Write any seven Comparison Between Star (Y) and Delta (Δ) Connected Systems
In three-phase electrical AC engineering, Star and Delta configurations differ across connections, voltage-current ratios, insulation demands, and distribution applications.
1. Structural Topologies
2. Seven Core Differences
| No. | Parameter | Star (Y) Connection | Delta (Δ) Connection |
|---|---|---|---|
| 1 | Basic Formation | Formed by joining all similar terminals (starts or finishes) of the three phases at a common central point. | Formed by joining the finish of each winding to the start of the next winding, creating a closed loop/mesh. |
| 2 | Neutral Point & Wiring | A neutral point (\(N\)) is present. Supports both 3-phase 3-wire and 3-phase 4-wire configurations. | No neutral point exists. Restricted strictly to a 3-phase 3-wire configuration. |
| 3 | Voltage Relationship |
Line voltage is \(\sqrt{3}\) times phase voltage:
$$V_L = \sqrt{3} \cdot V_{ph} \approx 1.732 \, V_{ph}$$ |
Line voltage equals phase voltage:
$$V_L = V_{ph}$$ |
| 4 | Current Relationship |
Line current equals phase current:
$$I_L = I_{ph}$$ |
Line current is \(\sqrt{3}\) times phase current:
$$I_L = \sqrt{3} \cdot I_{ph} \approx 1.732 \, I_{ph}$$ |
| 5 | Insulation & Turns Required | Because phase voltage is reduced (\(V_{ph} = V_L / \sqrt{3}\)), less insulation and fewer turns per coil are required. | Because each phase handles full line voltage (\(V_{ph} = V_L\)), more insulation and a higher number of turns per phase are required. |
| 6 | Load Compatibility & Voltage Levels | Provides two distinct voltage levels (\(415\text{ V}\) line-to-line and \(240\text{ V}\) line-to-neutral). Handles both balanced and unbalanced loads easily. | Provides only a single line voltage level. Best suited for balanced loads; unbalanced loads can cause circulatory circulating currents. |
| 7 | Practical Applications | Used in low-voltage secondary distribution (residential/commercial lighting and power) and high-voltage alternator outputs. | Used in high-voltage transmission networks, industrial motor drives, and running windings of large 3-phase induction motors. |
- Star: Voltage divides by \(\sqrt{3}\) across phases (\(V_{ph} = V_L / \sqrt{3}\)), currents stay identical (\(I_L = I_{ph}\)).
- Delta: Voltages stay identical (\(V_L = V_{ph}\)), current divides by \(\sqrt{3}\) across branches (\(I_{ph} = I_L / \sqrt{3}\)).
- Total Active Power in terms of line values remains unchanged for both: \(P = \sqrt{3} V_L I_L \cos\phi\).
Explain with a diagram the generation of three phase ac voltages
1. Working Principle
Three-phase alternating voltages are generated based on Faraday’s Law of Electromagnetic Induction. When three identical coils or phase windings are placed in a uniform magnetic field and rotated at a constant angular speed \(\omega\) (or conversely, when a magnetic field rotates inside a stationary 3-phase stator armature), alternating electromotive forces (EMFs) are induced in each coil.
- The three identical coils are physically mounted on the armature core displaced by \(120^\circ\) electrical in space from each other.
- As the rotor completes rotations, each coil cuts magnetic flux lines at identical rates, inducing alternating EMFs of the same peak amplitude (\(E_m\)) and same frequency (\(f\)).
- Because of the \(120^\circ\) physical displacement in space, the induced voltages attain their maximum positive peaks with a time phase displacement of \(120^\circ\) (or \(\frac{2\pi}{3}\) radians / one-third of a time period \(T/3\)).
2. Generator Construction and Three-Phase Waveforms
3. Mathematical Equations of Induced EMFs
Taking the induced voltage in the R-phase as the reference wave:
$$e_R = E_m \sin(\omega t)$$
Since the Y-phase coil is displaced by \(120^\circ\) behind the R-phase coil, its voltage reaches the same cycle stage \(120^\circ\) later:
$$e_Y = E_m \sin(\omega t - 120^\circ) = E_m \sin\left(\omega t - \frac{2\pi}{3}\right)$$
Similarly, the B-phase coil lags behind the R-phase by \(240^\circ\) (or leads the R-phase by \(120^\circ\)):
$$e_B = E_m \sin(\omega t - 240^\circ) = E_m \sin(\omega t + 120^\circ) = E_m \sin\left(\omega t - \frac{4\pi}{3}\right)$$
where:
- \(e_R, e_Y, e_B\) = Instantaneous induced voltages of phases R, Y, and B.
