Single-Phase AC Circuits: Waveforms, Power, Resonance & Practical Applications

Core Concepts: Sinusoidal AC Waveforms and Their Properties

1. Sinusoidal Waveform and Its Parameters

A sinusoidal AC voltage/current varies as $v(t)=V_m\sin (\omega t+\varphi )$, where:

• $V_m$: Peak (maximum) voltage
• $\omega =2\pi f$: Angular frequency (rad/s)
• $\varphi$: Phase angle (degrees or radians)

Key Parameters:

Comparison of full-wave vs. half-wave rectification (Image: Omegatron, CC BY-SA 3.0, via Wikimedia Commons)

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2. Power in AC Circuits

a) Instantaneous, Average, and Reactive Power

• Instantaneous Power ($p(t)$): $p(t)=v(t)\cdot i(t)$. For purely resistive circuits, $p(t)$ is always positive. For inductive/capacitive circuits, power oscillates between source and load.

• Average Power ($P$): $P=V_{\text{rms}}{I}_{\text{rms}}\cos (\varphi )$, where $\cos (\varphi )$ is the power factor.

  • Unity Power Factor (UPF): $\varphi =0^{\circ }$ (purely resistive load).
  • Lagging PF: Inductive load ($\varphi >0$).
  • Leading PF: Capacitive load ($\varphi <0$).

• Reactive Power ($Q$): $Q=V_{\text{rms}}{I}_{\text{rms}}\sin (\varphi )$ (measured in VAR).

  • Represents energy oscillating between source and load (no real work done).
  • Apparent Power ($S$): $S=\sqrt{P^2+Q^2}$ (measured in VA).

Relationship between P, Q, and S (Image: Wikieditor4321, CC BY-SA 4.0, via Wikimedia Commons)

b) Worked Example: Power Calculation for a Fluorescent Lamp

A 40 W fluorescent lamp operates at 220 V, 50 Hz with a power factor of 0.6 (lagging).

1. Calculate RMS current: $P=V_{\text{rms}}{I}_{\text{rms}}\cos (\varphi )$ $40=220\times I\times 0.6$ $I=\frac{40}{220\times 0.6}=0.303\text{A}$.

2. Calculate Reactive Power: $Q=V_{\text{rms}}{I}_{\text{rms}}\sin (\varphi )$ $\varphi ={\cos }^{-1}(0.6)=53.13^{\circ }$ $Q=220\times 0.303\times \sin (53.13^{\circ })=40\text{VAR}$.

3. Apparent Power: $S=\sqrt{40^2+40^2}=56.57\text{VA}$.

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In the Real World

1. NTC Power Grid (Nepal):

   • Idea Used: Three-phase power distribution (though this unit covers single-phase, NTC’s low-voltage grids use single-phase AC for residential areas).
   • How: Single-phase transformers step down 11 kV to 230 V for homes. The power factor correction (adding capacitors to offset inductive loads like motors) improves efficiency, reducing reactive power losses.
   • Example: During peak hours (6–9 PM), NTC monitors lagging power factors in factories and penalizes low PF to discourage inefficient usage.

2. WhatsApp Servers (Global):

   • Idea Used: AC power conditioning and resonance in filters.
   • How: Data centers use active harmonic filters (RLC circuits tuned to specific frequencies) to suppress noise in power supplies. For example, a server’s power supply might have a series resonant circuit at 50/60 Hz to reject voltage spikes, ensuring stable operation for message routing.

3. Daraz Logistics Warehouses (Nepal):

   • Idea Used: Single-phase induction motors and power factor correction.
   • How: Conveyor belts in warehouses use single-phase capacitor-start motors (a type of AC motor). To improve efficiency, warehouses install power factor correction capacitors (e.g., 50 µF banks) to compensate for the inductive load of motors, reducing electricity bills by up to 15%.

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Series RLC Circuits and Resonance

1. Impedance in RLC Circuits

The total impedance ($Z$) of a series RLC circuit is: $$Z=\sqrt{R^2+(X_L-X_C{)}^2}$$ where:

• $X_L=\omega L$: Inductive reactance (Ω)
• $X_C=\frac{1}{\omega C}$: Capacitive reactance (Ω)

Components: resistor, inductor, capacitor, AC source (Image: Nkkoy, CC0, via Wikimedia Commons)

2. Resonance in Series RLC Circuits

• Resonance Condition: $X_L=X_C$ ⇒ ${\omega }_0=\frac{1}{\sqrt{LC}}$. At resonance:
  • Impedance is purely resistive ($Z=R$).
  • Current is maximum ($I=\frac{V}{R}$).
  • Bandwidth ($\Delta \omega$) = $\frac{R}{L}$.

Worked Example: Resonance in a Radio Tuner

A radio tuner has $L=2\text{mH}$ and $C=0.5\text{µF}$. Find:

1. Resonant Frequency: ${\omega }_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{2\times 10^{-3}\times 0.5\times 10^{-6}}}=100,000\text{rad/s}$ $f_0=\frac{{\omega }_0}{2\pi }=15.92\text{kHz}$.

2. Bandwidth if $R=10\Omega$: $\Delta \omega =\frac{R}{L}=\frac{10}{2\times 10^{-3}}=5,000\text{rad/s}$.

Real-World Tie-In: This principle is used in Nepal’s FM radio stations (e.g., Radio Nepal). The tuner in your radio has a variable capacitor that adjusts resonance to pick up specific frequencies (e.g., 90.8 MHz for Radio Nepal Kathmandu).

