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Basics of Electrical and Electronics Engineering•Unit 2• EEE1200

Module 2: AC Circuits (Single-Phase & Three-Phase Systems)

Sinusoidal voltage generation, AC waveform metrics (RMS, Average, Form & Peak factors), pure R, L, C dynamics, series R-L circuits, power triangle, and balanced three-phase Star (Y) vs. Delta (Δ) networks.

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Skills Built:Faraday Rotating Coil EMF FormulationRMS & Half-Cycle Average MetricsForm Factor (1.11) & Peak Factor (1.414)Pure Element Quadrature (±90°) Phase DynamicsSeries R-L Impedance Triangle (Z² = R² + X_L²)Power Triangle: Active (W), Reactive (VAR), Apparent (VA)Three-Phase Star (V_L = √3 V_ph) & Delta (I_L = √3 I_ph)3-Phase Power Equations & 3x Delta Power Advantage

1. 5 Key Ideas: Fundamental Principles of AC & Polyphase Systems

Idea 1: Sinusoidal EMF Generation & Faraday's Principle

Alternating voltage is generated mechanically in AC alternators by rotating a planar coil of NN turns and cross-sectional area AA with uniform angular velocity ω\omega (rad/s) inside a stationary magnetic field of flux density BB.

  1. When the coil's plane is aligned with the magnetic axis, the magnetic flux linking the coil is at its maximum: Φm=BmA\Phi_m = B_m A.
  2. In time tt seconds, the coil rotates through an angular displacement θ=ωt\theta = \omega t.
  3. The instantaneous perpendicular component of flux linking the coil becomes: Φ(t)=Φmcos⁡(ωt)\Phi(t) = \Phi_m \cos(\omega t)
  4. By Faraday's Law of Electromagnetic Induction, the instantaneous induced EMF across the coil's terminals is: e(t)=−ddt[NΦ(t)]=−Nddt[Φmcos⁡(ωt)]=NΦmωsin⁡(ωt)e(t) = -\frac{d}{dt}\left[N \Phi(t)\right] = -N \frac{d}{dt}\left[\Phi_m \cos(\omega t)\right] = N \Phi_m \omega \sin(\omega t)
  5. The maximum peak EMF occurs at ωt=90∘\omega t = 90^\circ (sin⁡ωt=1\sin\omega t = 1): Em=ωNΦm=2πfNBmA[Volts]E_m = \omega N \Phi_m = 2\pi f N B_m A \quad [\text{Volts}]   ⟹  e(t)=Emsin⁡(ωt),i(t)=Imsin⁡(ωt)\implies e(t) = E_m \sin(\omega t), \qquad i(t) = I_m \sin(\omega t)

Idea 2: AC Waveform Metrics — RMS, Average, Form & Peak Factors

Because AC sinusoidal quantities reverse direction periodically, their mathematical properties are quantified through standardized metrics:

  • Cycle: One complete set of positive and negative alternating values (360∘360^\circ or 2π radians2\pi\text{ radians}).
  • Time Period (TT): The duration required to complete one full cycle (T=1/fT = 1/f seconds).
  • Frequency (ff): Number of complete cycles executed per second (f=1/T Hertzf = 1/T\text{ Hertz}).
  • Root Mean Square (RMS) / Effective Value (Vrms,IrmsV_{\text{rms}}, I_{\text{rms}}): The steady direct current (DC) that produces the exact same amount of thermal heat in an identical resistor over the same duration: Irms=12π∫02πi2(θ) dθ=Im2≈0.7071 ImI_{\text{rms}} = \sqrt{\frac{1}{2\pi} \int_0^{2\pi} i^2(\theta) \, d\theta} = \frac{I_m}{\sqrt{2}} \approx 0.7071 \, I_m
  • Average Value (Vavg,IavgV_{\text{avg}}, I_{\text{avg}}): The arithmetic mean of mid-ordinates taken over one half-cycle (the full-cycle average of any symmetrical wave is zero): Iavg=1π∫0πImsin⁡θ dθ=2Imπ≈0.6366 ImI_{\text{avg}} = \frac{1}{\pi} \int_0^\pi I_m \sin\theta \, d\theta = \frac{2 I_m}{\pi} \approx 0.6366 \, I_m
  • Form Factor (KfK_f): The ratio of the RMS value to the half-cycle average value. For an ideal sinusoidal wave: Kf=RMS ValueAverage Value=0.7071 Im0.6366 Im=π22≈1.11K_f = \frac{\text{RMS Value}}{\text{Average Value}} = \frac{0.7071 \, I_m}{0.6366 \, I_m} = \frac{\pi}{2\sqrt{2}} \approx 1.11
  • Crest / Peak / Amplitude Factor (KpK_p): The ratio of the maximum peak amplitude to the RMS value: Kp=Maximum ValueRMS Value=ImIm/2=2≈1.414K_p = \frac{\text{Maximum Value}}{\text{RMS Value}} = \frac{I_m}{I_m / \sqrt{2}} = \sqrt{2} \approx 1.414

