Practice question
Question
A \( 180 \, \text{V} \) (rms) source supplies a \( 90 \, \Omega \) resistor. What is the peak current?
Explanation
**RMS value** I_rms = I_peak/√2, V_rms = V_peak/√2 for sinusoidal AC, significance rms gives equivalent DC value producing same heating power P = I_rms² R, average power over cycle, instruments measure rms, average over full cycle zero, half-cycle average 2 I_peak/π, peak = √2 rms. RMS current: I = (V/R) = (180/90) = 2 A . Peak current: i_m = √(2) I = 1.414 × 2 ≈ 2.828 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 2.828 A, consistent with phasor analysis and resonance condition X_L = X_C.
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