Introduction

Balanced Three-Phase Voltages

Balanced Wye-Wye Connection

Balanced Wye-Delta Connection

Balanced Delta-Wye Connection

Balanced Delta-Delta Connection

Introduction

🌐 Three-Phase Circuits — Overview

• A three-phase system is a type of polyphase system, where AC sources operate at the same frequency but are 120° out of phase with each other.
• It’s the most common and economical method to generate, transmit, and distribute electrical power.

🧠 Why Is It Important?

1. Nearly all electric power is generated and distributed using three-phase systems.
2. It can provide constant (non-pulsating) power to loads.
3. For the same amount of power, the three-phase system is more economical than the single-phase

🧲 How It Works

• A three-phase generator consists of:
  • A rotating magnet (rotor)
  • A stationary coil system (stator) with windings spaced 120° apart
• As the rotor spins, it induces AC voltages in the coils, each 120° apart in phase but equal in magnitude.
  

Balanced Three-Phase Voltages

Source and Load Connections

• Voltage sources can be connected in:
  • Wye (Y) configuration
  • Delta (Δ) configuration
• Loads can also be connected in:
  • Wye (Y) configuration
  • Delta (Δ) configuration

This results in four possible source-load combinations:

• Y-Y
• Y-Δ
• Δ-Y
• Δ-Δ

Balanced Conditions

• All phase voltages have the same magnitude: $$|V_{an}| = |V_{bn}| = |V_{cn}|$$
• The sum of phase voltages is zero: $$V_{an} + V_{bn} + V_{cn} = 0$$

Phase Sequence

• Defines the order in which voltages reach their peaks.

➕ Positive Sequence (abc)

➖ Negative Sequence (acb)

วิธีดูง่าย ๆ ถ้าให้กราฟมา ว่าเป็น Sequence อะไร คือขวาสุดจะเป็น ${V_{an}}$ เสมอ แล้วไล่ Clockwise!! ก็จะรู้ว่าเป็นอะไรเลย

Balanced Loads

Wye (Y) Load:

where $\mathbf{Z}_Y$ is the load impedance per phase.

Delta (Δ) Load:

where $\mathbf{Z}_\Delta$ is the load impedance per phase.

Conversion between Y and Δ:

Balanced Wye-Wye (Y-Y) Connection

🧠 What It Is

• A Balanced Y-Y system is when both the source and load are Wye (Y) connected and balanced.
• This configuration is the most important because:

Any balanced three-phase system can be reduced to an equivalent Y-Y system.

⚙️ Key Terms

In a Y-Y system:
Line current = Phase current

Relationship Between Voltages

Phase Voltage (line to neutral):

Assume positive sequence (abc):

• ตรงตัวเลยเนอะ Phase Voltage (คือแต่ละสายเทียบกับสาย Neutral)

Line Voltage (line to line):

From vector subtraction:

• Magnitude ของทั้งสามก็จะเท่ากันแหละ เข้าใจเนอะ แต่เป็น $\sqrt{3}$ เท่าของ $V_p$ แล้วก็นำหน้าอยู่ $30^\circ$
• เลยทำให้เรารู้ Relationship แรก

Line and Phase Current

Since the system is balanced:

• ซึ่งมันก็คือ Ohm’s Law ธรรมดาเนี่ยแหละ
• ส่วน $\mathbf{I}_b,\mathbf{I}_c$ ก็ค่อย Extend answer จาก $\mathbf{I}_a$ ไป ทีละ $120^\circ$ แบบปกติเลย
• Sum ของ Line Currents ทั้ง 3 Phase ก็เป็น 0 (นึกภาพแล้วอ่อเลยย) → $I_n=0$

• This method reduces 3-phase analysis to a single-phase problem.
• Only one phase is needed to solve the entire system due to symmetry and balance.
• Always confirm phase sequence (abc vs. acb) before applying angles.

Balanced Wye-Delta (Y-Δ) Connection

🧠 What It Is

• A Balanced Y-Δ system consists of a Wye-connected (Y) source supplying power to a Delta-connected (Δ) load.
• Still assumes positive sequence (abc).
• This configuration introduces a difference between line current and phase current on the load side.