- \(E_m\) = Maximum (peak) amplitude of the induced EMF (\(E_m = 2\pi f N \Phi_m\)).
- \(\omega = 2\pi f\) = Angular frequency in radians per second.
4. Phasor Representation and Phase Sum
In phasor notation, the three generated voltages form a symmetrical balanced star of vectors:
$$\vec{E}_R = E_{ph}\angle 0^\circ$$
$$\vec{E}_Y = E_{ph}\angle -120^\circ$$
$$\vec{E}_B = E_{ph}\angle -240^\circ = E_{ph}\angle +120^\circ$$
Sum of Instantaneous Voltages: In any balanced 3-phase system, the algebraic sum of the instantaneous voltages at any given instant of time is strictly zero:
$$e_R + e_Y + e_B = E_m \left[ \sin(\omega t) + \sin(\omega t - 120^\circ) + \sin(\omega t + 120^\circ) \right] = 0$$
With the help of a vector diagram derive the relation between line and phase values of current in a delta connected system
1. System Current Definitions & KCL at Junction Nodes
In a balanced three-phase delta (\(\Delta\)) connected load, the three phase impedances are connected end-to-end in a closed mesh. The transmission lines R, Y, and B connect directly to the three corner nodes.
- Phase currents: \(I_{RY}\), \(I_{YB}\), and \(I_{BR}\) circulating within the branches.
- Line currents: \(I_R\), \(I_Y\), and \(I_B\) entering from external conductors.
Applying Kirchhoff's Current Law (KCL) at each corner junction:
- At Node R: \(\vec{I}_R + \vec{I}_{BR} = \vec{I}_{RY} \implies \vec{I}_R = \vec{I}_{RY} - \vec{I}_{BR} = \vec{I}_{RY} + (-\vec{I}_{BR})\)
- At Node Y: \(\vec{I}_Y + \vec{I}_{RY} = \vec{I}_{YB} \implies \vec{I}_Y = \vec{I}_{YB} - \vec{I}_{RY} = \vec{I}_{YB} + (-\vec{I}_{RY})\)
- At Node B: \(\vec{I}_B + \vec{I}_{YB} = \vec{I}_{BR} \implies \vec{I}_B = \vec{I}_{BR} - \vec{I}_{YB} = \vec{I}_{BR} + (-\vec{I}_{YB})\)
For a balanced system, the phase currents are equal in magnitude and displaced from each other by \(120^\circ\):
$$|I_{RY}| = |I_{YB}| = |I_{BR}| = I_{ph}$$
$$|I_R| = |I_Y| = |I_B| = I_L$$
2. Vector (Phasor) and Connection Diagram
3. Step-by-Step Mathematical Derivation
To determine the line current \(I_R\), evaluate the vector difference between phase currents \(\vec{I}_{RY}\) and \(\vec{I}_{BR}\):
$$\vec{I}_R = \vec{I}_{RY} - \vec{I}_{BR} = \vec{I}_{RY} + (-\vec{I}_{BR})$$
Step 1: Angle Between Phasors
In a balanced 3-phase system, the angle between the phase currents \(\vec{I}_{RY}\) and \(\vec{I}_{BR}\) is \(120^\circ\).
Reversing the vector \(\vec{I}_{BR}\) produces \(-\vec{I}_{BR}\) with a \(180^\circ\) phase shift. The interior angle \(\theta\) between \(\vec{I}_{RY}\) and \(-\vec{I}_{BR}\) is therefore:
$$\theta = 180^\circ - 120^\circ = 60^\circ$$
Step 2: Apply the Parallelogram Law of Vectors
The magnitude of the resultant line current \(I_R\) is:
$$I_R = \sqrt{I_{RY}^2 + I_{BR}^2 + 2 \cdot I_{RY} \cdot I_{BR} \cdot \cos(60^\circ)}$$
For a balanced system, \(|I_{RY}| = |I_{BR}| = I_{ph}\) and \(|I_R| = I_L\):
$$I_L = \sqrt{I_{ph}^2 + I_{ph}^2 + 2 \cdot I_{ph} \cdot I_{ph} \cdot \cos(60^\circ)}$$
Since \(\cos(60^\circ) = 0.5\):
$$I_L = \sqrt{I_{ph}^2 + I_{ph}^2 + 2 \cdot I_{ph}^2 \cdot (0.5)}$$
$$I_L = \sqrt{I_{ph}^2 + I_{ph}^2 + I_{ph}^2}$$
$$I_L = \sqrt{3 \cdot I_{ph}^2}$$
4. Phase Angle & Voltage Relation Summary
- Phase Displacement: As seen from the vector diagram, the resultant line current \(\vec{I}_R\) lags the corresponding phase current \(\vec{I}_{RY}\) by \(30^\circ\).
