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Question

A series LCR circuit with \( R = 60 \, \Omega \) is at resonance with a \( 240 \, \text{V} \) (rms)
source. What is the rms current?

Options

Choose one · Correct answer highlighted

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 = 60 Ω . RMS current: I = (V/R) = (240/60) = 4 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 4 A, consistent with phasor analysis and resonance condition X_L = X_C.

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