3-Phase Servo AVR (AC Voltage Stabilizer) — Parts, Tests, Repair & Maintenance
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Objectives:
• Identify a.c. current and d.c. voltage.
• Differentiate between peak and r.m.s. values.
• Determine the phase difference between current and voltage.
• Interpret R, R-C, R-L, and R-L-C circuits.
• Analyze vector diagrams and calculate effective voltage, reactance, and impedance.
• Recognize resonance and calculate the resonant frequency (F₀ = 1⁄(2π√(LC))).
• Determine instantaneous power, average power, and power factor in a.c. circuits.
Alternating current (a.c.) is a type of electrical current in which the flow of electrons periodically reverses direction, whereas direct current (d.c.) flows steadily in one direction. In this e‑book, we explain key a.c. concepts and how they apply to simple circuits.
A.c. current alternates its direction, usually in a sinusoidal form, while d.c. voltage remains constant. In many circuits, a.c. voltage is expressed in terms of its peak (maximum) value and its r.m.s. (root mean square) value, which is the effective value.
V_rms = V_peak / √2
The peak value of an a.c. waveform is its maximum amplitude, while the r.m.s. value represents the equivalent d.c. value in terms of power. For a sinusoidal waveform:
I_rms = I_peak / √2
Understanding these values is crucial when designing and analyzing circuits.
The phase difference between current and voltage in an a.c. circuit indicates the time lag between their waveforms. It is often represented by vector diagrams where the voltage and current vectors are drawn at an angle φ.
Power Factor = cos φ
In a resistor (R) circuit, the a.c. voltage and current are in phase. In a capacitive circuit (R-C), the current leads the voltage by 90º, and in an inductive circuit (R-L), the current lags the voltage by 90º.
In an R-L-C circuit, the resistor (R), inductor (L), and capacitor (C) work together. The total opposition to current, called impedance (Z), is given by:
Z = √(R² + (X_L - X_C)²)
where X_L = 2πfL is the inductive reactance and X_C = 1/(2πfC) is the capacitive reactance.
Resonance occurs when the inductive and capacitive reactances cancel each other (X_L = X_C), minimizing impedance. The resonant frequency is given by:
F₀ = 1⁄(2π√(LC))
At resonance, the circuit’s current is at a maximum, and the power factor is unity.
In a.c. circuits, the instantaneous power varies with time. The average power is given by:
P_avg = V_rms × I_rms × cos φ
The power factor (cos φ) indicates the phase difference between voltage and current.
Example 1: A sinusoidal a.c. voltage has a peak value of 170 V. Calculate its r.m.s. value.
Solution:
V_rms = V_peak / √2 = 170 / 1.414 ≈ 120.2 V.
Example 2: In an R-L circuit with R = 10 Ω and L = 0.05 H at f = 50 Hz, determine the inductive reactance (X_L).
Solution:
X_L = 2πfL = 2π×50×0.05 = 15.71 Ω.
Example 3: For an R-C circuit with R = 20 Ω and C = 100 μF at f = 60 Hz, calculate the capacitive reactance (X_C).
Solution:
X_C = 1/(2πfC) = 1/(2π×60×100×10⁻⁶) ≈ 26.53 Ω.
Example 4: In an R-L-C circuit with R = 15 Ω, L = 0.1 H, and C = 50 μF at 50 Hz, calculate the impedance (Z).
Solution:
X_L = 2π×50×0.1 = 31.42 Ω; X_C = 1/(2π×50×50×10⁻⁶) ≈ 63.66 Ω.
Net reactance, X = |X_L - X_C| = |31.42 - 63.66| = 32.24 Ω.
Impedance Z = √(R² + X²) = √(15² + 32.24²) ≈ √(225 + 1039) ≈ √1264 ≈ 35.56 Ω.
Example 5: Determine the resonant frequency of an R-L-C circuit with L = 0.2 H and C = 20 μF.
Solution:
F₀ = 1/(2π√(LC)) = 1/(2π√(0.2×20×10⁻⁶)) = 1/(2π√(4×10⁻⁶)) = 1/(2π×0.002) ≈ 79.58 Hz.
Example 6: A.c. current and voltage in a circuit have a phase difference of 30º. If V_rms = 120 V and I_rms = 10 A, calculate the average power and power factor.
Solution:
Power Factor = cos(30º) ≈ 0.866.
Average Power = V_rms × I_rms × cos(30º) = 120 × 10 × 0.866 ≈ 1039.2 W.
Example 7: In an a.c. circuit, if the peak voltage is 200 V, determine the r.m.s. voltage.
Solution:
V_rms = 200/√2 ≈ 141.42 V.
Example 8: An a.c. circuit has a phase angle of 45º. If the effective current is 8 A, what is the power factor?
Solution:
Power Factor = cos(45º) ≈ 0.707.
Example 9: For an R-L circuit with R = 12 Ω and L = 0.08 H at 60 Hz, calculate the phase difference between current and voltage.
Solution:
X_L = 2π×60×0.08 ≈ 30.16 Ω.
Phase angle φ = tan⁻¹(X_L/R) = tan⁻¹(30.16/12) ≈ tan⁻¹(2.513) ≈ 68.7º.
Example 10: In an R-L-C circuit at resonance, the impedance equals the resistance. If R = 25 Ω and the circuit is resonant, what is the effective voltage across the resistor when I_rms = 5 A?
Solution:
At resonance, impedance Z = R = 25 Ω.
Effective voltage, V_rms = I_rms × R = 5 A × 25 Ω = 125 V.
Alternating current circuits are used in power distribution and many electronic devices. Understanding vector diagrams and phase relationships is key to designing efficient circuits. Resonant circuits are crucial in radio, filtering, and signal processing.
Always check your calculations for peak-to-r.m.s. conversions and verify phase angles using vector diagrams. Remember, at resonance, the impedance of an R-L-C circuit is minimum and equals the resistance, maximizing current flow.
Review these key concepts and test your knowledge with the quiz below.
Click the "Start Quiz" button to begin. You will have 15 minutes to answer 30 questions.
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