-
Voltage Relation: Since each phase branch is directly connected between two external transmission lines, line voltage is identical to phase voltage:
$$V_L = V_{ph}$$
Write the Equations of Various Three-Phase Powers
In a balanced three-phase AC system, power is categorized into three types: Active Power (True/Real Power), Reactive Power, and Apparent Power. These formulas remain identical in terms of line quantities for both Star (\(\text{Y}\)) and Delta (\(\Delta\)) connections.
1. Power Triangle Representation
2. Detailed Power Equations
A. Active Power (\(P\)) — Real / True Power
The actual power converted into useful work (such as mechanical motion or heat).
-
In terms of phase quantities:
$$P = 3 \cdot V_{ph} \cdot I_{ph} \cdot \cos\phi$$
-
In terms of line quantities:
$$P = \sqrt{3} \cdot V_L \cdot I_L \cdot \cos\phi$$
- Standard Unit: Watts (\(\text{W}\)) or Kilowatts (\(\text{kW}\))
B. Reactive Power (\(Q\)) — Quadrature Power
The power that oscillates back and forth between the source and reactive components (inductors and capacitors) to establish and sustain magnetic or electric fields.
-
In terms of phase quantities:
$$Q = 3 \cdot V_{ph} \cdot I_{ph} \cdot \sin\phi$$
-
In terms of line quantities:
$$Q = \sqrt{3} \cdot V_L \cdot I_L \cdot \sin\phi$$
- Standard Unit: Volt-Ampere Reactive (\(\text{VAR}\)) or Kilovolt-Ampere Reactive (\(\text{kVAR}\))
C. Apparent Power (\(S\)) — Total Power
The product of total RMS voltage and RMS current, representing the overall power-handling capacity of electrical machinery (such as transformers and alternators).
-
In terms of phase quantities:
$$S = 3 \cdot V_{ph} \cdot I_{ph}$$
-
In terms of line quantities:
$$S = \sqrt{3} \cdot V_L \cdot I_L$$
- Standard Unit: Volt-Ampere (\(\text{VA}\)) or Kilovolt-Ampere (\(\text{kVA}\))
3. Interrelations in the Power Triangle
From the right-angled power triangle:
-
Total Apparent Power Magnitude:
$$S = \sqrt{P^2 + Q^2}$$
-
Complex Power (\(\vec{S}\)):
$$\vec{S} = P \pm jQ = S\angle\phi$$
(\(+jQ\) for inductive loads with lagging power factor, \(-jQ\) for capacitive loads with leading power factor) -
Power Factor (\(\cos\phi\)):
$$\cos\phi = \frac{P}{S} = \frac{\text{Active Power}}{\text{Apparent Power}}$$
-
Reactive Factor (\(\sin\phi\)):
$$\sin\phi = \frac{Q}{S} = \frac{\text{Reactive Power}}{\text{Apparent Power}}$$
4. Summary Table
| Type of Power | Symbol | Phase Formula | Line Formula | Units |
|---|---|---|---|---|
| Active (Real) Power | \(P\) | \(3 \, V_{ph} I_{ph} \cos\phi\) | \(\sqrt{3} \, V_L I_L \cos\phi\) | \(\text{W}, \text{kW}, \text{MW}\) |
| Reactive Power | \(Q\) | \(3 \, V_{ph} I_{ph} \sin\phi\) | \(\sqrt{3} \, V_L I_L \sin\phi\) | \(\text{VAR}, \text{kVAR}, \text{MVAR}\) |
| Apparent Power | \(S\) | \(3 \, V_{ph} I_{ph}\) | \(\sqrt{3} \, V_L I_L\) | \(\text{VA}, \text{kVA}, \text{MVA}\) |
Relation Between Line and Phase Voltage in a Star (Y) Connected System
1. System Definitions and Kirchhoff's Voltage Law
In a balanced star-connected 3-phase system, three identical windings or load impedances are connected together at a common terminal called the Neutral Point (\(N\)).
