AC Through Inductor - Inductive Reactance

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 \( 65 \, \text{mH} \) 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 = 65 × 10⁻³ H . X_L = 314 × 0.065 = 20.41 Ω . RMS current: I = (V/X_L) = (230/20.41) ≈ 11.27 A . Peak current: i_m = √(2) I = 1.414 × 11.27 ≈ 15.94 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms =

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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 Through Inductor - Inductive Reactance

A \( 80 \, \Omega \) resistor is connected to a \( 160 \, \text{V} \) (rms) AC source. What is the rms current?

**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. RMS current: I = (V/R) . Given: V = 160 V , R = 80 Ω . I = (160/80) = 2 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

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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 Through Inductor - Inductive Reactance

A \( 65 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the r

**AC through inductor** voltage leads current by 90°, V = L dI/dt, V(t)=V_peak sin(ωt+90°), I(t)=I_peak sin ωt, instantaneous power P= V I =½ V_peak I_peak sin2ωt, average zero over cycle because energy stored in magnetic field ½ L I² returned to source each quarter cycle, no net dissipation. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 65 × 10⁻³ H . X_L = 376.8 × 0.065 = 24.49 Ω . RMS current: I = (V/X_L) = (110/24.49) ≈ 4.49 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms =

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AC Through Inductor - Inductive Reactance

A \( 22 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) 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 = 22 × 10⁻⁶ F . X_C = (1/314 × 22 × 10⁻⁶) ≈ 144.7 Ω . RMS current: I = (V/X_C) = (220/144.7) ≈ 1.52 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 1.52

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AC Through Inductor - Inductive Reactance

A \( 95 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the i

**AC through inductor** voltage leads current by 90°, V = L dI/dt, V(t)=V_peak sin(ωt+90°), I(t)=I_peak sin ωt, instantaneous power P= V I =½ V_peak I_peak sin2ωt, average zero over cycle because energy stored in magnetic field ½ L I² returned to source each quarter cycle, no net dissipation. X_L = ω L , ω = 2π f . f = 50 Hz , L = 95 × 10⁻³ H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.095 = 29.83 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms

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AC Through Inductor - Inductive Reactance

A \( 16 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. 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 = 16 × 10⁻⁶ F . X_C = (1/314 × 16 × 10⁻⁶) ≈ 199 Ω . RMS current: I = (V/X_C) = (230/199) ≈ 1.156 A . Peak current: i_m = √(2) I = 1.414 ×

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AC Through Inductor - Inductive Reactance

A \( 90 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the i

**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 f . f = 50 Hz , L = 90 × 10⁻³ H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.09 = 28.26 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 25 Ω,

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AC Through Inductor - Inductive Reactance

A \( 90 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the r

**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 × 60 = 376.8 rad/s . L = 90 × 10⁻³ H . X_L = 376.8 × 0.09 = 33.91 Ω . RMS current: I = (V/X_L) = (220/33.91) ≈ 6.49 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 6.49 A, consistent

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AC Through Inductor - Inductive Reactance

A \( 55 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the peak

**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 × 60 = 376.8 rad/s . L = 55 × 10⁻³ H . X_L = 376.8 × 0.055 = 20.72 Ω . RMS current: I = (V/X_L) = (110/20.72) ≈ 5.31 A . Peak current: i_m = √(2) I = 1.414 × 5.31 ≈

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