Idea 3: Single-Element AC Circuit Dynamics (RR, LL, CC)

  • Pure Resistance (RR): The voltage and current waves pass through zero and peak at the exact same instant: v(t)=Vmsin⁡(ωt),i(t)=Imsin⁡(ωt)  ⟹  ϕ=0∘    (In-Phase)v(t) = V_m \sin(\omega t), \quad i(t) = I_m \sin(\omega t) \implies \phi = 0^\circ \;\; (\text{In-Phase}) The instantaneous power p(t)=v(t)⋅i(t)=VmIm2−VmIm2cos⁡(2ωt)p(t) = v(t) \cdot i(t) = \frac{V_m I_m}{2} - \frac{V_m I_m}{2}\cos(2\omega t) pulsates at twice the supply frequency (2f2f). Its average value is strictly positive: Pavg=VrmsIrms=I2RP_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} = I^2 R [Watts].
  • Pure Inductance (LL): The induced back-EMF (e=−L di/dte = -L \, di/dt) opposes changes in current. By integration, the current lags the applied voltage by 90∘90^\circ (π/2 rad\pi/2\text{ rad}): i(t)=VmωLsin⁡(ωt−π2)=Imsin⁡(ωt−π2),XL=ωL=2πfL  [Ω]i(t) = \frac{V_m}{\omega L} \sin\left(\omega t - \frac{\pi}{2}\right) = I_m \sin\left(\omega t - \frac{\pi}{2}\right), \quad X_L = \omega L = 2\pi f L \; [\Omega] The instantaneous power p(t)=−VmIm2sin⁡(2ωt)p(t) = -\frac{V_m I_m}{2}\sin(2\omega t) is a pure sine wave oscillating at 2f2f. Energy is alternately absorbed into and released from the magnetic field; average real power consumed is strictly zero (Pavg=0 WP_{\text{avg}} = 0\text{ W}).
  • Pure Capacitance (CC): The current is governed by charge accumulation (i=dq/dt=C dv/dti = dq/dt = C \, dv/dt). The current leads the applied voltage by 90∘90^\circ (π/2 rad\pi/2\text{ rad}): i(t)=ωCVmcos⁡(ωt)=Imsin⁡(ωt+π2),XC=1ωC=12πfC  [Ω]i(t) = \omega C V_m \cos(\omega t) = I_m \sin\left(\omega t + \frac{\pi}{2}\right), \quad X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \; [\Omega] Energy is alternately stored in and discharged from the electrostatic field; average real power is zero (Pavg=0 WP_{\text{avg}} = 0\text{ W}).

Idea 4: Series RR-LL Circuit & The Power Triangle

When a resistance RR and an inductance LL are connected in series across an AC supply v(t)=Vmsin⁡(ωt)v(t) = V_m \sin(\omega t):