- Phase voltages (measured from each line to neutral \(N\)): \(V_{RN}\), \(V_{YN}\), and \(V_{BN}\).
- Line voltages (measured between any two line terminals): \(V_{RY}\), \(V_{YB}\), and \(V_{BR}\).
Applying Kirchhoff's Voltage Law (KVL) between the line terminals:
- Between lines R and Y: \(\vec{V}_{RY} = \vec{V}_{RN} - \vec{V}_{YN} = \vec{V}_{RN} + (-\vec{V}_{YN})\)
- Between lines Y and B: \(\vec{V}_{YB} = \vec{V}_{YN} - \vec{V}_{BN} = \vec{V}_{YN} + (-\vec{V}_{BN})\)
- Between lines B and R: \(\vec{V}_{BR} = \vec{V}_{BN} - \vec{V}_{RN} = \vec{V}_{BN} + (-\vec{V}_{RN})\)
For a balanced system, the phase voltages have equal magnitudes and are mutually displaced by \(120^\circ\):
$$|V_{RN}| = |V_{YN}| = |V_{BN}| = V_{ph}$$
$$|V_{RY}| = |V_{YB}| = |V_{BR}| = V_L$$
2. Vector (Phasor) Diagram
3. Step-by-Step Mathematical Derivation
To determine the line voltage \(V_{RY}\), calculate the phasor difference between \(\vec{V}_{RN}\) and \(\vec{V}_{YN}\):
$$\vec{V}_{RY} = \vec{V}_{RN} - \vec{V}_{YN} = \vec{V}_{RN} + (-\vec{V}_{YN})$$
Step 1: Angle Between Vectors
In a balanced 3-phase supply, the angle between the phase voltages \(\vec{V}_{RN}\) and \(\vec{V}_{YN}\) is \(120^\circ\).
Reversing the phasor \(\vec{V}_{YN}\) to obtain \(-\vec{V}_{YN}\) introduces a \(180^\circ\) phase reversal. Consequently, the angle \(\theta\) between \(\vec{V}_{RN}\) and \(-\vec{V}_{YN}\) is:
$$\theta = 180^\circ - 120^\circ = 60^\circ$$
Step 2: Apply the Parallelogram Law of Vectors
The magnitude of the resultant line voltage \(V_{RY}\) is given by:
$$V_{RY} = \sqrt{V_{RN}^2 + V_{YN}^2 + 2 \cdot V_{RN} \cdot V_{YN} \cdot \cos(60^\circ)}$$
For a balanced system, \(|V_{RN}| = |V_{YN}| = V_{ph}\) and \(|V_{RY}| = V_L\):
$$V_L = \sqrt{V_{ph}^2 + V_{ph}^2 + 2 \cdot V_{ph} \cdot V_{ph} \cdot \cos(60^\circ)}$$
Substituting \(\cos(60^\circ) = 0.5\):
$$V_L = \sqrt{V_{ph}^2 + V_{ph}^2 + 2 \cdot V_{ph}^2 \cdot (0.5)}$$
$$V_L = \sqrt{V_{ph}^2 + V_{ph}^2 + V_{ph}^2}$$
$$V_L = \sqrt{3 \cdot V_{ph}^2}$$
4. Phase Relationship & Current Summary
- Phase Angle: The line voltage \(\vec{V}_{RY}\) leads the corresponding phase voltage \(\vec{V}_{RN}\) by \(30^\circ\).
