JEE Main):
In a capacitor being charged, no real current flows through the gap between the plates. Yet the changing electric field E in the gap (due to changing charge on plates) creates an equivalent "current" — the displacement current.
Displacement current: I_d = ε₀ dΦ_E/dt
where Φ_E = electric flux through the gap = EA (between plates). dΦ_E/dt = A(dE/dt) = (A/ε₀d)dσ/dt... = dQ/dt = I (the actual conduction current).
Key result: displacement current I_d = conduction current I in the circuit. At every instant: I_d (through capacitor gap) = I (through connecting wires). Continuity of current is maintained.
Direction of B due to displacement current: same rules as for conduction current (right-hand rule with I_d).
Maxwell's Four Equations (from Aakash PDF — AC and EM Waves JEE Main):
(1) Gauss's Law for Electricity: ∮E·dA = Q_enc/ε₀ (electric flux ∝ enclosed charge)
(2) Gauss's Law for Magnetism: ∮B·dA = 0 (no magnetic monopoles; magnetic field lines are closed loops)
(3) Faraday's Law: ∮E·dl = –dΦ_B/dt (changing B creates E)
(4) Ampere-Maxwell Law: ∮B·dl = μ₀(I_c + I_d) = μ₀I_c + μ₀ε₀ dΦ_E/dt (current + changing E creates B)
Together, Maxwell's four equations completely describe all electromagnetic phenomena. They predict EM waves travelling at c=1/√(μ₀ε₀). Download the Free PDF for AC and Electromagnetic Waves for all displacement current and Maxwell examples for JEE Main.
Electromagnetic Waves — Properties, Speed, Intensity, Radiation Pressure, EM Spectrum
Why c=1/√(μ₀ε₀), E₀/B₀=c, Intensity I=½cε₀E₀², and Radiation Pressure Are the Most-Tested EM Wave JEE Main Formulas
Properties of Electromagnetic Waves (from Aakash PDF — AC and EM Waves JEE Main):
(1) Speed in vacuum: c = 1/√(μ₀ε₀) = 3×10⁸ m/s
μ₀ = 4π×10⁻⁷ T·m/A; ε₀ = 8.85×10⁻¹² C²/N·m². Numerically: 1/√(4π×10⁻⁷×8.85×10⁻¹²) = 3×10⁸ m/s ✓
(2) E and B are perpendicular to each other and both are perpendicular to the direction of propagation (transverse wave).
(3) E and B are in phase (both reach maxima and minima at the same time and place).
(4) Ratio of peak fields: E₀/B₀ = c or E_rms/B_rms = c
(5) Speed in a medium: v = c/n where n = refractive index (n ≥ 1 for any medium).
(6) EM waves carry energy, momentum, and angular momentum. They do NOT require a medium.
(7) EM waves can be polarised (transverse waves — can be linearly/circularly polarised).
Intensity of EM Wave (from Aakash PDF — AC and EM Waves JEE Main):
Intensity I = average energy crossing unit area per unit time (W/m²).
I = ½cε₀E₀² = cε₀E_rms²
Also: I = E₀B₀/(2μ₀) = cB₀²/(2μ₀) = E_rms B_rms/μ₀
Poynting vector: S = (1/μ₀)(E×B) = E×H (energy flux density; magnitude = intensity at that instant)
|S| = E×B/μ₀ = E×H
Average intensity: ⟨S⟩ = I = ½cε₀E₀² = ½E₀B₀/μ₀
For a point source of power P: I = P/(4πr²) (inverse square law).
Radiation Pressure (from Aakash PDF — AC and EM Waves JEE Main):
EM waves carry momentum. When absorbed or reflected, they exert pressure on surfaces:
Radiation pressure (absorbed): P_rad = I/c
Radiation pressure (perfectly reflected): P_rad = 2I/c
Momentum delivered per second (absorbed): dp/dt = I×A/c. For reflected: dp/dt = 2IA/c.
Energy density: u = ε₀E² = B²/μ₀ (instantaneous); average u_avg = ½ε₀E₀² = B₀²/2μ₀.
Relation: I = u_avg × c (intensity = energy density × wave speed).
Electromagnetic Spectrum (from Aakash PDF — AC and EM Waves JEE Main):
All EM waves travel at c=3×10⁸ m/s in vacuum. c=fλ. Classified by wavelength (or frequency):
(1) Radio waves: λ > 0.1 m (f < 3×10⁹ Hz). Used in AM/FM radio, TV broadcasting, long-range communication.
