#AC

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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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( R = 100 \, \Omega \), \( X_L = 80 \, \Omega \), \( X_C = 60 \, \Omega \). What is the impeda

**LCR example** R=100 Ω, X_L=130 Ω, X_C=70 Ω, X_L-X_C=60 Ω, Z=√(100²+60²)=116.6 Ω, V_rms=300 V, I_rms=2.573 A, power P= I_rms² R =662 W? Actually P= V_rms I_rms cos φ, cos φ=R/Z=0.857, P=300×2.573×0.857=661.6 W, illustrating power factor. Z = √(R² + (X_L - X_C)²) . Z = √(100² + (80 - 60)²) = √(10000 + 400) = √(10400) ≈ 102 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 102 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Through Resistor - Phasor and Power

In an AC circuit with a series combination of resistor and capacitor, what happens to the phase difference if the freque

**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. In an RC series circuit, Φ = tan⁻¹ ( (X_C/R) ) , where X_C = (1/ω C) . As frequency ( ω ) approaches infinity, X_C approaches zero, making Φ approach 0°, so the circuit becomes nearly resistive. 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 It approaches 0°, consistent with

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AC Fundamentals - RMS, Average and Peak Values

What is the significance of the rms value in specifying AC quantities?

**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. The rms (root mean square) value of an AC quantity (e.g., voltage or current) is the equivalent DC value that produces the same average power in a resistive load. It accounts for the time-varying nature of AC, making it a standard measure for power calculations. 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

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