Practice question
Question
In a series LCR circuit at resonance, the power dissipated is \( 4800 \, \text{W} \) with \( R = 3 \,
\Omega \). What is the rms current?
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. At resonance, Z = R , and power P = I² R . 4800 = I² × 3 . I² = (4800/3) = 1600 . I = √(1600) = 40 A . 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
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.