(2) Microwaves: 0.1 m > λ > 1 mm (f = 3×10⁹ to 3×10¹¹ Hz). Used in RADAR, satellite communication, microwave ovens, mobile phones.
(3) Infrared: 1 mm > λ > 700 nm. Produced by hot objects. Used in TV remotes, night-vision cameras, heating (greenhouses), medical therapy.
(4) Visible light: 700 nm to 400 nm (VIBGYOR — violet 400nm to red 700nm). Detected by human eye. f ≈ 4×10¹⁴ to 7×10¹⁴ Hz.
(5) Ultraviolet (UV): 400 nm to 1 nm. Produced by sun and arc lamps. Used in sterilisation, LASIK surgery, detecting forged banknotes, vitamin D synthesis.
(6) X-rays: 1 nm to 0.001 nm (10 pm). Produced by X-ray tubes (high-energy electron bombardment of metals). Used in medical imaging, crystallography, cancer therapy.
(7) Gamma rays (γ-rays): λ < 0.001 nm (< 10 pm). Produced by nuclear transitions and radioactive decay. Highest frequency, highest energy, most penetrating. Used in cancer treatment, sterilising medical equipment.
Memory for decreasing wavelength (increasing frequency): Radio Micro InfraRed Visible UltraViolet X Gamma = RM IR VU XG. Download the Free PDF for AC and Electromagnetic Waves for all EM wave and spectrum examples for JEE Main.
Download Free PDF — AC and Electromagnetic Waves JEE Main Formula Sheet
All AC and Electromagnetic Waves formulas from the Aakash Rapid Revision PDF: V=V₀sinωt; I=I₀sin(ωt+φ); V_rms=V₀/√2; I_rms=I₀/√2; average (half cycle)=2V₀/π; peak factor=√2; R(in phase cosφ=1 P=I²R); L(V leads I 90° X_L=ωL P=0 blocks high f); C(I leads V 90° X_C=1/ωC P=0 blocks DC); CIVIL mnemonic; Z=√(R²+(X_L–X_C)²); tanφ=(X_L–X_C)/R; I_rms=V_rms/Z; resonance ω₀=1/√(LC) f₀=1/2π√(LC) Z=R I=max φ=0; V_L=V_C=QV; Q=ω₀L/R=1/ω₀CR=(1/R)√(L/C); bandwidth Δω=R/L=ω₀/Q; P=V_rms I_rms cosφ; cosφ=R/Z; pure L/C P=0; transformer V₁/V₂=N₁/N₂ I₁/I₂=N₂/N₁; ideal V₁I₁=V₂I₂; step-up N₂>N₁; step-down N₂
Why AC and Electromagnetic Waves Is a Guaranteed Scoring Chapter in JEE Main Physics
The LCR resonance formula Z=R at ω₀=1/√(LC) is a 4-mark guarantee in JEE Main AC questions. At resonance: impedance Z=R (minimum), current I=V/R (maximum), power factor cosφ=1 (maximum power), and both V_L and V_C equal Q times the supply voltage. A single problem can test all four simultaneously. The quality factor Q=ω₀L/R also gives the bandwidth Δω=R/L=ω₀/Q — so knowing Q tells you how "sharp" or "selective" the circuit is.
The EM wave result c=E₀/B₀ and intensity I=½cε₀E₀² are the two most-tested electromagnetic wave formulas in JEE Main. The ratio c=E₀/B₀ gives a direct way to find B₀ if E₀ is given (B₀=E₀/c). The intensity formula I=½cε₀E₀² connects amplitude to power per area. If E₀ doubles, I quadruples (I∝E₀²). Radiation pressure P=I/c (absorbed) and P=2I/c (reflected) follow directly from the momentum carried by EM waves.
Transformer formula V₁/V₂=N₁/N₂=I₂/I₁ is the simplest calculation in AC JEE Main — zero-error guarantee if the formula is known. Step-up (N₂>N₁): voltage increases, current decreases. Step-down: opposite. The power equation V₁I₁=V₂I₂ (ideal) is the energy conservation condition. For real efficiency η: output power = η × input power. Download the Free PDF for AC and Electromagnetic Waves to have all formulas ready.
Who Should Use This AC and Electromagnetic Waves Formula Sheet?
