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#reactive current

2 public questions tagged with this topic.

A \( 23 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the r

**Average power in LCR** P_avg = V_rms I_rms cos φ, cos φ = R/Z power factor, φ phase between V and I, tan φ = (X_L - X_C)/R. For R=80 Ω, X_L=100 Ω, X_C=40 Ω, X_L-X_C=60 Ω, Z=√(80²+60²)=100 Ω, cos φ=0.8, V_rms=240 V, I_rms=2.4 A, P=240×2.4×0.8=460.8 W, only R dissipates. X_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 23 × 10⁻⁶ F . X_C = (1/376.8 × 23 × 10⁻⁶) ≈ 115.4 Ω . RMS current: I = (V/X_C) = (110/115.4) ≈ 0.953 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L

Ref: NCERT > Physics Book > Alternating Currents > Power in AC Circuits - Power Factor and Wattless Current

A \( 30 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the r

**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 = 30 × 10⁻⁶ F . X_C = (1/314 × 30 × 10⁻⁶) ≈ 106.1 Ω . RMS current: I = (V/X_C) = (230/106.1) ≈ 2.17 A . Applying X_L = ωL, X_C = 1/ωC, Z =

Ref: NCERT > Physics Book > Alternating Currents > AC Through Inductor - Inductive Reactance