Skip to content

Question

A light bulb rated at \( 100 \, \text{W} \) operates on a \( 220 \, \text{V} \) (rms) AC supply. What
is the peak voltage across the bulb?

Options

Choose one · Correct answer highlighted

Explanation

**AC through resistor** voltage and current in phase, φ=0°, I = V/R instantaneously, I(t)=I_peak sin ωt, V(t)=V_peak sin ωt, phasor diagram V and I same direction, power instantaneous P = V I = V_peak I_peak sin² ωt, average P_avg = V_rms I_rms = V_rms²/R, always positive, energy dissipated as heat. Peak voltage: v_m = √(2) V . Given: V = 220 V (rms). v_m = 1.414 × 220 = 311.08 V ≈ 311 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 311 V, consistent

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.