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LCR Series Circuit - Impedance and Phasor Diagram

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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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A series LCR circuit with \( R = 50 \, \Omega \), \( X_L = 70 \, \Omega \), \( X_C = 30 \, \Omega \) has a \( 200 \, \te

**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) = √(50² + (70 - 30)²) = √(2500 + 1600) = √(4100) ≈ 64 Ω . RMS current: I = (V/Z) = (200/64) ≈ 3.125 A . Power: P = I² R = (3.125)² × 50 ≈ 488.28 W . Applying X_L = ωL, X_C =

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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.

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What is the primary factor determining the resonant frequency in a series LCR circuit?

**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). The resonant frequency in a series LCR circuit is determined by the inductance ( L ) and capacitance ( C ), given by f₀ = (1/2π √(L C)) . Resistance affects damping but not the resonant frequency itself. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms

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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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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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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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A series LCR circuit has \( R = 10 \, \Omega \), \( X_L = 20 \, \Omega \), \( X_C = 30 \, \Omega \). 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) = (30 - 20/10) = 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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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 =

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Why does the impedance of an AC circuit with only an inductor increase with 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. The impedance in a purely inductive circuit is the inductive reactance ( X_L = ω L ), where ω = 2π f . As frequency ( f ) increases, ω increases linearly, causing X_L (and thus impedance) to increase proportionally. 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 Because inductive reactance

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A series LCR circuit has \( R = 25 \, \Omega \), \( X_L = 50 \, \Omega \), \( X_C = 70 \, \Omega \). 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) = (70 - 50/25) = (20/25) = 0.8 . Φ = tan⁻¹(0.8) ≈ 38.66° . 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 38.66°, consistent with phasor analysis and resonance condition X_L = X_C.

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A series LCR circuit has \( L = 5 \, \text{H} \), \( C = 20 \, \mu\text{F} \). 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 = 5 H , C = 20 × 10⁻⁶ F . ω₀ = (1/√(5 × 20 × 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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