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#AC source

39 public questions tagged with this topic.

Power in AC Circuits - Power Factor and Wattless Current

A \( 45 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the peak

**Wattless current** occurs in pure inductor or capacitor, I_rms non-zero but average power zero because φ=±90°, cos φ=0, energy oscillates between source and field, no dissipation, used in choke coil to limit current without heating, unlike resistor where power dissipated. X_L = ω L , ω = 2π × 50 = 314 rad/s . L = 45 × 10⁻³ H . X_L = 314 × 0.045 = 14.13 Ω . RMS current: I = (V/X_L) = (230/14.13) ≈ 16.28 A . Peak current: i_m = √(2) I = 1.414 × 16.28 ≈ 23.02 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R²

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Power in AC Circuits - Power Factor and Wattless Current

A \( 23 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the r

**Average power in LCR** P_avg = V_rms I_rms cos φ, cos φ = R/Z power factor, φ phase between V and I, tan φ = (X_L - X_C)/R. For R=80 Ω, X_L=100 Ω, X_C=40 Ω, X_L-X_C=60 Ω, Z=√(80²+60²)=100 Ω, cos φ=0.8, V_rms=240 V, I_rms=2.4 A, P=240×2.4×0.8=460.8 W, only R dissipates. X_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 23 × 10⁻⁶ F . X_C = (1/376.8 × 23 × 10⁻⁶) ≈ 115.4 Ω . RMS current: I = (V/X_C) = (110/115.4) ≈ 0.953 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L

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Resonance in LCR Circuit and Q-Factor

A \( 17 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is th

**At resonance** V_L = I X_L = I X_C = V_C, may be larger than source voltage Q times, Q-factor = ω₀ L/R =1/(ω₀ C R)= V_L/V = V_C/V, measures sharpness, higher Q sharper resonance, bandwidth Δω = R/L = ω₀/Q, resonant frequency independent of R. X_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 17 × 10⁻⁶ F . X_C = (1/376.8 × 17 × 10⁻⁶) ≈ 156 Ω . RMS current: I = (V/X_C) = (110/156) ≈ 0.705 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms =

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Power in AC Circuits - Power Factor and Wattless Current

A \( 268.7 \, \text{V} \) (peak) AC source is connected to a \( 95 \, \Omega \) resistor. What is the average power cons

**Wattless current** occurs in pure inductor or capacitor, I_rms non-zero but average power zero because φ=±90°, cos φ=0, energy oscillates between source and field, no dissipation, used in choke coil to limit current without heating, unlike resistor where power dissipated. RMS voltage: V = (v_m/√(2)) = (268.7/1.414) ≈ 190 V . RMS current: I = (V/R) = (190/95) = 2 A . Average power: P = I² R = 2² × 95 = 380 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 380 W,

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Resonance in LCR Circuit and Q-Factor

A \( 220 \, \text{V} \) (rms), \( 50 \, \text{Hz} \) AC source is connected to a \( 44 \, \text{mH} \) inductor. Calcula

**Condition for resistive behavior** X_L = X_C, so ωL=1/ωC, ω²=1/LC, this is resonance, impedance purely resistive Z=R, current maximum, power maximum P= V²/R, inductive and capacitive reactances equal magnitude opposite sign cancel, circuit appears resistive. Inductive reactance: X_L = ω L , where ω = 2π f . Given: f = 50 Hz , L = 44 mH = 0.044 H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.044 = 13.816 Ω ≈ 13.82 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms

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Resonance in LCR Circuit and Q-Factor

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

**At resonance** V_L = I X_L = I X_C = V_C, may be larger than source voltage Q times, Q-factor = ω₀ L/R =1/(ω₀ C R)= V_L/V = V_C/V, measures sharpness, higher Q sharper resonance, bandwidth Δω = R/L = ω₀/Q, resonant frequency independent of R. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 60 × 10⁻³ H . X_L = 376.8 × 0.06 = 22.61 Ω . RMS current: I = (V/X_L) = (110/22.61) ≈ 4.87 A . Peak current: i_m = √(2) I = 1.414 × 4.87 ≈ 6.88 A . Applying X_L = ωL,

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AC Through Resistor - Phasor and Power

A \( 13 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the c

**Resistor in AC** behaves as DC, no reactance, impedance Z=R, current follows voltage exactly, average power over cycle V_rms I_rms, for 200 V rms, 80 Ω, P=500 W? Actually 200²/80=500 W, peak current √2×2.5=3.535 A, average power ½ V_peak I_peak. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 13 × 10⁻⁶ F . X_C = (1/314 × 13 × 10⁻⁶) ≈ 245.1 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 245.1 Ω, consistent with phasor analysis

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AC Through Resistor - Phasor and Power

A \( 50 \, \Omega \) resistor and \( 20 \, \mu\text{F} \) capacitor are in series with a \( 110 \, \text{V} \), \( 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 × 20 × 10⁻⁶) ≈ 159.2 Ω . Z = √(R² + X_C²) = √(50² + 159.2²) ≈ 166.6 Ω . RMS current: I = (V/Z) = (110/166.6) ≈ 0.66 A . Voltage across resistor: V_R = I R = 0.66 × 50 ≈ 33 V . Applying X_L

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AC Through Resistor - Phasor and Power

A \( 15 \, \Omega \) resistor is connected to a \( 75 \, \text{V} \) (rms) AC source. What is the rms current?

**Resistor in AC** behaves as DC, no reactance, impedance Z=R, current follows voltage exactly, average power over cycle V_rms I_rms, for 200 V rms, 80 Ω, P=500 W? Actually 200²/80=500 W, peak current √2×2.5=3.535 A, average power ½ V_peak I_peak. RMS current: I = (V/R) . Given: V = 75 V , R = 15 Ω . I = (75/15) = 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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Power in AC Circuits - Power Factor and Wattless Current

A \( 175 \, \text{V} \) (rms) AC source supplies a \( 35 \, \Omega \) resistor. What is the average power consumed?

**Average power in LCR** P_avg = V_rms I_rms cos φ, cos φ = R/Z power factor, φ phase between V and I, tan φ = (X_L - X_C)/R. For R=80 Ω, X_L=100 Ω, X_C=40 Ω, X_L-X_C=60 Ω, Z=√(80²+60²)=100 Ω, cos φ=0.8, V_rms=240 V, I_rms=2.4 A, P=240×2.4×0.8=460.8 W, only R dissipates. RMS current: I = (V/R) = (175/35) = 5 A . Average power: P = I² R = 5² × 35 = 875 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 875 W, consistent

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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 Capacitor - Capacitive Reactance

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

**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_L = ω L , ω = 2π × 50 = 314 rad/s . L = 50 × 10⁻³ H . X_L = 314 × 0.05 = 15.7 Ω . RMS current: I = (V/X_L) = (220/15.7) ≈ 14.01 A . Peak current: i_m = √(2) I = 1.414 × 14.01 ≈ 19.81 A . Applying X_L = ωL,

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