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#impedance calculation

4 public questions tagged with this topic.

A series LCR circuit has \( R = 200 \, \Omega \), \( C = 15 \, \mu\text{F} \), and is connected to a \( 220 \, \text{V}

**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. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s , C = 15 × 10⁻⁶ F . X_C = (1/314 × 15 × 10⁻⁶) ≈ 212.3 Ω . No inductor, so X_L = 0 . Impedance: Z = √(R² + (X_C - X_L)²) = √(200² + 212.3²) ≈ 291.5 Ω . Applying X_L = ωL, X_C =

Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor

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.

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

A series LCR circuit has \( R = 15 \, \Omega \), \( X_L = 25 \, \Omega \), \( X_C = 40 \, \Omega \). What is the phase a

**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). tan Φ = (X_C - X_L/R) = (40 - 25/15) = 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

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