12.3 Balanced & Unbalanced Three-Phase AC Circuits

12.3 Balanced & Unbalanced Three-Phase AC Circuits

Three-phase AC power generation, transmission, and utilization dominate industrial power systems due to constant instantaneous power delivery, higher conductor efficiency, and self-starting rotating magnetic fields in AC motors. This section covers balanced and unbalanced Wye ($Y$) and Delta ($\Delta$) configurations, neutral currents, and power measurement via the Two-Wattmeter Method.

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1. Three-Phase Voltage Generation & Phase Sequences

A balanced three-phase AC generator produces three equal sinusoidal phase voltages displaced in time by $120^\circ$ ($2\pi/3$ radians):

Positive (ABC) Phase Sequence:

$\mathbf{V}_{AN} = V_p \angle 0^\circ$

$\mathbf{V}_{BN} = V_p \angle -120^\circ$

$\mathbf{V}_{CN} = V_p \angle -240^\circ = V_p \angle +120^\circ$

$\text{Sum of Balanced Phase Voltages: } \mathbf{V}_{AN} + \mathbf{V}_{BN} + \mathbf{V}_{CN} = 0$

Negative (CBA) Phase Sequence:

$\mathbf{V}_{AN} = V_p \angle 0^\circ, \quad \mathbf{V}_{BN} = V_p \angle +120^\circ, \quad \mathbf{V}_{CN} = V_p \angle -120^\circ$

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2. Wye ($Y$) and Delta ($\Delta$) Phase-Line Voltage & Current Relations

Balanced Wye ($Y$) Connection

In a Wye connection, line conductors connect to outer terminals, and neutral ($N$) connects to the central star point.

                    BALANCED WYE (Y) CONNECTION
                     Phase A ───────► Line Current I_A
                        │
                       [Z]
                        │
       Neutral N ───────┼─────── [Z] ─────── Phase B ───► Line Current I_B
                        │
                       [Z]
                        │
                     Phase C ───────► Line Current I_C

• Line Voltage vs. Phase Voltage: Line-to-line voltages $\mathbf{V}{AB} = \mathbf{V}{AN} - \mathbf{V}_{BN}$ lead phase-to-neutral voltages by $30^\circ$:

$V_{L-L} = \sqrt{3} V_{L-N} \quad \implies \quad \mathbf{V}_{AB} = \sqrt{3} V_p \angle +30^\circ$

• Line Current vs. Phase Current: Line current equals phase current:

$I_L = I_p$

Balanced Delta ($\Delta$) Connection

In a Delta connection, load impedances connect directly across pairs of line phase conductors ($A\text{--}B$, $B\text{--}C$, $C\text{--}A$).

• Line Voltage vs. Phase Voltage: Line voltage equals phase load voltage:

$V_L = V_p$

• Line Current vs. Phase Current: Line current is $\sqrt{3}$ times phase current and lags phase current by $30^\circ$ (for positive sequence):

$I_L = \sqrt{3} I_p \quad \implies \quad \mathbf{I}_A = \mathbf{I}_{AB} - \mathbf{I}_{CA} = \sqrt{3} I_p \angle -30^\circ$

Wye-Delta Impedance Equivalence

To convert a balanced Delta load to an equivalent Wye load for per-phase single-line analysis:

$\mathbf{Z}_Y = \frac{\mathbf{Z}_\Delta}{3}$

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3. Three-Phase Power Formulas & Complex Power

For any balanced three-phase load (whether Wye or Delta connected), the total power expressions are identical when written in terms of line quantities ($V_L, I_L$):

where $\theta$ is the impedance phase angle of individual phase load $\mathbf{Z}_p = |Z_p| \angle \theta$.

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4. Unbalanced Three-Phase Systems & Neutral Current Analysis

When load impedances across phases are unequal ($\mathbf{Z}_A \neq \mathbf{Z}_B \neq \mathbf{Z}_C$), symmetry breaks.

Unbalanced 4-Wire Wye Systems

In a 4-wire Wye system with a solidly grounded neutral, phase voltages remain balanced at $V_p$. Line currents equal individual phase load currents:

$\mathbf{I}_A = \frac{\mathbf{V}_{AN}}{\mathbf{Z}_A}, \quad \mathbf{I}_B = \frac{\mathbf{V}_{BN}}{\mathbf{Z}_B}, \quad \mathbf{I}_C = \frac{\mathbf{V}_{CN}}{\mathbf{Z}_C}$

Neutral return current $\mathbf{I}_N$ is the phasor sum of line currents:

$\mathbf{I}_N = \mathbf{I}_A + \mathbf{I}_B + \mathbf{I}_C$

Board Exam Fact: In a balanced 4-wire system, $\mathbf{I}_N = 0$. Unbalance creates non-zero neutral return current, increasing neutral conductor heating.

Unbalanced 3-Wire Wye Systems (Neutral Shift Voltage)

When an ungrounded Wye load is unbalanced, the load neutral point $n$ shifts away from the source neutral point $N$. The Neutral Shift Voltage $\mathbf{V}_{nN}$ is calculated via Millman's Theorem:

$\mathbf{V}_{nN} = \frac{\mathbf{Y}_A \mathbf{V}_{AN} + \mathbf{Y}_B \mathbf{V}_{BN} + \mathbf{Y}_C \mathbf{V}_{CN}}{\mathbf{Y}_A + \mathbf{Y}_B + \mathbf{Y}_C}$

Actual phase voltages across load impedances become distorted:

$\mathbf{V}_{an} = \mathbf{V}_{AN} - \mathbf{V}_{nN}, \quad \mathbf{V}_{bn} = \mathbf{V}_{BN} - \mathbf{V}_{nN}, \quad \mathbf{V}_{cn} = \mathbf{V}_{CN} - \mathbf{V}_{nN}$

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5. Three-Phase Power Measurement: The Two-Wattmeter Method

Blondel's Theorem proves that total power in an $N$-wire system can be measured using $(N-1)$ wattmeters. For any 3-wire three-phase system (balanced or unbalanced, Wye or Delta), two wattmeters suffice.