⚙️ Key Terms

In a Y-Δ system:
Line current ≠ Phase current
But: $I_L=3⋅I_p$

🔌 Voltage Relationships

Assume positive sequence (abc):

Van=Vp∠0∘Vbn=Vp∠−120∘Vcn=Vp∠+120∘

Line voltages (using vector subtraction):

Vab=Van−Vbn=3Vp∠30∘Vbc=Vbn−Vcn=3Vp∠−90∘Vca=Vcn−Van=3Vp∠150∘

🔁 Phase & Line Current Relationships

Phase Currents (in Δ load):

Each phase current flows across each leg of the Delta:

IAB=VABZΔIBC=VBCZΔICA=VCAZΔ

Line Currents (KCL at nodes):

Apply Kirchhoff's Current Law (KCL) to get line currents:

Ia=IAB−ICAIb=IBC−IABIc=ICA−IBC

From phasor relationships:

Ia=IAB⋅3∠−30∘

So:

🔗 Line current = $\sqrt{3}$ × Phase current and lags by $30^\circ$

🔄 Delta-to-Wye Transformation (Optional Trick)

To simplify the circuit, transform the Delta load into a Y equivalent:

Now it's a standard Y-Y system, and you can:

• Use single-phase equivalent analysis

• Solve for line current as:

  Ia=VanZY

Then:

• Phase current:

  IAB=13⋅Ia∠+30∘

🪜 Step-by-Step Analysis Procedure

🧭 Final Notes

• Line currents are larger than phase currents.

• Phase current leads line current by $30^\circ$.

• Using the Δ-to-Y transformation simplifies analysis a lot, especially in exams.

Let me know if you want the diagram (e.g. from Figure 13 or 14) included as an image placeholder like:

![Figure 13: Balanced Y-Δ connection](insert-image-here)

Balanced Delta-Wye (Δ-Y) Connection

🧠 What It Is

• A Balanced Δ-Y system is when the source is Delta-connected and the load is Wye-connected.
• This setup is very common in power transmission and distribution systems.
• The delta source can be converted into a wye equivalent to simplify analysis → turns it into a Y-Y system.

⚙️ Key Terms

In Delta connection:
Line Voltage = Phase Voltage

In Wye load:
Line Current = Phase Current

🔁 Delta to Wye Source Conversion

To simplify analysis:

• Convert the delta source to an equivalent wye source.
• Use this identity:

⚡ Relationship Between Voltages

Assuming abc sequence:

After conversion:

🔌 Line Currents Derivation (Without Delta-Wye conversion)

Apply KVL in loop aANBba:

Assume abc sequence ⇒ $I_b = I_a \angle -120^\circ$, then:

Solve:

🔍 Step-by-Step Analysis Procedure

1. Transform Delta source to an equivalent Wye:

   • $V_{an} = \frac{V_{ab}}{\sqrt{3}}\angle(-30^\circ)$
   • $Z_s$ becomes $\frac{Z_s}{3}$

2. Draw the single-phase equivalent circuit (typically phase A).

3. Solve for the line/phase current:

   $$\mathbf{I}_a = \frac{\mathbf{V}_{an}}{Z_Y}$$

4. Extend solution to other phases using 120° phase shifts:

   $$\mathbf{I}_b = \mathbf{I}_a \angle -120^\circ \quad \mathbf{I}_c = \mathbf{I}_a \angle +120^\circ$$

Takeaways

• Transforming Delta to Wye makes analysis simpler.
• In Δ-Y:
  • Voltage transformation involves $\sqrt{3}$ scaling and -30° phase shift.
  • Current calculation can follow delta loop KVL or convert to Y-Y for standard method.

Balanced Delta-Delta (Δ-Δ) Connection

🧠 What It Is

• A Balanced Δ-Δ system means both source and load are Delta-connected and balanced.
• It's less common than Y-Y or Y-Δ, but useful for specific transformer configurations or motor loads.
• If needed, you can always convert Δ to Y (for either source or load) to simplify the analysis.

✅ Many problems can be reduced to Y-Y format by using Δ-to-Y transformation.

⚙️ Key Concepts and Terms

🔁 Relationship Between Line and Phase Currents

• In Δ, phase voltages = line voltages directly:

• Phase currents:

• Line currents (from KCL):

• Magnitude relationship:

🔄 Δ-to-Y Transformation (For Easier Analysis)

• To simplify the analysis, convert both Δ source/load to Y:

• Once transformed to Y-Y:
  • Use normal Y-Y analysis like in your previous notes.
  • Solve for single-phase equivalent.
  • Then extend by $120^\circ$ rotation.

⚠️ Don't forget to convert both the source and the load to Y if the problem allows it, to simplify the system into a single-phase equivalent model.

🧪 Step-by-Step Analysis (Optional Shortcut via Y-Y)

1. Convert Delta load to equivalent Wye:

   $$Z_Y = \frac{Z_\Delta}{3}$$

2. If needed, convert source to Wye too.

3. Draw single-phase equivalent circuit (usually for phase A).

4. Solve using:

   $$I_a = \frac{V_{an}}{Z_Y + Z_{line} + Z_s}$$

5. Extend solution to $I_b$ and $I_c$ by rotating $-120^\circ$ and $+120^\circ$.

6. Convert back (if necessary) to get actual Δ current values.

![Insert textbook figure here if needed]