  • The resistive drop VR=IRV_R = IR is in phase with current II.
  • The inductive drop VL=IXLV_L = I X_L leads current II by 90∘90^\circ.
  • By phasor vector addition, total applied voltage is: V=VR2+VL2=IR2+XL2=IZV = \sqrt{V_R^2 + V_L^2} = I \sqrt{R^2 + X_L^2} = I Z
  • Circuit Impedance (ZZ): Z=R2+XL2[Ω],ϕ=arctan⁡(XLR)Z = \sqrt{R^2 + X_L^2} \quad [\Omega], \qquad \phi = \arctan\left(\frac{X_L}{R}\right)
  • The current lags the applied voltage by phase angle ϕ\phi: i(t)=Imsin⁡(ωt−ϕ)i(t) = I_m \sin(\omega t - \phi).
  • The Three Powers:
    1. Active / Real Power (PP): Rate of electrical energy permanently converted to heat and mechanical work: P=VIcos⁡ϕ=I2R[Watts (W)]P = V I \cos\phi = I^2 R \quad [\text{Watts (W)}]
    2. Reactive Power (QQ): Energy oscillating between supply and magnetic/electric fields without doing real work: Q=VIsin⁡ϕ=I2XL[Volt-Amperes Reactive (VAR)]Q = V I \sin\phi = I^2 X_L \quad [\text{Volt-Amperes Reactive (VAR)}]
    3. Apparent Power (SS): The total capacity of the electrical supply apparatus (transformers and cables): S=VI=I2Z=P2+Q2[Volt-Amperes (VA)]S = V I = I^2 Z = \sqrt{P^2 + Q^2} \quad [\text{Volt-Amperes (VA)}]
    4. Power Factor (p.f.\text{p.f.}): p.f.=cos⁡ϕ=RZ=PS(Lagging for inductive circuits)\text{p.f.} = \cos\phi = \frac{R}{Z} = \frac{P}{S} \quad (\text{Lagging for inductive circuits})

Idea 5: Polyphase Systems & The Star (YY) vs. Delta (Δ\Delta) Topologies

A three-phase alternator generates three identical sinusoidal EMFs mutually displaced by 120∘120^\circ (2π/3 rad2\pi/3\text{ rad}): eR=Emsin⁡(ωt),eY=Emsin⁡(ωt−120∘),eB=Emsin⁡(ωt+120∘)e_R = E_m \sin(\omega t), \quad e_Y = E_m \sin(\omega t - 120^\circ), \quad e_B = E_m \sin(\omega t + 120^\circ)

Seven Advantages of 3-Phase Systems Over Single-Phase:

  1. Conductor Economy: Requires less conductor mass to transmit equal power over equal distance at identical loss.
  2. Uniform Constant Torque: Produces a smooth rotating magnetic field (single-phase machines produce pulsating torque and vibration).
  3. Inherently Self-Starting: 3-phase induction motors start autonomously without auxiliary starting switches or capacitors.
  4. Higher Power Density & Lower Cost: 3-phase machines are physically smaller and less costly for the same kVA rating.
  5. No Harmonics in Parallel Alternators: Third harmonics cancel out in closed delta windings or balanced star connections.
  6. Dual-Voltage Availability: Star networks provide two voltages: Line-to-Line (400 V400\text{ V} for motors) and Line-to-Neutral (230 V230\text{ V} for domestic appliances).
  7. Balanced Neutral Cancellation: In a balanced star system, the vector sum of currents in the neutral wire is zero (IR+IY+IB=0I_R + I_Y + I_B = 0).

2. Formula & Unit Quick-Reference Strip

Formula & Unit Quick-Reference Strip
Peak Generated EMFVolts (V)
E_m = 2π * f * N * B_m * A

Faraday peak induced EMF in a coil of N turns and area A rotating at speed f in flux density B_m.

RMS & Average ValuesVolts (V)
V_rms = V_m / √2, V_avg = (2 * V_m) / π

Effective heating equivalent (0.7071 * V_m) and half-cycle arithmetic mean (0.6366 * V_m).

Form & Peak FactorsDimensionless
K_f = V_rms / V_avg = 1.11, K_p = V_m / V_rms = 1.414

Form factor (shape ratio) and Crest/Peak factor (peak ratio) for ideal sinusoidal AC waveforms.

Inductive & Capacitive ReactanceOhms (Ω)
X_L = 2π * f * L, X_C = 1 / (2π * f * C)

Opposition offered to sinusoidal current by inductor (proportional to f) and capacitor (inverse to f).

Series R-L Circuit ImpedanceOhms (Ω), rad/deg
Z = √(R² + X_L²), tan(φ) = X_L / R

Phasor hypotenuse impedance and lagging current phase angle in series R-L branch.

