#AC circuit analysis

LCR Series Circuit - Impedance and Phasor Diagram

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 indu

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LCR Series Circuit - Impedance and Phasor Diagram

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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AC Through Resistor - Phasor and Power

A \( 80 \, \Omega \) resistor and \( 25 \, \mu\text{F} \) capacitor are in series with a \( 220 \, \text{V} \), \( 50 \,

**AC through resistor** voltage and current in phase, φ=0°, I = V/R instantaneously, I(t)=I_peak sin ωt, V(t)=V_peak sin ωt, phasor diagram V and I same direction, power instantaneous P = V I = V_peak I_peak sin² ωt, average P_avg = V_rms I_rms = V_rms²/R, always positive, energy dissipated as heat. X_C = (1/ω C) = (1/314 × 25 × 10⁻⁶) ≈ 127.4 Ω . Z = √(R² + X_C²) = √(80² + 127.4²) = √(6400 + 16230.76) ≈ 150.5 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 =

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AC Fundamentals - RMS, Average and Peak Values

A series LCR circuit with \( R = 70 \, \Omega \), \( X_L = 90 \, \Omega \), \( X_C = 40 \, \Omega \) has a \( 210 \, \te

**Average power** in resistor P_avg = V_rms I_rms = V_rms²/R = I_rms² R, for 254.6 V peak, V_rms =180 V, R=90 Ω, P=180²/90=360 W, for 226.3 V peak, V_rms=160 V, R=80 Ω, P=320 W, illustrating rms use for power. Z = √(R² + (X_L - X_C)²) = √(70² + (90 - 40)²) = √(4900 + 2500) = √(7400) ≈ 86 Ω . RMS current: I = (V/Z) = (210/86) ≈ 2.442 A . Power: P = I² R = (2.442)² × 70 ≈ 417.6 W . 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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