#230 V

AC Through Inductor - Inductive Reactance

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 =

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AC Fundamentals - RMS, Average and Peak Values

A \( 230 \, \text{V} \) (rms) source supplies a \( 115 \, \Omega \) resistor. What is the peak current?

**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. RMS current: I = (V/R) = (230/115) = 2 A . Peak current: i_m = √(2) I = 1.414 × 2 ≈ 2.828 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 2.828 A, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Through Inductor - Inductive Reactance

A \( 85 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the i

**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_L = ω L , ω = 2π f . f = 50 Hz , L = 85 × 10⁻³ H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.085 = 26.69 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R²

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AC Through Inductor - Inductive Reactance

A \( 85 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the peak

**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_L = ω L , ω = 2π × 50 = 314 rad/s . L = 85 × 10⁻³ H . X_L = 314 × 0.085 = 26.69 Ω . RMS current: I = (V/X_L) = (230/26.69) ≈ 8.62 A . Peak current: i_m = √(2) I = 1.414 × 8.62 ≈ 12.19 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms =

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