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Question

A \( 60 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source.
What is the peak current?

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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. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 60 × 10⁻³ H . X_L = 376.8 × 0.06 = 22.61 Ω . RMS current: I = (V/X_L) = (110/22.61) ≈ 4.87 A . Peak current: i_m = √(2) I = 1.414 × 4.87 ≈ 6.88 A . Applying X_L = ωL,

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