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Parallel RLC Circuits

1. Admittance and Resonance

• Admittance ($Y$): $Y=\sqrt{G^2+(B_L-B_C{)}^2}$, where $G=\frac{1}{R}$, $B_L=\frac{1}{\omega L}$, $B_C=\omega C$.
• Resonance Condition: $B_L=B_C$ ⇒ ${\omega }_0=\frac{1}{\sqrt{LC}}$ (same as series). At resonance:
  • Admittance is purely conductive ($Y=G$).
  • Impedance is maximum ($Z=R$).

Components: resistor, inductor, capacitor, AC source (Image: SpinningSpark, CC BY-SA 3.0, via Wikimedia Commons)

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Power Measurement in AC Circuits

1. Wattmeter Method for Single-Phase Power

• Single Wattmeter Method: Measures average power directly: $P=VI\cos (\varphi )$.

  • Limitation: Cannot distinguish between active and reactive power.

• Two-Wattmeter Method (for 3-phase, but concept applies to single-phase analysis): Uses two wattmeters to measure: $P=W_1+W_2$ $Q=\sqrt{3}(W_1-W_2)$ Difference from One-Wattmeter Method:

  Feature	One-Wattmeter Method	Two-Wattmeter Method
  Accuracy	Measures only active power	Measures both P and Q
  Complexity	Simple	Requires two instruments
  Application	Single-phase circuits	Three-phase circuits

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Practical Applications

1. Transformers (Single-Phase)

• Idea: Step-up/down AC voltages using mutual inductance.
• Example: Nepal’s 11 kV/400 V distribution transformers (used by NTC).
  • Turns Ratio: $\frac{V_p}{V_s}=\frac{N_p}{N_s}$.
  • Worked Example: A transformer steps down 11,000 V to 230 V. If the primary has 5,500 turns, find secondary turns: $\frac{11,000}{230}=\frac{5,500}{N_s}$ $N_s=110$ turns.

2. AC Motors (Induction Motors)

• Idea: Rotating magnetic field induces current in the rotor (Lenz’s law).
• Example: Single-phase induction motors in Daraz’s warehouse fans.
  • Starting Method: Capacitor-start (creates phase shift for initial torque).
  • Power Factor: Typically 0.7–0.8 lagging (requires PF correction).

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Exam Tip

What Examiners Look For

1. Waveform Analysis:

   • Always sketch the waveform when asked about half-wave/full-wave rectification.
   • Calculate RMS, average, and form factor from the given equation or graph.
   • Common Mistake: Forgetting to multiply by $\frac{2}{\pi }$ for average value of half-wave rectified sine waves.

2. Power Calculations:

   • Memorize the power triangle ($P,Q,S$) and use it to derive relationships.
   • For lagging/leading PF, draw the phasor diagram to visualize phase angles.
   • Shortcut: If $\cos (\varphi )$ is given, $\sin (\varphi )=\sqrt{1-{\cos }^2(\varphi )}$.

3. Resonance:

   • Series resonance: $X_L=X_C$, $Z=R$, current peaks.
   • Parallel resonance: $B_L=B_C$, $Z$ peaks.
   • Exam Question: Always check if the circuit is series or parallel before applying resonance conditions.

4. Practical Scenarios:

   • NTC Grid: Relate power factor correction to reducing losses.
   • Motors: Explain why capacitors are used in single-phase motors (to create a rotating field).
   • Transformers: Link turns ratio to voltage step-up/down in real systems (e.g., NTC transformers).

5. Diagrams:

   • Mandatory: Draw phasor diagrams for RLC circuits to show phase relationships.
   • Label everything: Resistors, inductors, capacitors, and angles in phasor diagrams.

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Common Pitfalls

• Assuming all AC circuits are resistive: Always consider $X_L$ and $X_C$.
• Ignoring units: Power in watts (W), reactive power in VAR, apparent power in VA.
• Mixing series/parallel formulas: $Z$ for series, $Y$ for parallel.
• Forgetting to convert frequencies: $\omega =2\pi f$ before plugging into $X_L$ or $X_C$.

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Quick Revision Checklist

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Final Worked Example: Exam-Style Question

Question: A single-phase load consists of a resistor $R=30\Omega$ in series with an inductor $L=0.1\text{H}$. The supply is $v(t)=200\sin (314t)$ V. Find: a) The inductive reactance $X_L$. b) The impedance $Z$. c) The RMS current. d) The average power dissipated.

Solution: a) Inductive Reactance: $\omega =314\text{rad/s}$ $X_L=\omega L=314\times 0.1=31.4\Omega$.

b) Impedance: $Z=\sqrt{R^2+X_L^2}=\sqrt{30^2+31.4^2}=43.5\Omega$.

c) RMS Current: $V_{\text{rms}}=\frac{200}{\sqrt{2}}=141.4\text{V}$ $I_{\text{rms}}=\frac{V_{\text{rms}}}{Z}=\frac{141.4}{43.5}=3.25\text{A}$.

d) Average Power: $\varphi ={\tan }^{-1}\left(\frac{X_L}{R}\right)={\tan }^{-1}\left(\frac{31.4}{30}\right)=46.4^{\circ }$ $P=V_{\text{rms}}{I}_{\text{rms}}\cos (\varphi )=141.4\times 3.25\times \cos (46.4^{\circ })=325\text{W}$.

Real-World Tie-In: This is analogous to a single-phase motor (e.g., a water pump in a rural Nepalese household). The inductive load (motor winding) causes a lagging power factor, and the average power calculated here represents the actual mechanical work done by the pump.