#60 Hz

AC Through Capacitor - Capacitive Reactance

A \( 11 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the c

**AC through capacitor** voltage lags current by 90°, Q= C V, I= dQ/dt = C dV/dt, V(t)=V_peak sin ωt, I(t)=I_peak sin(ωt+90°), average power zero because energy stored in electric field ½ C V² returned each cycle, capacitor blocks DC but passes AC, X_C decreases with f. X_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 11 × 10⁻⁶ F . X_C = (1/376.8 × 11 × 10⁻⁶) ≈ 241.2 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

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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 \( 18 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is th

**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_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 18 × 10⁻⁶ F . X_C = (1/376.8 × 18 × 10⁻⁶) ≈ 147.3 Ω . RMS current: I = (V/X_C) = (110/147.3) ≈ 0.747 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²),

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AC Through Inductor - Inductive Reactance

A \( 10 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC supply. 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 × 60 = 376.8 rad/s . C = 10 × 10⁻⁶ F . X_C = (1/376.8 × 10 × 10⁻⁶) ≈ 265.4 Ω . RMS current: I = (V/X_C) = (110/265.4) ≈ 0.414 A . Peak current: i_m = √(2) I = 1.414 × 0.414 ≈ 0.585 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²),

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AC Through Inductor - Inductive Reactance

A \( 45 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) 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 × 60 = 376.8 rad/s . C = 45 × 10⁻⁶ F . X_C = (1/376.8 × 45 × 10⁻⁶) ≈ 59 Ω . RMS current: I = (V/X_C) = (110/59) ≈ 1.864 A . Peak current: i_m = √(2) I = 1.414 × 1.864 ≈ 2.64 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²),

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