-
Line vs Phase Current: Because each line conductor connects directly in series with a phase winding, no current division occurs:
$$I_L = I_{ph}$$
- \(V_L = \sqrt{3} \cdot V_{ph}\) (Line voltage is \(\sqrt{3}\) times phase voltage and leads it by \(30^\circ\))
- \(I_L = I_{ph}\)
- Total Active Power: \(P = \sqrt{3} \cdot V_L \cdot I_L \cdot \cos\phi = 3 \cdot V_{ph} \cdot I_{ph} \cdot \cos\phi\)
Write any three advantages of three phase systems
Compared to a single-phase AC system, a three-phase system offers significant economic and operational advantages in power generation, transmission, and motor operation:
-
Economy in Transmission (Savings in Conductor Material):
To transmit a given amount of electric power over a given distance at a specified line voltage and power loss, a 3-phase system requires only 75% (0.75 times) of the conductor material compared to an equivalent single-phase system. This results in substantial savings in copper or aluminium costs and supporting tower structures. -
Production of Rotating Magnetic Field (Self-Starting Motors):
When three-phase currents pass through a three-phase stator winding, they generate a constant-magnitude Rotating Magnetic Field (RMF) rotating at synchronous speed:$$\Phi_R = 1.5 \, \Phi_m$$
This enables three-phase induction motors to produce starting torque and be inherently self-starting, eliminating the need for auxiliary windings, centrifugal switches, or starting capacitors required by single-phase motors. -
Constant (Pulsation-Free) Power Output:
While instantaneous power in a single-phase circuit pulsates at twice the supply frequency (\(2f\)), the total instantaneous power delivered by a balanced three-phase system is strictly constant and continuous:$$P(t) = 3 \, V_{ph} I_{ph} \cos\phi = \text{Constant}$$
Because power never drops to zero, three-phase motors run with uniform torque, minimal vibration, and smoother mechanical operation.
Comparison Summary
| Feature | Single-Phase System | Three-Phase System |
|---|---|---|
| Conductor Material | 100% (Baseline) | 75% of single-phase material |
| Motor Starting | Not self-starting (needs starter/capacitor) | Inherently self-starting (due to RMF) |
| Instantaneous Power | Pulsating (causes vibrations) | Strictly constant and steady |
| Machine Output (Rating) | Lower output for given frame size | ~1.5 times higher output for same frame size |
A balanced star connected load of 8+j6 ohm per phase isconnected to a 3- phase,230V.Find ( i)Line current (ii)Power(iii)Reactive power
Problem Statement
A balanced star-connected load of \((8 + j6)\ \Omega\) per phase is connected to a \(3\)-phase, \(230\text{ V}\) supply.
Find:
- Line current (\(I_L\))
- Active power (\(P\))
- Reactive power (\(Q\))
Given Data
- Line voltage (\(V_L\)) = \(230\text{ V}\)
- Impedance per phase (\(Z_{ph}\)) = \(8 + j6\ \Omega\)
- Resistance per phase (\(R_{ph}\)) = \(8\ \Omega\)
- Inductive reactance per phase (\(X_{ph}\)) = \(6\ \Omega\)
- Connection type = Star (\(\text{Y}\)) connection
Step 1: Impedance and Power Factor per Phase
Magnitude of phase impedance (\(Z_{ph}\)):
$$Z_{ph} = \sqrt{R_{ph}^2 + X_{ph}^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = \mathbf{10\ \Omega}$$
Power factor of the load (\(\cos\phi\)):
$$\cos\phi = \frac{R_{ph}}{Z_{ph}} = \frac{8}{10} = \mathbf{0.8} \quad (\text{lagging})$$
Reactive factor (\(\sin\phi\)):
$$\sin\phi = \frac{X_{ph}}{Z_{ph}} = \frac{6}{10} = \mathbf{0.6}$$
Step 2: Line Current (\(I_L\))
In a star connection, phase voltage (\(V_{ph}\)) is:
$$V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{230}{\sqrt{3}} \approx \mathbf{132.79\text{ V}}$$
Phase current (\(I_{ph}\)):
$$I_{ph} = \frac{V_{ph}}{Z_{ph}} = \frac{132.79}{10} = \mathbf{13.28\text{ A}}$$
For a star connection, the line current equals the phase current (\(I_L = I_{ph}\)):
$$I_L = I_{ph} = \mathbf{13.28\text{ A}}$$
Step 3: Active Power (\(P\))
Active (true) power taken by a balanced 3-phase load:
$$P = \sqrt{3} \cdot V_L \cdot I_L \cdot \cos\phi$$
$$P = \sqrt{3} \times 230 \times 13.28 \times 0.8 \approx \mathbf{4232\text{ W}} \quad (\text{or } 4.232\text{ kW})$$
(Verification: \(P = 3 \cdot I_{ph}^2 \cdot R_{ph} = 3 \times (13.28)^2 \times 8 \approx 4232\text{ W}\))
Step 4: Reactive Power (\(Q\))
Reactive power taken by the circuit:
$$Q = \sqrt{3} \cdot V_L \cdot I_L \cdot \sin\phi$$
$$Q = \sqrt{3} \times 230 \times 13.28 \times 0.6 \approx \mathbf{3174\text{ VAR}} \quad (\text{or } 3.174\text{ kVAR})$$
(Verification: \(Q = 3 \cdot I_{ph}^2 \cdot X_{ph} = 3 \times (13.28)^2 \times 6 \approx 3174\text{ VAR}\))
- Line Current (\(I_L\)): \(13.28\text{ A}\)
- Active Power (\(P\)): \(4232\text{ W}\) (or \(4.232\text{ kW}\))
- Reactive Power (\(Q\)): \(3174\text{ VAR}\) (or \(3.174\text{ kVAR}\))
Derive the equation of active power in three phase systems.
