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#resonance condition

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

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Resonance in LCR Circuit and Q-Factor

In an AC circuit with a series combination of resistor, inductor, and capacitor, what is the condition for the circuit t

**Condition for resistive behavior** X_L = X_C, so ωL=1/ωC, ω²=1/LC, this is resonance, impedance purely resistive Z=R, current maximum, power maximum P= V²/R, inductive and capacitive reactances equal magnitude opposite sign cancel, circuit appears resistive. For an LCR series circuit to be purely resistive, the net reactance must be zero ( X_L - X_C = 0 ), which occurs at resonance when X_L = X_C . This makes the impedance equal to the resistance ( Z = R ), and the circuit behaves as if only resistance is present. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms =

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

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