#50 Hz

AC Through Resistor - Phasor and Power

A \( 95 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the peak

**Current relative to voltage** in resistor in phase, φ=0°, power factor cos φ=1, maximum power, unlike inductor/capacitor where average power zero due to 90° phase shift, explaining why resistor heats while pure L/C does not. X_L = ω L , ω = 2π × 50 = 314 rad/s . L = 95 × 10⁻³ H . X_L = 314 × 0.095 = 29.83 Ω . RMS current: I = (V/X_L) = (220/29.83) ≈ 7.375 A . Peak current: i_m = √(2) I = 1.414 × 7.375 ≈ 10.43 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms

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AC Through Resistor - Phasor and Power

A \( 45 \, \Omega \) resistor and \( 18 \, \mu\text{F} \) capacitor are in series with a \( 220 \, \text{V} \), \( 50 \,

**Current relative to voltage** in resistor in phase, φ=0°, power factor cos φ=1, maximum power, unlike inductor/capacitor where average power zero due to 90° phase shift, explaining why resistor heats while pure L/C does not. X_C = (1/ω C) = (1/314 × 18 × 10⁻⁶) ≈ 176.8 Ω . Z = √(R² + X_C²) = √(45² + 176.8²) = √(2026 + 31258.24) ≈ 182.4 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 182.4 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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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 \( 21 \, \mu\text{F} \) 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 \( 22 \, \mu\text{F} \) 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 Resistor - Phasor and Power

A \( 70 \, \Omega \) resistor and \( 30 \, \mu\text{F} \) 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 × 30 × 10⁻⁶) ≈ 106.1 Ω . Z = √(R² + X_C²) = √(70² + 106.1²) = √(4900 + 11257.21) ≈ 127.3 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 =

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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 \( 55 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) 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 × 50 = 314 rad/s . L = 55 × 10⁻³ H . X_L = 314 × 0.055 = 17.27 Ω . RMS current: I = (V/X_L) = (230/17.27) ≈ 13.32 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 \( 70 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 50 \, \text{Hz} \) 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 = 70 × 10⁻³ H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.07 = 21.98 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R²

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AC Through Inductor - Inductive Reactance

A \( 60 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) 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 × 50 = 314 rad/s . L = 60 × 10⁻³ H . X_L = 314 × 0.06 = 18.84 Ω . RMS current: I = (V/X_L) = (220/18.84) ≈ 11.68 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 11.68 A, consistent

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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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