Learning Outcomes After Completing AC and Electromagnetic Waves
After working through AC and Electromagnetic Waves using this formula sheet, a student should accomplish: On AC fundamentals: write V=V₀sinωt; compute V_rms=V₀/√2; compute I_rms=I₀/√2; state average over full cycle=0 and half cycle=2V₀/π; identify peak factor=√2. On AC elements: state phase relations (R in phase; L leads 90°; C lags 90°); compute X_L=ωL and X_C=1/ωC; state DC and high-frequency behavior; apply CIVIL mnemonic; state P=0 for pure L and C.
On LCR series: compute Z=√(R²+(X_L–X_C)²); compute tanφ=(X_L–X_C)/R; find I_rms=V_rms/Z; state resonance condition ω₀=1/√(LC); compute f₀=1/2π√(LC); state Z_min=R, I_max, φ=0 at resonance; compute Q=ω₀L/R=(1/R)√(L/C); state V_L=V_C=QV; compute bandwidth Δω=R/L. On power: compute P=V_rms I_rms cosφ with cosφ=R/Z; state wattless current condition. On transformer: apply V₁/V₂=N₁/N₂=I₂/I₁; identify step-up and step-down; compute efficiency; state why AC (not DC) is needed; name and explain all four losses.
On EM waves: state I_d=ε₀dΦ_E/dt; write all four Maxwell's equations; state c=1/√(μ₀ε₀)=3×10⁸ m/s; state E₀/B₀=c; state E and B are perpendicular and in phase; compute v=c/n; compute I=½cε₀E₀²; compute radiation pressure I/c (absorbed) and 2I/c (reflected); identify all seven EM wave types in order with wavelength ranges and applications. Download the Free PDF for AC and Electromagnetic Waves to test all outcomes before your JEE Main exam.
Get the Free PDF for AC and Electromagnetic Waves — JEE Main Quick Revision
The Aakash Rapid Revision & Formula Bank PDF for Alternating Current and Electromagnetic Waves contains all AC fundamentals, circuit element phase relations, LCR series impedance and resonance, quality factor, bandwidth, power and power factor, transformer, displacement current, Maxwell's equations, EM wave properties, intensity, radiation pressure, and complete EM spectrum table in one structured JEE Main physics reference.
Conclusion — AC and Electromagnetic Waves: The Modern Physics Gateway in JEE Main
Alternating Current and Electromagnetic Waves is where classical electromagnetism reaches its most practical and most fundamental conclusions simultaneously. On the practical side, AC circuits power the modern world — LCR resonance is the principle behind radio tuning, LCR filters are used in signal processing, and transformers enable efficient long-distance power transmission. On the fundamental side, Maxwell's equations predict that c=1/√(μ₀ε₀)=3×10⁸ m/s — the speed of light — making light itself an electromagnetic wave and unifying optics with electromagnetism in one extraordinary synthesis.
Five most JEE Main-tested results: (1) Z=√(R²+(X_L–X_C)²) and resonance ω₀=1/√(LC) — the core LCR results; (2) Q=ω₀L/R with bandwidth Δω=ω₀/Q — quality factor; (3) P=V_rms I_rms cosφ with cosφ=R/Z — power factor; (4) c=1/√(μ₀ε₀)=E₀/B₀=3×10⁸ m/s — speed of EM waves; (5) I=½cε₀E₀² and radiation pressure I/c or 2I/c — EM wave intensity. Use this page and the Free PDF Download for AC and Electromagnetic Waves as your complete JEE Main revision foundation.
Frequently Asked Questions — AC and Electromagnetic Waves Formulas
What is the series LCR circuit impedance formula in AC and how is resonance achieved?
In AC and Electromagnetic Waves, series LCR circuit from Aakash PDF: R, L, C in series with AC source V=V₀sinωt. All carry the same current I. Phasor voltages: V_R=IR (in phase with I); V_L=IX_L (90° ahead); V_C=IX_C (90° behind). Net voltage phasor: V=√(V_R²+(V_L–V_C)²)=I√(R²+(X_L–X_C)²). Impedance Z=V/I=√(R²+(X_L–X_C)²)=√(R²+(ωL–1/ωC)²). Phase: tanφ=(X_L–X_C)/R. Current I_rms=V_rms/Z. At low ω: X_C>>X_L → capacitive (I leads V). At high ω: X_L>>X_C → inductive (V leads I). At resonance: X_L=X_C → ωL=1/ωC → ω₀=1/√(LC) → Z=R (minimum), I=V/R (maximum), φ=0 (in phase), cosφ=1, P=I²R (maximum). V_L=V_C=IX_L₀=I(ω₀L)=Q×V where Q=ω₀L/R. JEE Main AC: "R=10Ω, L=0.1H, C=100μF, V_rms=100V, f=50Hz. Find Z" → X_L=2π×50×0.1=31.4Ω; X_C=1/(2π×50×100×10⁻⁶)=31.8Ω; Z=√(100+(31.4–31.8)²)=√(100+0.16)≈10Ω (near resonance).
