Skip to content
Login Register

#reactive components

2 public questions tagged with this topic.

LCR Series Circuit - Impedance and Phasor Diagram

In a series LCR circuit, why does resonance not occur if either the inductor or capacitor is absent?

**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. Resonance in a series LCR circuit requires both an inductor and capacitor to create a condition where X_L = X_C , canceling the reactive components. Without either, there’s no opposing reactance to balance, preventing resonance. 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

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

View Details
AC Fundamentals - RMS, Average and Peak Values

A series LCR circuit with \( R = 100 \, \Omega \), \( X_L = 130 \, \Omega \), \( X_C = 70 \, \Omega \) has a \( 300 \, \

**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) = √(100² + (130 - 70)²) = √(10000 + 3600) = √(13600) ≈ 116.62 Ω . RMS current: I = (V/Z) = (300/116.62) ≈ 2.573 A . Power: P = I² R = (2.573)² × 100 ≈ 661.8 W . 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 > AC Fundamentals - RMS, Average and Peak Values

View Details