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#electrical engineering

82 public questions tagged with this topic.

Power in AC Circuits - Power Factor and Wattless Current

A series LCR circuit has \( R = 40 \, \Omega \), \( X_L = 60 \, \Omega \), \( X_C = 20 \, \Omega \). What is the power f

**Power factor** cos φ = R/Z, 0≤cos φ≤1, at resonance cos φ=1 maximum power, pure L or C cos φ=0 zero average power, current wattless because energy stored returned. Instantaneous power when current maximum in pure L: I=I_peak, V=0 because V leads 90°, so P= V I =0 at that instant, average zero. Z = √(R² + (X_L - X_C)²) = √(40² + (60 - 20)²) = √(1600 + 1600) = √(3200) ≈ 56.57 Ω . Power factor: cos Φ = (R/Z) = (40/56.57) ≈ 0.707 . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P

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AC Through Inductor - Inductive Reactance

A \( 25 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) 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 × 60 = 376.8 rad/s . C = 25 × 10⁻⁶ F . X_C = (1/376.8 × 25 × 10⁻⁶) ≈ 106.2 Ω . RMS current: I = (V/X_C) = (110/106.2) ≈ 1.036 A . Applying X_L = ωL, X_C = 1/ωC, Z =

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AC Through Inductor - Inductive Reactance

A \( 27 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) 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_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 27 × 10⁻⁶ F . X_C = (1/314 × 27 × 10⁻⁶) ≈ 117.8 Ω . RMS current: I = (V/X_C) = (220/117.8) ≈ 1.867 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 1.867

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AC Through Capacitor - Capacitive Reactance

A \( 9 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) 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 = 9 × 10⁻⁶ F . X_C = (1/314 × 9 × 10⁻⁶) ≈ 353.7 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 \( 14 \, \mu\text{F} \) 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 = 14 × 10⁻⁶ F . X_C = (1/314 × 14 × 10⁻⁶) ≈ 227.5 Ω . RMS current: I = (V/X_C) = (230/227.5) ≈ 1.01 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 1.01

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AC Through Resistor - Phasor and Power

A \( 60 \, \Omega \) resistor and \( 40 \, \mu\text{F} \) capacitor are in series with a \( 230 \, \text{V} \), \( 50 \,

**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) = (1/314 × 40 × 10⁻⁶) ≈ 79.6 Ω . Z = √(R² + X_C²) = √(60² + 79.6²) = √(3600 + 6336.16) ≈ 99.8 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 99.8 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Fundamentals - RMS, Average and Peak Values

A series LCR circuit with \( R = 40 \, \Omega \), \( X_L = 50 \, \Omega \), \( X_C = 30 \, \Omega \) has a \( 120 \, \te

**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. Z = √(R² + (X_L - X_C)²) = √(40² + (50 - 30)²) = √(1600 + 400) = √(2000) ≈ 44.72 Ω . RMS current: I = (V/Z) = (120/44.72) ≈ 2.68 A . Power: P = I² R = (2.68)² × 40 ≈ 287.3 W . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L -

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AC Fundamentals - RMS, Average and Peak Values

A \( 100 \, \Omega \) resistor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC supply. What is the ave

**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. Average power: P = I² R , where I = (V/R) . I = (220/100) = 2.2 A . P = (2.2)² × 100 = 4.84 × 100 = 484 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 484 W, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Through Inductor - Inductive Reactance

A \( 120 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the

**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 = 120 × 10⁻³ H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.12 = 37.68 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 \( 55 \, \text{mH} \) 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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