#angular frequency

Resonance in LCR Circuit and Q-Factor

A series LCR circuit has \( L = 4 \, \text{H} \), \( C = 25 \, \mu\text{F} \). What is the resonant angular frequency?

**Condition for resistive behavior** X_L = X_C, so ωL=1/ωC, ω²=1/LC, this is resonance, impedance purely resistive Z=R, current maximum, power maximum P= V²/R, inductive and capacitive reactances equal magnitude opposite sign cancel, circuit appears resistive. ω₀ = (1/√(L C)) . L = 4 H , C = 25 × 10⁻⁶ F . ω₀ = (1/√(4 × 25 × 10⁻⁶)) = (1/√(10⁻⁴)) = 100 rad/s . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 100 rad/s, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Through Resistor - Phasor and Power

A series LCR circuit has \( L = 2.5 \, \text{H} \), \( C = 40 \, \mu\text{F} \). What is the resonant angular frequency?

**Current relative to voltage** in resistor in phase, φ=0°, power factor cos φ=1, maximum power, unlike inductor/capacitor where average power zero due to 90° phase shift, explaining why resistor heats while pure L/C does not. ω₀ = (1/√(L C)) . L = 2.5 H , C = 40 × 10⁻⁶ F . ω₀ = (1/√(2.5 × 40 × 10⁻⁶)) = (1/√(10⁻⁴)) = 100 rad/s . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 100 rad/s, consistent with phasor analysis and resonance condition X_L = X_C.

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( L = 6 \, \text{H} \), \( C = 10 \, \mu\text{F} \). What is the resonant angular frequency?

**LCR example** R=100 Ω, X_L=130 Ω, X_C=70 Ω, X_L-X_C=60 Ω, Z=√(100²+60²)=116.6 Ω, V_rms=300 V, I_rms=2.573 A, power P= I_rms² R =662 W? Actually P= V_rms I_rms cos φ, cos φ=R/Z=0.857, P=300×2.573×0.857=661.6 W, illustrating power factor. ω₀ = (1/√(L C)) . L = 6 H , C = 10 × 10⁻⁶ F . ω₀ = (1/√(6 × 10 × 10⁻⁶)) = (1/√(6 × 10⁻⁵)) ≈ 129.1 rad/s . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 129.1 rad/s, consistent with phasor analysis and resonance condition X_L = X_C.

Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( L = 3 \, \text{H} \), \( C = 12 \, \mu\text{F} \). What is the resonant angular frequency?

**LCR series impedance** Z = √(R² + (X_L - X_C)²), R resistance (Ω), X_L=ωL, X_C=1/ωC, phase angle φ = tan⁻¹((X_L-X_C)/R), current I_rms = V_rms/Z, voltage across R in phase with I, across L leads by 90°, across C lags by 90°, phasor diagram vector sum V = √(V_R² + (V_L - V_C)²). ω₀ = (1/√(L C)) . L = 3 H , C = 12 × 10⁻⁶ F . ω₀ = (1/√(3 × 12 × 10⁻⁶)) = (1/√(36 × 10⁻⁶)) = (10³/6) ≈ 166.67 rad/s . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P =

Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( L = 2 \, \text{H} \), \( C = 50 \, \mu\text{F} \). What is the resonant angular frequency?

**Impedance behavior at high frequencies** X_L=ωL dominates ∝ f, X_C=1/ωC →0, so Z≈√(R²+X_L²)≈X_L large, current small, circuit inductive, φ→90°, at low frequencies X_C large, Z≈X_C, capacitive, φ→-90°, at intermediate resonance Z minimal =R. Resonant frequency: ω₀ = (1/√(L C)) . L = 2 H , C = 50 × 10⁻⁶ F . ω₀ = (1/√(2 × 50 × 10⁻⁶)) = (1/√(10⁻⁴)) = 100 rad/s . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 100 rad/s, consistent with phasor analysis and resonance condition X_L = X_C.

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( L = 8 \, \text{H} \), \( C = 5 \, \mu\text{F} \). What is the resonant angular frequency?

**LCR series impedance** Z = √(R² + (X_L - X_C)²), R resistance (Ω), X_L=ωL, X_C=1/ωC, phase angle φ = tan⁻¹((X_L-X_C)/R), current I_rms = V_rms/Z, voltage across R in phase with I, across L leads by 90°, across C lags by 90°, phasor diagram vector sum V = √(V_R² + (V_L - V_C)²). ω₀ = (1/√(L C)) . L = 8 H , C = 5 × 10⁻⁶ F . ω₀ = (1/√(8 × 5 × 10⁻⁶)) = (1/√(40 × 10⁻⁶)) ≈ 158.1 rad/s . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms

Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( L = 3.5 \, \text{H} \), \( C = 8 \, \mu\text{F} \). What is the resonant angular frequency?