                      TWO-WATTMETER METHOD SCHEMATIC
       Phase A ───────┬───[ Current Coil W1 ]───────────► Load
                      │
                  [ Potential Coil W1 ]
                      │
       Phase B ───────┴─────────────────────────────────► Load
                      │
                  [ Potential Coil W2 ]
                      │
       Phase C ───────┬───[ Current Coil W2 ]───────────► Load

Wattmeter Readings & Equations

For a balanced load with line voltage $V_L$, line current $I_L$, and power factor angle $\theta$:

$W_1 = V_L I_L \cos(30^\circ - \theta)$

$W_2 = V_L I_L \cos(30^\circ + \theta)$

Sum and Difference Relations

• Total Active Power ($P_{3\phi}$):

$P_{3\phi} = W_1 + W_2 = V_L I_L \left[ \cos(30^\circ - \theta) + \cos(30^\circ + \theta) \right] = \sqrt{3} V_L I_L \cos\theta$

• Total Reactive Power ($Q_{3\phi}$):

$Q_{3\phi} = \sqrt{3} (W_1 - W_2) = \sqrt{3} V_L I_L \left[ \cos(30^\circ - \theta) - \cos(30^\circ + \theta) \right] = \sqrt{3} V_L I_L \sin\theta$

• Power Factor Angle ($\theta$) & Power Factor ($ ext{PF}$):

$\tan\theta = \sqrt{3} \left( \frac{W_1 - W_2}{W_1 + W_2} \right) \quad \implies \quad \text{PF} = \cos\left[ \arctan\left( \sqrt{3} \frac{W_1 - W_2}{W_1 + W_2} \right) \right]$

Key Wattmeter Reading Cases:

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Solved Board Exam Examples

Example 1: Balanced Delta Load Analysis

Problem: A balanced 3-phase Delta-connected load with impedance $\mathbf{Z}\Delta = 18 + j 24\ \Omega$ per phase is connected to a $440\ \text{V}$, $60\ \text{Hz}$ 3-phase supply. Calculate: (a) phase current $I_p$, (b) line current $I_L$, (c) total active power $P{3\phi}$, and (d) total reactive power $Q_{3\phi}$.

Solution:

1. Calculate phase impedance magnitude $|Z_p|$ and phase angle $\theta$: $|Z_p| = \sqrt{18^2 + 24^2} = \sqrt{324 + 576} = \sqrt{900} = 30\ \Omega$ $\theta = \arctan\left(\frac{24}{18}\right) = 53.13^circ \implies \cos\theta = 0.60, \quad \sin\theta = 0.80$
2. In Delta connection, $V_p = V_L = 440\ \text{V}$. Compute phase current $I_p$: $I_p = \frac{V_p}{|Z_p|} = \frac{440}{30} = 14.67\ \text{A}$
3. Compute line current $I_L$ for Delta connection: $I_L = \sqrt{3} I_p = \sqrt{3} \times 14.67 = 1.73205 \times 14.67 = 25.40\ \text{A}$
4. Compute Total 3-Phase Active and Reactive Power: $P_{3\phi} = \sqrt{3} V_L I_L \cos\theta = \sqrt{3} \times 440 \times 25.40 \times 0.60 = 11,616\ \text{W} = 11.62\ \text{kW}$ $Q_{3\phi} = \sqrt{3} V_L I_L \sin\theta = \sqrt{3} \times 440 \times 25.40 \times 0.80 = 15,488\ \text{VAR} = 15.49\ \text{kVAR}$

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Example 2: Two-Wattmeter Method Analysis

Problem: Two wattmeters connected to measure power in a $230\ \text{V}$, 3-phase balanced system yield readings of $W_1 = 8000\ \text{W}$ and $W_2 = 4000\ \text{W}$. Calculate: (a) total active power $P_{3\phi}$, (b) total reactive power $Q_{3\phi}$, (c) load power factor $\text{PF}$, and (d) line current $I_L$.

Solution:

1. Calculate total active power $P_{3\phi}$: $P_{3\phi} = W_1 + W_2 = 8000 + 4000 = 12,000\ \text{W} = 12\ \text{kW}$
2. Calculate total reactive power $Q_{3\phi}$: $Q_{3\phi} = \sqrt{3} (W_1 - W_2) = \sqrt{3} \times (8000 - 4000) = 1.73205 \times 4000 = 6928.2\ \text{VAR} = 6.93\ \text{kVAR}$
3. Calculate tangent of phase angle $\theta$ and power factor: $\tan\theta = \frac{Q_{3\phi}}{P_{3\phi}} = \frac{6928.2}{12,000} = 0.57735 \quad \implies \quad \theta = \arctan(0.57735) = 30^circ$ $\text{PF} = \cos(30^circ) = \frac{\sqrt{3}}{2} \approx 0.8660$
4. Calculate line current $I_L$ from active power formula: $P_{3\phi} = \sqrt{3} V_L I_L \cos\theta \quad \implies \quad I_L = \frac{P_{3\phi}}{\sqrt{3} V_L \cos\theta}$ $I_L = \frac{12,000}{\sqrt{3} \times 230 \times 0.8660} = \frac{12,000}{345.0} = 34.78\ \text{A}$