Practice question
Question
In an LCR series circuit, if the frequency is below the resonant frequency, which component dominates
the circuit's behavior?
Explanation
**At resonance** V_L = I X_L = I X_C = V_C, may be larger than source voltage Q times, Q-factor = ω₀ L/R =1/(ω₀ C R)= V_L/V = V_C/V, measures sharpness, higher Q sharper resonance, bandwidth Δω = R/L = ω₀/Q, resonant frequency independent of R. Below resonance, ω is low, making X_C = (1/ω C) larger than X_L = ω L . Thus, the capacitive reactance dominates, and the circuit behaves as predominantly capacitive, with current leading 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
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.