**LCR example** R=100 Ω, X_L=130 Ω, X_C=70 Ω, X_L-X_C=60 Ω, Z=√(100²+60²)=116.6 Ω, V_rms=300 V, I_rms=2.573 A, power P= I_rms² R =662 W? Actually P= V_rms I_rms cos φ, cos φ=R/Z=0.857, P=300×2.573×0.857=661.6 W, illustrating power factor. ω₀ = (1/√(L C)) . L = 3.5 H , C = 8 × 10⁻⁶ F . ω₀ = (1/√(3.5 × 8 × 10⁻⁶)) = (1/√(28 × 10⁻⁶)) ≈ 188.98 rad/s . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 188.98 rad/s, consistent with phasor analysis and resonance condition X_L = X_C.

Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram

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Energy, Intensity and Momentum of EM Waves

An electromagnetic wave in vacuum has an angular frequency \( \omega = 4 \times 10^{11} \, \text{rad/s} \). What is its

**Poynting vector** S = E×B/μ₀ gives energy flow (W/m²), magnitude S = E B/μ₀, average = E₀ B₀/(2μ₀) = intensity, direction of propagation, showing energy transport perpendicular to E and B. Frequency v = (ω/2π) . Given ω = 4 × 10¹¹ rad/s , v = (4 × 10¹¹/2 π) ≈ 6.37 × 10¹⁰ Hz . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 6.37 × 10¹⁰ Hz, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

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EM Wave Characteristics - Transverse Nature and E/B Ratio

An electromagnetic wave has an angular frequency \( \omega = 6 \times 10^{11} \, \text{rad/s} \). What is its frequency

**Relationship E and B** in EM wave E₀ = c B₀, B₀ = E₀/c, for vacuum. Fields sustain each other via Maxwell's equations ∇×E = -∂B/∂t, ∇×B = μ₀ ε₀ ∂E/∂t, time-varying E produces B and vice versa, self-sustaining propagation without medium, speed c. Frequency v = (ω/2π) . Given ω = 6 × 10¹¹ rad/s , v = (6 × 10¹¹/2 π) ≈ 9.55 × 10¹⁰ Hz . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 9.55 × 10¹⁰ Hz, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > EM Wave Characteristics - Transverse Nature and E/B Ratio

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Displacement Current and Ampere-Maxwell Law

An electromagnetic wave in vacuum has an electric field given by \( E_x = 150 \sin(5 \times 10^3 z - 1.5 \times 10^{12}

**Ampere-Maxwell law** ∮ B·dl = μ₀(I_c + ε₀ dΦ_E/dt) generalizes Ampere's law, displacement current arises from time-varying electric field, source of magnetic field like conduction current. For rate of change of flux 2×10¹¹ V·m/s, I_d = ε₀×2×10¹¹ =8.85×10⁻¹²×2×10¹¹=1.77 A. Comparing with E_x = E₀ sin(kz - ω t) , we have ω = 1.5 × 10¹² rad/s . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 1.5 × 10¹² rad/s, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Displacement Current and Ampere-Maxwell Law

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Displacement Current and Ampere-Maxwell Law

An electromagnetic wave has an angular frequency \( \omega = 8 \times 10^{11} \, \text{rad/s} \). What is its frequency

**Displacement current** I_d = ε₀ dΦ_E/dt, Φ_E = ∫ E·dA electric flux (V·m), ε₀=8.85×10⁻¹² F/m, ensures continuity of current in charging capacitor where conduction current stops between plates, I_d equals conduction current in wires, 3 A conduction ⇒ 3 A displacement, maintaining Ampere's law ∮ B·dl = μ₀(I_c+I_d). Frequency v = (ω/2π) . Given ω = 8 × 10¹¹ rad/s , v = (8 × 10¹¹/2 π) ≈ 1.27 × 10¹¹ Hz . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 1.27 × 10¹¹ Hz, illustrating EM wave transverse nature and Maxwell's displacement current concept

Ref: NCERT > Physics Book > Electromagnetic Waves > Displacement Current and Ampere-Maxwell Law

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Forced Oscillations and Resonance

A pendulum has \( L = 1.4 \, \text{m}, g = 9.8 \, \text{m/s}^2 \). What is its angular frequency?

**Driven system** exhibits amplitude-frequency response peaking at resonance, phase shift between drive and displacement varying 0° to 180° across resonance. Forced oscillations sustain motion against damping, amplitude controlled by detuning |ω_d - ω₀| and damping strength b. ω = √((g/L)) = √((9.8/1.4)) ≈ √(7) ≈ 2.65 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.65 rad/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Forced Oscillations and Resonance

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