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Question

A \( 200 \, \text{V} \) (rms) AC source supplies a \( 100 \, \Omega \) resistor. What is the peak
voltage?

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Explanation

**Current relative to voltage** in resistor in phase, φ=0°, power factor cos φ=1, maximum power, unlike inductor/capacitor where average power zero due to 90° phase shift, explaining why resistor heats while pure L/C does not. Peak voltage: v_m = √(2) V . V = 200 V . v_m = 1.414 × 200 = 282.8 V ≈ 283 V . 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 283 V, consistent with phasor analysis and resonance condition X_L = X_C.

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