#AC supply

Power in AC Circuits - Power Factor and Wattless Current

A \( 60 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC supply. What is th

**Power factor** cos φ = R/Z, 0≤cos φ≤1, at resonance cos φ=1 maximum power, pure L or C cos φ=0 zero average power, current wattless because energy stored returned. Instantaneous power when current maximum in pure L: I=I_peak, V=0 because V leads 90°, so P= V I =0 at that instant, average zero. Capacitive reactance: X_C = (1/ω C) , where ω = 2π f . f = 60 Hz , C = 60 × 10⁻⁶ F . ω = 2 × 3.14 × 60 = 376.8 rad/s . X_C = (1/376.8 × 60 × 10⁻⁶) = 44.24 Ω . RMS current: I = (V/X_C)

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AC Through Capacitor - Capacitive Reactance

A \( 25 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is th

**Capacitive reactance** X_C =1/(ω C)=1/(2π f C) (Ω), C capacitance (F), current leads voltage by 90°, I_rms = V_rms/X_C = V_rms ω C, I_peak = V_peak ω C, impedance Z = X_C for pure C. For 45 μF, 60 Hz, X_C=1/(2π×60×45×10⁻⁶)=58.9 Ω, V_rms=110 V, I_rms=1.867 A, I_peak=2.64 A. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 25 × 10⁻⁶ F . X_C = (1/314 × 25 × 10⁻⁶) ≈ 127.4 Ω . RMS current: I = (V/X_C) = (230/127.4) ≈ 1.805 A . Peak current: i_m = √(2) I = 1.414 × 1.805 ≈ 2.55 A .

Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance

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AC Through Inductor - Inductive Reactance

A \( 20 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC supply. What is th

**Inductive reactance** X_L = ω L =2π f L (Ω), L inductance (H), f frequency (Hz), ω=2πf angular frequency (rad/s), opposes AC, impedance Z = X_L for pure L, current lags voltage by 90°, I_rms = V_rms/X_L, I_peak = V_peak/X_L. For 85 mH, 50 Hz, X_L=2π×50×0.085=26.7 Ω, V_rms=230 V, I_rms=8.61 A, I_peak=12.18 A. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 20 × 10⁻⁶ F . X_C = (1/314 × 20 × 10⁻⁶) ≈ 159.2 Ω . RMS current: I = (V/X_C) = (220/159.2) ≈ 1.38 A . Applying X_L = ωL, X_C = 1/ωC, Z =

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AC Through Inductor - Inductive Reactance

A \( 21 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is th

**Inductive reactance calculation** for 60 mH, 50 Hz, X_L=2π×50×0.06=18.85 Ω, V_rms=220 V, I_rms=220/18.85=11.67 A, for 70 mH, 50 Hz, X_L=21.99 Ω, I=110/21.99=5 A, showing X_L ∝ f L, higher f or L increases opposition. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 21 × 10⁻⁶ F . X_C = (1/314 × 21 × 10⁻⁶) ≈ 151.6 Ω . RMS current: I = (V/X_C) = (230/151.6) ≈ 1.517 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 1.517

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AC Fundamentals - RMS, Average and Peak Values

A \( 100 \, \Omega \) resistor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC supply. What is the ave

**Average power** in resistor P_avg = V_rms I_rms = V_rms²/R = I_rms² R, for 254.6 V peak, V_rms =180 V, R=90 Ω, P=180²/90=360 W, for 226.3 V peak, V_rms=160 V, R=80 Ω, P=320 W, illustrating rms use for power. Average power: P = I² R , where I = (V/R) . I = (220/100) = 2.2 A . P = (2.2)² × 100 = 4.84 × 100 = 484 W . 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 484 W, consistent with phasor analysis and resonance condition X_L = X_C.

Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values

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