What is the quality factor Q in AC circuits and how does it relate to bandwidth?
In AC and Electromagnetic Waves, quality factor Q from Aakash PDF: Q=ω₀L/R=1/ω₀CR=(1/R)√(L/C). Three physical meanings: (1) Voltage magnification: at resonance, V_L=V_C=Q×V_supply. Circuit amplifies voltage by factor Q. High Q circuit can have V_L hundreds of times larger than input voltage. (2) Sharpness of resonance: sharp peak at ω₀ with narrow bandwidth. (3) Q=2π×(energy stored)/(energy dissipated per cycle). Bandwidth: Δω=R/L=ω₀/Q (half-power bandwidth). At ω₁=ω₀–R/2L and ω₂=ω₀+R/2L: I=I_max/√2; P=P_max/2; Z=√2×R. Q=ω₀/Δω → higher Q → narrower bandwidth → more selective circuit. Radio tuner uses high-Q LCR to select one station (narrow bandwidth) from many. JEE Main AC: "R=2Ω, L=0.1H, C=250pF. Find ω₀, Q, bandwidth" → ω₀=1/√(0.1×250×10⁻¹²)=1/√(25×10⁻¹²)=2×10⁵ rad/s. Q=ω₀L/R=2×10⁵×0.1/2=10⁴. Bandwidth=ω₀/Q=2×10⁵/10⁴=20 rad/s. V_L at resonance = QV = 10⁴ times supply voltage!
What is power factor in AC circuits and when is it maximum?
In AC and Electromagnetic Waves, power factor from Aakash PDF: cosφ where φ=phase angle between V and I. P=V_rms I_rms cosφ. cosφ=R/Z (always; derived from phasor diagram: R is the adjacent side, Z is hypotenuse). Range: 0≤cosφ≤1. cosφ=1 (φ=0): pure resistive or LCR at resonance → maximum real power P=V_rms I_rms. cosφ=0 (φ=90°): pure L or pure C → zero real power (wattless current). Wattless component of current: I_rms sinφ (90° out of phase with V, produces zero power). Active component: I_rms cosφ (in phase with V, produces real power). Apparent power S=V_rms I_rms (VA). Real power P=S cosφ (W). Reactive power Q_r=S sinφ (VAR). For series LCR: cosφ=R/Z=R/√(R²+(X_L–X_C)²). Purely resistive: Z=R, cosφ=1. At resonance: X_L=X_C so Z=R, cosφ=1. Far from resonance: Z>>R, cosφ≈0. JEE Main AC: "V_rms=220V, I_rms=2A, P=300W. Find cosφ" → cosφ=P/(V_rms×I_rms)=300/(220×2)=300/440≈0.68. R=P/I²_rms=300/4=75Ω; Z=V_rms/I_rms=110Ω; check cosφ=75/110≈0.68 ✓.
What is displacement current and why did Maxwell introduce it in Electromagnetic Waves?
In AC and Electromagnetic Waves, displacement current from Aakash PDF: when a capacitor is being charged through a circuit, real current I flows through the wires but NOT through the gap between capacitor plates. Maxwell noticed that Ampere's law ∮B·dl=μ₀I_c was inconsistent: different Amperian surfaces for the same path gave different currents → different B values. To resolve this, Maxwell added displacement current I_d=ε₀dΦ_E/dt. Physical meaning: the changing electric flux in the capacitor gap (due to changing charge on plates) creates B just as a real current would. In the capacitor: I_d=ε₀dΦ_E/dt=ε₀A dE/dt=ε₀A d(σ/ε₀)/dt=A dσ/dt=dQ/dt=I (equals the conduction current in the wire). Modified Ampere-Maxwell law: ∮B·dl=μ₀(I_c+I_d)=μ₀I_c+μ₀ε₀dΦ_E/dt. Direction of B from I_d: same rule as I_c (right-hand rule with displacement current direction). Physical consequence: changing E→B (by displacement current) and changing B→E (by Faraday) → self-sustaining EM wave at speed c=1/√(μ₀ε₀). JEE Main EM waves: "find displacement current between capacitor plates if dV/dt=10⁸ V/s, C=1μF" → I_d=C×dV/dt=10⁻⁶×10⁸=100A (equals charging current I in wire).