1. Total Power in Terms of Phase Quantities
In any balanced 3-phase system (whether connected in Star or Delta), the total electrical active power consumed is the algebraic sum of the active powers consumed by the three individual phases.
For a single phase, the active (real) power is:
$$P_{\text{phase}} = V_{ph} \cdot I_{ph} \cdot \cos\phi$$
where:
- \(V_{ph}\) = RMS value of phase voltage
- \(I_{ph}\) = RMS value of phase current
- \(\cos\phi\) = Power factor of the load (\(\phi\) is the phase angle between \(V_{ph}\) and \(I_{ph}\))
Since the load is balanced, the power consumed in each of the three phases is identical. Therefore, the total active power (\(P\)) is:
$$P = 3 \times P_{\text{phase}} = 3 \cdot V_{ph} \cdot I_{ph} \cdot \cos\phi \quad \text{--- (Equation 1)}$$
2. Derivation for Star (Y) Connected System
In a balanced Star-connected system, the relationships between line and phase quantities are:
$$V_{ph} = \frac{V_L}{\sqrt{3}}$$
$$I_{ph} = I_L$$
where \(V_L\) is the line-to-line voltage and \(I_L\) is the line current.
Substituting these values into Equation (1):
$$P = 3 \cdot \left(\frac{V_L}{\sqrt{3}}\right) \cdot I_L \cdot \cos\phi$$
Simplifying \(\frac{3}{\sqrt{3}} = \sqrt{3}\):
$$P = \sqrt{3} \cdot V_L \cdot I_L \cdot \cos\phi \quad \text{Watts (W)}$$
3. Derivation for Delta (Δ) Connected System
In a balanced Delta-connected system, the relationships between line and phase quantities are:
$$V_{ph} = V_L$$
$$I_{ph} = \frac{I_L}{\sqrt{3}}$$
Substituting these values into Equation (1):
$$P = 3 \cdot V_L \cdot \left(\frac{I_L}{\sqrt{3}}\right) \cdot \cos\phi$$
Simplifying \(\frac{3}{\sqrt{3}} = \sqrt{3}\):
$$P = \sqrt{3} \cdot V_L \cdot I_L \cdot \cos\phi \quad \text{Watts (W)}$$
4. Summary of All 3-Phase Power Equations
| Power Type | In Terms of Phase Values | In Terms of Line Values | Unit |
|---|---|---|---|
| Active (Real) Power (\(P\)) | $$3 \cdot V_{ph} \cdot I_{ph} \cdot \cos\phi$$ | $$\sqrt{3} \cdot V_L \cdot I_L \cdot \cos\phi$$ | Watts (W) / kW |
| Reactive Power (\(Q\)) | $$3 \cdot V_{ph} \cdot I_{ph} \cdot \sin\phi$$ | $$\sqrt{3} \cdot V_L \cdot I_L \cdot \sin\phi$$ | VAR / kVAR |
| Apparent Power (\(S\)) | $$3 \cdot V_{ph} \cdot I_{ph}$$ | $$\sqrt{3} \cdot V_L \cdot I_L$$ | VA / kVA |
$$P = \sqrt{3} \, V_L \, I_L \cos\phi$$
Notice that the phase angle \(\phi\) is always the angle between the phase voltage and phase current, never the angle between line quantities.How star and delta connections are formed in three phase system
A balanced three-phase system has three identical independent coils or windings (one for each phase: R, Y, and B). Each winding has two terminals:
- R-Phase winding: Start terminal \(R_1\), Finish terminal \(R_2\)
- Y-Phase winding: Start terminal \(Y_1\), Finish terminal \(Y_2\)
- B-Phase winding: Start terminal \(B_1\), Finish terminal \(B_2\)
Instead of using six separate transmission conductors (two per winding), these terminals are interconnected in two primary configurations: Star (Wye) Connection and Delta (Mesh) Connection.