Active, Reactive & Apparent PowerW (Watts), VAR, VA
P = V*I*cos(φ), Q = V*I*sin(φ), S = V*I

Power triangle relationship: S = √(P² + Q²) with power factor pf = cos(φ) = P / S = R / Z.

Star (Y) Connection RelationsVolts (V), Amperes (A)
V_L = √3 * V_ph, I_L = I_ph

Line voltage leads phase voltage by 30°; line current equals phase winding current.

Delta (Δ) Connection RelationsVolts (V), Amperes (A)
V_L = V_ph, I_L = √3 * I_ph

Line voltage equals phase winding voltage; line current lags phase current by 30°.

Total Three-Phase PowerWatts (W)
P = √3 * V_L * I_L * cos(φ) = 3 * V_ph * I_ph * cos(φ)

Universal total active power consumed in balanced 3-phase load (P_Δ = 3 * P_Y for same load).


3. Mathematical Derivations: Sinusoidal Generation, Reactance & 3-Phase Systems


4. Typical Laboratory Hardware Values & Grid Standards

Typical Laboratory Hardware Ranges
Parameter / PartTypical ValueLab RangePractical Notes
Indian Utility Grid Frequency50.0 Hz (±1%)49.5 Hz – 50.5 HzNational grid standard; T = 20.0 ms per cycle.
Domestic Single-Phase Voltage230 V RMS (±10%)207 V – 253 VIndian standard IS 12360; peak voltage V_m = 325.3 V.
Three-Phase Distribution Voltage400 V / 415 V Line-to-Line380 V – 440 VStandard 3-phase 4-wire industrial/commercial service.
Utility Power Factor Penalty Threshold0.85 to 0.90 lagging0.80 – 0.95Power distribution utilities penalize industrial customers operating below 0.85 pf.
Ballast / Choke Coil Inductance0.1 H to 2.5 H0.05 H – 5.0 HLaminated silicon steel core to limit alternating current.
P.F. Correction Capacitor Bank10 µF to 200 µF per kVAR1 µF – 1,000 µFMetallized polypropylene film capacitors rated for continuous AC voltage.

5. Worked Examples: Presidency University Curriculum Solutions

Step-by-Step Worked Problem

[Presidency University Slide 20] A supply voltage has a mean value of 150 V. Determine its maximum peak value and its RMS value. Assume a symmetrical sinusoidal waveform.

Step 1: Relate Mean (Average) Value to Maximum Value
For an ideal sinusoidal voltage, the mean (average) value over one half-cycle is given by: V_avg = 0.637 * V_max. Rearranging to solve for V_max: V_max = V_avg / 0.637 = 150 / 0.637 ≈ 235.48 V.
Step 2: Calculate RMS (Effective) Value
The RMS value of a symmetrical sine wave is: V_rms = V_max / √2 = 0.7071 * V_max. Substituting V_max = 235.48 V: V_rms = 0.7071 * 235.48 ≈ 166.5 V.
Step-by-Step Worked Problem

[Presidency University Slide 26] An AC voltage waveform has an RMS value of 115.6 V, an average value of 100 V, and a maximum value of 200 V. Determine the form factor and the peak factor of the waveform.

Step 1: Calculate Form Factor (K_f)
Form factor is defined as the ratio of RMS value to Average value: Form Factor K_f = V_rms / V_avg = 115.6 V / 100 V = 1.156.
Step 2: Calculate Peak Factor (K_p)
Peak (crest) factor is defined as the ratio of Maximum value to RMS value: Peak Factor K_p = V_max / V_rms = 200 V / 115.6 V ≈ 1.73.
Step-by-Step Worked Problem

[Presidency University Slide 54–55] A circuit consists of a resistance of 20 Ω and an inductance of 0.05 H connected in series across a 230 V, 50 Hz supply. Find: (1) Current drawn, (2) Power factor, (3) Power consumed, and (4) Phase angle with vector diagram drops.