What is the speed of electromagnetic waves and how is it derived from Maxwell's equations?
In AC and Electromagnetic Waves, speed of EM waves from Aakash PDF: Maxwell's equations (3) and (4) can be combined to give wave equations for E and B: ∂²E/∂x²=μ₀ε₀∂²E/∂t² and ∂²B/∂x²=μ₀ε₀∂²B/∂t². Comparing with standard wave equation ∂²y/∂x²=(1/v²)∂²y/∂t²: wave speed c=1/√(μ₀ε₀). Numerically: c=1/√(4π×10⁻⁷×8.85×10⁻¹²)=1/√(1.11×10⁻¹⁷)=1/(3.33×10⁻⁹)=3×10⁸ m/s. This was Maxwell's greatest achievement — it showed that the speed of EM waves (predicted purely from electrical measurements of μ₀ and ε₀) exactly equals the measured speed of light → LIGHT IS AN EM WAVE. In a medium of relative permittivity ε_r and relative permeability μ_r: v=1/√(μ₀μ_rε₀ε_r)=c/√(μ_rε_r)=c/n where n=√(μ_rε_r) (refractive index). For non-magnetic materials μ_r≈1: n=√ε_r. Ratio of peak fields: E₀/B₀=c (derived from Maxwell's equations). JEE Main EM: "E₀=300V/m for EM wave in vacuum. Find B₀" → B₀=E₀/c=300/3×10⁸=10⁻⁶T=1μT.
What is the intensity of an electromagnetic wave and how is it related to electric field amplitude?
In AC and Electromagnetic Waves, EM wave intensity from Aakash PDF: intensity I = power per unit area = average energy transported per unit time per unit area. For a plane EM wave with peak electric field E₀: Instantaneous energy density: u=ε₀E²+B²/μ₀=ε₀E²+ε₀c²×B²/μ₀=... average: u_avg=½ε₀E₀²+½B₀²/μ₀=½ε₀E₀²+½ε₀c²×B₀²=½ε₀E₀²+½ε₀E₀²=ε₀E₀² (Wait: B₀=E₀/c → B₀²/μ₀=E₀²/c²μ₀=ε₀E₀²). So u_avg=½ε₀E₀²+½ε₀E₀²=ε₀E₀². Intensity I=u_avg×c=cε₀E₀²... but I=½cε₀E₀² (using E_rms=E₀/√2: I=cε₀E_rms²=cε₀E₀²/2=½cε₀E₀²). Check: I=E₀B₀/2μ₀=E₀(E₀/c)/2μ₀=E₀²/2cμ₀=E₀²c²ε₀/2c=½cε₀E₀² ✓. For point source power P: I=P/4πr² (inverse square). If E₀ doubles: I quadruples (I∝E₀²). JEE Main: "intensity of sunlight at Earth=1400W/m². Find E₀" → E₀=√(2I/cε₀)=√(2×1400/3×10⁸×8.85×10⁻¹²)=√(2800/2.655×10⁻³)=√(1054415)≈1027V/m. B₀=E₀/c≈3.4×10⁻⁶T.
What is radiation pressure of electromagnetic waves in AC and EM Waves?
In AC and Electromagnetic Waves, radiation pressure from Aakash PDF: EM waves carry momentum p=U/c per unit time per unit area (where U=energy). This momentum transfer to a surface exerts radiation pressure. Absorbed surface: momentum change = I×A×Δt/c per second. Force = I×A/c. Pressure P_rad=Force/Area=I/c. Perfectly reflected surface: incoming momentum = I×A×Δt/c; outgoing = same magnitude but reversed direction. Change = 2×I×A×Δt/c. Force = 2IA/c. Pressure P_rad=2I/c. Partially absorbed (reflectivity r): P_rad=I(1+r)/c. Total absorbed: r=0 → P=I/c. Perfect mirror r=1 → P=2I/c. Physical: the pressure is extremely small for ordinary intensities. For sunlight (I=1400W/m²): P=I/c=1400/3×10⁸=4.7×10⁻⁶ Pa (very small compared to atmospheric pressure 10⁵ Pa). But: radiation pressure is important for (1) Comet tails (solar wind pushes dust outward); (2) Radiation-pressure-driven spacecraft (solar sails); (3) Optical tweezers (trapping cells with laser). Force on a surface: F=PA=IA/c (absorbed) or F=2IA/c (reflected). These radiation pressure formulas are tested directly in JEE Main as "find radiation pressure/force" questions.
