A \( 80 \, \Omega \) resistor and \( 25 \, \mu\text{F} \) capacitor are in series with a \( 220 \, \text{V} \), \( 50 \,
**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. X_C = (1/ω C) = (1/314 × 25 × 10⁻⁶) ≈ 127.4 Ω . Z = √(R² + X_C²) = √(80² + 127.4²) = √(6400 + 16230.76) ≈ 150.5 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 =
Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power