Step 1: Calculate Inductive Reactance (X_L)
X_L = 2π f L = 2 * 3.1416 * 50 * 0.05 = 15.71 Ω.
Step 2: Calculate Total Circuit Impedance (Z)
Z = √(R² + X_L²) = √(20² + 15.71²) = √(400 + 246.8) = √646.8 ≈ 25.43 Ω.
Step 3: Calculate Circuit Current (I)
I = V / Z = 230 V / 25.43 Ω = 9.045 A ≈ 9.04 A.
Step 4: Calculate Power Factor (p.f.) & Active Power (P)
Power factor = cos(φ) = R / Z = 20 / 25.43 = 0.786 (lagging). Active Power P = V * I * cos(φ) = 230 * 9.045 * 0.786 ≈ 1634.4 W.
Step 5: Voltage Drops & Phase Angle
Resistive drop V_R = I * R = 9.04 * 20 = 180.8 V. Inductive drop V_L = I * X_L = 9.04 * 15.71 = 142.0 V. Phase angle φ = arctan(X_L / R) = arctan(15.71 / 20) = 38.15° ≈ 38° lagging.
Step-by-Step Worked Problem

[Presidency University Slide 59–60] A coil takes a current of 2 A from a 12 V DC supply. When connected to a 240 V, 50 Hz AC supply, the current is 20 A. Calculate: (1) Resistance, (2) Impedance, (3) Inductive reactance, and (4) Inductance of the coil.

Step 1: Calculate DC Resistance (R)
On DC supply, frequency is zero (f = 0), so inductive reactance is zero (X_L = 0). The opposition is purely resistive: R = V_dc / I_dc = 12 V / 2 A = 6 Ω.
Step 2: Calculate AC Impedance (Z)
When connected to 240 V AC: Z = V_ac / I_ac = 240 V / 20 A = 12 Ω.
Step 3: Calculate Inductive Reactance (X_L)
Since Z = √(R² + X_L²), we have: X_L = √(Z² - R²) = √(12² - 6²) = √(144 - 36) = √108 ≈ 10.39 Ω.
Step 4: Calculate Inductance (L)
Since X_L = 2π f L, we have: L = X_L / (2π f) = 10.39 / (2 * 3.1416 * 50) = 10.39 / 314.16 ≈ 0.0331 H = 33.1 mH.
Step-by-Step Worked Problem

[Presidency University Slide 82–83] A 400 V, 3-phase supply is connected across a balanced load of three impedances, each consisting of a 32 Ω resistance and a 24 Ω inductive reactance in series. Determine the current drawn from the power mains, the power factor, and the power consumed if the three impedances are: (a) Star (Y) connected, and (b) Delta (Δ) connected.

Step 1: Calculate Phase Impedance (Z_ph) and Power Factor
For each phase branch: Z_ph = √(R² + X_L²) = √(32² + 24²) = √(1024 + 576) = √1600 = 40 Ω. Power factor = cos(φ) = R_ph / Z_ph = 32 / 40 = 0.8 (lagging). Notice that Z_ph and power factor are completely independent of whether the coils are wired in Star or Delta!
Step 2: Case (a) Star (Y) Connection
Phase voltage: V_ph = V_L / √3 = 400 / √3 ≈ 230.94 V (231 V). Phase current: I_ph = V_ph / Z_ph = 230.94 / 40 = 5.774 A ≈ 5.78 A. In Star, line current equals phase current: I_L = I_ph = 5.78 A. Total active power: P_Y = √3 * V_L * I_L * cos(φ) = √3 * 400 * 5.774 * 0.8 ≈ 3200 W (3204.4 W with rounded 231V).
Step 3: Case (b) Delta (Δ) Connection
Phase voltage equals line voltage: V_ph = V_L = 400 V. Phase current: I_ph = V_ph / Z_ph = 400 / 40 = 10 A. In Delta, line current is: I_L = √3 * I_ph = √3 * 10 ≈ 17.32 A. Total active power: P_Δ = √3 * V_L * I_L * cos(φ) = √3 * 400 * 17.32 * 0.8 = 9600 W.
Step 4: Critical Comparative Takeaway
Comparing the two connection powers: P_Δ / P_Y = 9600 W / 3200 W = 3.0. The Delta configuration draws exactly 3 times the power and draws √3 times the line current of the Star configuration!