What is the electromagnetic spectrum in AC and EM Waves and what are the wavelength ranges?
In AC and Electromagnetic Waves, EM spectrum from Aakash PDF (decreasing wavelength = increasing frequency = increasing energy): (1) Radio waves: λ>0.1m. Sources: oscillating circuits. Uses: AM radio (medium wave), FM radio, TV, mobile phones, long-range communication. (2) Microwaves: 1mm<λ<0.1m (f=3×10⁹ to 3×10¹¹ Hz). Sources: magnetrons, Gunn diodes. Uses: RADAR (range and direction of aircraft), satellite TV, mobile phones, microwave cooking (water molecules absorb at 2.45GHz). (3) Infrared: 700nm<λ<1mm. Sources: hot objects, sun. Uses: heating (greenhouses), TV remotes, night-vision cameras, medical thermography, infrared spectroscopy, fire alarms. (4) Visible: 400nm<λ<700nm. VIBGYOR (Violet=400nm to Red=700nm). Only type detected by human eye. Sources: hot bodies, lasers, LEDs. (5) Ultraviolet: 1nm<λ<400nm. Sources: sun (blocked by ozone layer), mercury lamps. Uses: sterilisation/disinfection (kills bacteria), fluorescent lighting, vitamin D synthesis, forgery detection, LASIK eye surgery. (6) X-rays: 0.001nm<λ<1nm. Sources: X-ray tubes, Bremsstrahlung. Uses: medical diagnosis (bones), airport security, crystal structure determination (Bragg diffraction), radiation therapy for cancer. (7) Gamma rays: λ<0.001nm. Sources: radioactive nuclei, supernovae. Uses: cancer treatment, sterilising medical instruments, nuclear physics. All travel at c=3×10⁸m/s in vacuum. c=fλ. Energy E=hf (photon energy).
What is the transformer ratio formula in AC and what are the sources of energy loss?
In AC and Electromagnetic Waves, transformer from Aakash PDF: ideal transformer: V₁/V₂=N₁/N₂ (turn ratio). Derived: same core → same dΦ/dt → EMF per turn same → V₁=N₁dΦ/dt; V₂=N₂dΦ/dt → V₁/V₂=N₁/N₂. Current ratio: ideal (no loss): P₁=P₂ → V₁I₁=V₂I₂ → I₁/I₂=V₂/V₁=N₂/N₁. Turn ratio k=N₂/N₁. Step-up: k>1 → V₂>V₁; I₂
What are the properties of electromagnetic waves in AC and EM Waves for JEE Main?
In AC and Electromagnetic Waves, EM wave properties from Aakash PDF: (1) Transverse wave: E and B perpendicular to direction of propagation and to each other. E, B, and propagation direction form right-handed triad: if propagation is +x, E could be +y, B must be +z. (2) Speed in vacuum: c=1/√(μ₀ε₀)=3×10⁸m/s. In medium v=1/√(με)=c/n where n=√(μ_rε_r). For optical media μ_r≈1: n=√ε_r. (3) E and B are in phase: both are zero and maximum at the same point and same time. (4) Ratio: E₀=cB₀ or Eₘ=cBₘ (peak values); E_rms=cB_rms. (5) No medium required: propagate through vacuum. (6) All EM waves (different frequencies) travel at same speed c in vacuum. (7) Carry energy (Poynting vector S=E×B/μ₀; ⟨|S|⟩=I=½cε₀E₀²). (8) Carry momentum: p=U/c per unit time per unit area. (9) Can be polarised (being transverse). (10) Obey superposition (can interfere and diffract). (11) Not deflected by electric or magnetic fields (no charge). (12) Photon energy E=hf=hc/λ (from quantum perspective). JEE Main EM wave questions: "an EM wave has E-field 6V/m. Find B-field" → B=E/c=6/3×10⁸=2×10⁻⁸T. "Find intensity" → I=½cε₀E₀²=½×3×10⁸×8.85×10⁻¹²×36=4.78×10⁻² W/m²≈0.0478 W/m².
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