1. Star (Y) Connection
How it is Formed:
A Star connection is formed by connecting together either all the finish terminals (\(R_2, Y_2, B_2\)) or all the start terminals (\(R_1, Y_1, B_1\)) of the three coils at a common junction point.
- This common junction point is known as the Neutral Point (\(N\)) or Star Point.
- The remaining three open terminals (\(R_1, Y_1, B_1\)) are brought out as the three line conductors (R, Y, B) for power transmission.
- If a wire is taken out from the neutral point, it is called a 3-phase 4-wire system; if omitted, it is a 3-phase 3-wire system.
Key Characteristics of Star:
- Line Voltage: \(V_L = \sqrt{3} \cdot V_{ph}\)
- Line Current: \(I_L = I_{ph}\)
- Neutral point: Present (ideal for unbalanced loads and domestic lighting distribution).
2. Delta (Δ) Connection
How it is Formed:
A Delta connection is formed by connecting the three coils in an end-to-end series sequence to make a closed loop or mesh:
- The finish of the first coil (\(R_2\)) is connected to the start of the second coil (\(Y_1\)).
- The finish of the second coil (\(Y_2\)) is connected to the start of the third coil (\(B_1\)).
- The finish of the third coil (\(B_2\)) is connected back to the start of the first coil (\(R_1\)).
The three junction points formed by these connections (\(R_1-B_2\), \(Y_1-R_2\), \(B_1-Y_2\)) are then brought out as the three line conductors (R, Y, B).
Comparison of Line and Phase Quantities in a Delta (Δ) Connected System
In a balanced 3-phase delta (\(\Delta\)) connection (mesh connection), the three phase windings or impedances are connected end-to-end to form a closed loop. The external lines are tapped directly from the three junction nodes.
1. Delta Connection Schematic
2. Key Governing Relations
A. Voltage Relationship
Because each transmission line terminal is directly connected across a phase branch, the voltage across any pair of lines is equal to the voltage across that respective phase:
$$V_L = V_{ph}$$
There is no phase difference between the line voltage and the corresponding phase voltage.
B. Current Relationship
By applying Kirchhoff's Current Law (KCL) at each node, every line current is the vector difference of two adjacent phase currents:
$$\vec{I}_R = \vec{I}_{RY} - \vec{I}_{BR}$$
Solving this vector difference yields the relationship:
Relation Between Line and Phase Current in a Delta-Connected System
1. Circuit Connection and Kirchhoff's Current Law (KCL)
In a balanced delta (\(\Delta\)) connected 3-phase system, the three phase windings (or loads) are connected end-to-end to form a closed mesh.
- Line terminals are labeled as R, Y, and B.
- Line currents entering the terminals: \(I_R\), \(I_Y\), \(I_B\)
- Phase currents flowing through the branches: \(I_{RY}\), \(I_{YB}\), \(I_{BR}\)
Applying Kirchhoff's Current Law (KCL) at each line terminal node:
- At Node R: \(\vec{I}_R = \vec{I}_{RY} - \vec{I}_{BR}\)
- At Node Y: \(\vec{I}_Y = \vec{I}_{YB} - \vec{I}_{RY}\)
- At Node B: \(\vec{I}_B = \vec{I}_{BR} - \vec{I}_{YB}\)
For a balanced system, the phase currents are equal in magnitude and displaced from each other by \(120^\circ\):
$$\vert{}I_{RY}\vert{} = \vert{}I_{YB}\vert{} = \vert{}I_{BR}\vert{} = I_{ph}$$
Similarly, line currents are equal in magnitude:
$$\vert{}I_R\vert{} = \vert{}I_Y\vert{} = \vert{}I_B\vert{} = I_L$$