6. Common Exam Traps & Where Marks Are Lost

Common Exam Traps & Where Marks Are Lost

Avoid these classic calculation errors frequently spotted on semester examination answer scripts:

  • Peak vs. RMS in Power Calculations: Never use peak voltage (V_m) in standard power formulas without dividing by 2! Active power is P = V_rms * I_rms * cos(φ) = (V_m * I_m / 2) * cos(φ). Using P = V_m * I_m * cos(φ) produces an answer that is exactly 200% too large.
  • Full-Cycle vs. Half-Cycle Average Value Trap: The average value of a symmetrical AC sine wave over one complete cycle (0 to 2π) is mathematically ZERO because positive and negative half-cycles cancel out. When an exam asks for 'Average Value', it refers strictly to the half-cycle mean: V_avg = (2 / π) * V_m ≈ 0.637 V_m.
  • Star vs. Delta √3 Factor Confusion: In Star (Y), voltage is stepped down across phases (V_ph = V_L / √3) while currents are equal (I_L = I_ph). In Delta (Δ), voltages are equal (V_ph = V_L) while line current is magnified (I_L = √3 * I_ph). Inverting these two rules is the single most common mistake in three-phase numericals.
  • Units of AC Power Confusion: Never mix up electrical power units! Active/Real power is strictly in WATTS (W), Reactive power in VOLT-AMPERES REACTIVE (VAR), and Apparent power in VOLT-AMPERES (VA). Writing '1634 VA' for real power loses examination marks.
  • Constant Phase Impedance Fallacy: Power factor depends solely on the internal resistance and reactance of the per-phase load impedance (cos φ = R_ph / Z_ph). Reconnecting a balanced load from Star to Delta does NOT change its power factor!

7. Interactive Circuit Simulator: AC Waveforms & Three-Phase Lab

Experiment with rotating coil generation, pure element phase dynamics, series RR-LL impedance triangles, and live Star vs. Delta three-phase power scaling in the bespoke simulator below:

Interactive Engineering Simulator
ac-circuits-3phase

Simulator will mount when scrolled into view (conserving CPU & memory)

Try Breaking It: Predict Before You Test

In a 400V three-phase system, what happens to the total active power drawn if you switch the load coils from Star (Y) to Delta (Δ)?

Interactive AC Rotating Phasor Vector & Power Triangle

Inspect how voltage and current rotating vectors generate active, reactive, and apparent power:

Interactive Engineering Simulator
phasor-diagram

Simulator will mount when scrolled into view (conserving CPU & memory)


8. Parametrised Practice Problem

Parametrised Practice Problem

Thévenin Equivalent of a Loaded Voltage Divider

A 48 V DC voltage source powers a divider with R1 = 60 Ω and R2 = 40 Ω. Calculate the open-circuit Thévenin voltage V_th (in Volts) appearing across terminals of R2.

V
Progressive Hints:

9. Unit Mastery Quiz

5-Question Mastery Check

Unit Quiz: Single-Phase & Three-Phase AC Circuits

Q1.What is the form factor of a purely sinusoidal alternating voltage waveform?

Q2.Why is the net average power consumed by a pure inductor or pure capacitor across an AC cycle equal to zero?

Q3.In a series R-L circuit with R = 20 Ω and X_L = 15 Ω, what is the circuit power factor?

Q4.In a balanced three-phase Star (Y) connected system, what is the mathematical relationship between Line Voltage (V_L) and Phase Voltage (V_ph)?

Q5.A balanced three-phase load connected in Star (Y) consumes 3,000 W. If the exact same three load coils are reconnected in Delta (Δ) across the same 400 V supply, what is the new total power consumed?

0 of 5 answered

10. References & Academic Attribution

  1. Dr. PL. Somasundaram, Assistant Professor – Senior Scale, Department of Electrical & Electronics Engineering, School of Engineering, Presidency University, Bengaluru — Lecture Slides for EEE1200: Basics of Electrical & Electronics Engineering (Module 2: AC Circuits).
  2. B.L. Theraja & A.K. Theraja, A Textbook of Electrical Technology: Volume I (Basic Electrical Engineering), S. Chand & Company Ltd.
  3. Vincent Del Toro, Electrical Engineering Fundamentals, 2nd Edition, Prentice-Hall International.
  4. Bureau of Indian Standards (BIS), IS 12360: Voltage bands for electrical installations including standard voltages and frequencies.