A \( 100 \, \text{W} \) bulb is connected to a \( 110 \, \text{V} \) (rms) AC supply. What is the resistance of the bulb
**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. Average power: P = (V²/R) . R = (V²/P) = ((110)²/100) = (12100/100) = 121 Ω . 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 121 Ω, consistent with phasor analysis and resonance condition X_L = X_C.
Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values