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Resonance in LCR Circuit and Q-Factor

A \( 200 \, \text{V} \) (rms) AC source is connected to a series LCR circuit with \( R = 20 \, \Omega \) at resonance. W

**Resonance in LCR** occurs when X_L = X_C, ω₀ =1/√(LC), f₀=1/(2π√(LC)), impedance Z=R minimum, current maximum I₀=V/R, circuit behaves as if only resistance because inductive and capacitive voltages equal opposite cancel, net reactance zero, phase φ=0°, power factor 1, current in phase with voltage. At resonance, Z = R = 20 Ω . RMS current: I = (V/R) = (200/20) = 10 A . Power: P = I² R = 10² × 20 = 2000 W . 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 2000

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Power in AC Circuits - Power Factor and Wattless Current

In an LCR circuit with \( R = 3 \, \Omega \), \( X_L = 8 \, \Omega \), \( X_C = 4 \, \Omega \), what is the power factor

**Average power in LCR** P_avg = V_rms I_rms cos φ, cos φ = R/Z power factor, φ phase between V and I, tan φ = (X_L - X_C)/R. For R=80 Ω, X_L=100 Ω, X_C=40 Ω, X_L-X_C=60 Ω, Z=√(80²+60²)=100 Ω, cos φ=0.8, V_rms=240 V, I_rms=2.4 A, P=240×2.4×0.8=460.8 W, only R dissipates. Impedance: Z = √(R² + (X_L - X_C)²) = √(3² + (8 - 4)²) = √(9 + 16) = 5 Ω . Power factor: cos Φ = (R/Z) = (3/5) = 0.6 . 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,

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Power in AC Circuits - Power Factor and Wattless Current

In an AC circuit with a series LCR combination, why does the power dissipated depend only on the resistive component?

**Average power in LCR** P_avg = V_rms I_rms cos φ, cos φ = R/Z power factor, φ phase between V and I, tan φ = (X_L - X_C)/R. For R=80 Ω, X_L=100 Ω, X_C=40 Ω, X_L-X_C=60 Ω, Z=√(80²+60²)=100 Ω, cos φ=0.8, V_rms=240 V, I_rms=2.4 A, P=240×2.4×0.8=460.8 W, only R dissipates. Power dissipation in an AC circuit ( P = I² R cos Φ ) occurs only through resistance, as inductors and capacitors store and release energy without converting it to heat. The reactive components affect the current and phase, but only R dissipates power. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² +

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( R = 80 \, \Omega \), \( X_L = 60 \, \Omega \), \( X_C = 40 \, \Omega \). What is the impedan

**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. Z = √(R² + (X_L - X_C)²) . Z = √(80² + (60 - 40)²) = √(6400 + 400) = √(6800) ≈ 82.46 Ω . 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 82.46 Ω, 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 \( R = 120 \, \Omega \), \( X_L = 100 \, \Omega \), \( X_C = 80 \, \Omega \). What is the imped

**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). Z = √(R² + (X_L - X_C)²) . Z = √(120² + (100 - 80)²) = √(14400 + 400) = √(14800) ≈ 121.66 Ω . 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 121.66

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( R = 15 \, \Omega \), \( X_L = 30 \, \Omega \), \( X_C = 45 \, \Omega \). What is the phase a

**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. tan Φ = (X_C - X_L/R) = (45 - 30/15) = 1 . Φ = tan⁻¹(1) = 45° . Since X_C > X_L , current leads voltage. 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 45°, 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 with \( R = 110 \, \Omega \), \( X_L = 140 \, \Omega \), \( X_C = 80 \, \Omega \) has a \( 330 \, \

**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. Z = √(R² + (X_L - X_C)²) = √(110² + (140 - 80)²) = √(12100 + 3600) = √(15700) ≈ 125.3 Ω . RMS current: I = (V/Z) = (330/125.3) ≈ 2.634 A . Power: P = I² R = (2.634)² × 110 ≈ 763.2 W . 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,

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( R = 130 \, \Omega \), \( X_L = 110 \, \Omega \), \( X_C = 90 \, \Omega \). What is the imped

**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. Z = √(R² + (X_L - X_C)²) . Z = √(130² + (110 - 90)²) = √(16900 + 400) = √(17300) ≈ 131.5 Ω . 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 131.5 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Fundamentals - RMS, Average and Peak Values

A \( 100 \, \text{W} \) bulb is connected to a \( 110 \, \text{V} \) (rms) AC supply. What is the resistance of the bulb

**RMS value** I_rms = I_peak/√2, V_rms = V_peak/√2 for sinusoidal AC, significance rms gives equivalent DC value producing same heating power P = I_rms² R, average power over cycle, instruments measure rms, average over full cycle zero, half-cycle average 2 I_peak/π, peak = √2 rms. Average power: P = (V²/R) . R = (V²/P) = ((110)²/100) = (12100/100) = 121 Ω . 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 121 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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Power in AC Circuits - Power Factor and Wattless Current

A series LCR circuit has \( R = 35 \, \Omega \), \( X_L = 60 \, \Omega \), \( X_C = 25 \, \Omega \). What is the power f

**Average power in LCR** P_avg = V_rms I_rms cos φ, cos φ = R/Z power factor, φ phase between V and I, tan φ = (X_L - X_C)/R. For R=80 Ω, X_L=100 Ω, X_C=40 Ω, X_L-X_C=60 Ω, Z=√(80²+60²)=100 Ω, cos φ=0.8, V_rms=240 V, I_rms=2.4 A, P=240×2.4×0.8=460.8 W, only R dissipates. Z = √(R² + (X_L - X_C)²) = √(35² + (60 - 25)²) = √(1225 + 1225) = √(2450) ≈ 49.5 Ω . Power factor: cos Φ = (R/Z) = (35/49.5) ≈ 0.707 . 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

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Resonance in LCR Circuit and Q-Factor

A series LCR circuit with \( R = 60 \, \Omega \) is at resonance with a \( 240 \, \text{V} \) (rms) source. What is the

**At resonance** V_L = I X_L = I X_C = V_C, may be larger than source voltage Q times, Q-factor = ω₀ L/R =1/(ω₀ C R)= V_L/V = V_C/V, measures sharpness, higher Q sharper resonance, bandwidth Δω = R/L = ω₀/Q, resonant frequency independent of R. At resonance, Z = R = 60 Ω . RMS current: I = (V/R) = (240/60) = 4 A . 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 4 A, 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 \( R = 90 \, \Omega \), \( X_L = 70 \, \Omega \), \( X_C = 50 \, \Omega \). What is the impedan

**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. Z = √(R² + (X_L - X_C)²) . Z = √(90² + (70 - 50)²) = √(8100 + 400) = √(8500) ≈ 92.2 Ω . 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 92.2 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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