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#AC source

26 public questions tagged with this topic.

AC Through Capacitor - Capacitive Reactance

A \( 6 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the ca

**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 × 50 = 314 rad/s . C = 6 × 10⁻⁶ F . X_C = (1/314 × 6 × 10⁻⁶) ≈ 530.5 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 Resistor - Phasor and Power

A \( 200 \, \text{V} \) (rms) AC source supplies a \( 100 \, \Omega \) resistor. What is the peak voltage?

**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. Peak voltage: v_m = √(2) V . V = 200 V . v_m = 1.414 × 200 = 282.8 V ≈ 283 V . 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 283 V, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Through Capacitor - Capacitive Reactance

A \( 60 \, \Omega \) resistor is connected to a \( 180 \, \text{V} \) (rms) AC source. What is the rms current?

**Capacitive reactance** X_C =1/(ω C)=1/(2π f C) (Ω), C capacitance (F), current leads voltage by 90°, I_rms = V_rms/X_C = V_rms ω C, I_peak = V_peak ω C, impedance Z = X_C for pure C. For 45 μF, 60 Hz, X_C=1/(2π×60×45×10⁻⁶)=58.9 Ω, V_rms=110 V, I_rms=1.867 A, I_peak=2.64 A. RMS current: I = (V/R) . Given: V = 180 V , R = 60 Ω . I = (180/60) = 3 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 3 A, consistent with phasor analysis

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AC Through Capacitor - Capacitive Reactance

A \( 20 \, \Omega \) resistor is connected to a \( 100 \, \text{V} \) (rms) AC source. What is the rms current?

**Capacitor average power** zero over complete cycle because P=V I =½ V_peak I_peak sin2ωt average zero, energy stored in field, not dissipated, unlike resistor. For 15 μF, 60 Hz, X_C=176.8 Ω, V_rms=110 V, I_rms=0.622 A, illustrating lower C higher X_C. RMS current: I = (V/R) . Given: V = 100 V , R = 20 Ω . I = (100/20) = 5 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 5 A, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Through Capacitor - Capacitive Reactance

A \( 75 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the r

**Capacitive reactance** X_C =1/(ω C)=1/(2π f C) (Ω), C capacitance (F), current leads voltage by 90°, I_rms = V_rms/X_C = V_rms ω C, I_peak = V_peak ω C, impedance Z = X_C for pure C. For 45 μF, 60 Hz, X_C=1/(2π×60×45×10⁻⁶)=58.9 Ω, V_rms=110 V, I_rms=1.867 A, I_peak=2.64 A. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 75 × 10⁻³ H . X_L = 376.8 × 0.075 = 28.26 Ω . RMS current: I = (V/X_L) = (110/28.26) ≈ 3.89 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms =

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AC Through Capacitor - Capacitive Reactance

A \( 20 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is th

**Capacitor average power** zero over complete cycle because P=V I =½ V_peak I_peak sin2ωt average zero, energy stored in field, not dissipated, unlike resistor. For 15 μF, 60 Hz, X_C=176.8 Ω, V_rms=110 V, I_rms=0.622 A, illustrating lower C higher X_C. X_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 20 × 10⁻⁶ F . X_C = (1/376.8 × 20 × 10⁻⁶) ≈ 132.7 Ω . RMS current: I = (V/X_C) = (110/132.7) ≈ 0.829 A . Peak current: i_m = √(2) I = 1.414 × 0.829 ≈ 1.17 A . Applying X_L = ωL, X_C = 1/ωC, Z

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AC Fundamentals - RMS, Average and Peak Values

A \( 212.1 \, \text{V} \) (peak) AC source is connected to a \( 75 \, \Omega \) resistor. What is the average power cons

**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 voltage: V = (v_m/√(2)) = (212.1/1.414) = 150 V . RMS current: I = (V/R) = (150/75) = 2 A . Average power: P = I² R = 2² × 75 = 300 W . 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 300 W, consistent

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AC Fundamentals - RMS, Average and Peak Values

A \( 20 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is th

**RMS value** I_rms = I_peak/√2, V_rms = V_peak/√2 for sinusoidal AC, significance rms gives equivalent DC value producing same heating power P = I_rms² R, average power over cycle, instruments measure rms, average over full cycle zero, half-cycle average 2 I_peak/π, peak = √2 rms. 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) = (230/159.2) ≈ 1.445 A . Peak current: i_m = √(2) I = 1.414 × 1.445 ≈ 2.04 A . Applying X_L =

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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 \( 80 \, \Omega \) 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 \( 65 \, \text{mH} \) 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 Capacitor - Capacitive Reactance

A \( 8 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the ca

**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 = 8 × 10⁻⁶ F . X_C = (1/376.8 × 8 × 10⁻⁶) ≈ 331.6 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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