Derive the expression of line voltage in terms of phase voltage in star connection.
2 Answers
To derive the expression for line voltage in terms of phase voltage in a star (Y) connection, we need to understand the configuration and relationships between the line voltages and phase voltages in a three-phase system.
### Definitions
1. **Phase Voltage (V_phase)**: This is the voltage across each individual load in the star connection. In a three-phase system, there are three phase voltages (V_A, V_B, V_C), corresponding to each of the three phases.
2. **Line Voltage (V_line)**: This is the voltage measured between any two of the three lines (conductors) in the system. In a three-phase system, there are three line voltages (V_AB, V_BC, V_CA).
### Star Connection Diagram
In a star connection:
- The three phases are connected at a common point (neutral).
- Each phase is connected to a load.
### Voltage Relationships
In a balanced three-phase system, the phase voltages and line voltages are related through geometry, specifically using vector relationships.
1. **Vector Representation**: The phase voltages can be represented as vectors in a plane:
- \( V_A \) is at 0 degrees.
- \( V_B \) is at 120 degrees.
- \( V_C \) is at 240 degrees (or -120 degrees).
2. **Line Voltage Calculation**:
To find the line voltage \( V_{AB} \), we consider the voltages:
\[
V_{AB} = V_A - V_B
\]
In complex notation, we can express:
\[
V_A = V_{phase}
\]
\[
V_B = V_{phase} \angle 120^\circ = V_{phase} \left( -\frac{1}{2} + j\frac{\sqrt{3}}{2} \right)
\]
3. **Calculating \( V_{AB} \)**:
Substituting the expressions:
\[
V_{AB} = V_{phase} - \left( V_{phase} \left( -\frac{1}{2} + j\frac{\sqrt{3}}{2} \right) \right)
\]
\[
V_{AB} = V_{phase} + \frac{1}{2} V_{phase} - j\frac{\sqrt{3}}{2} V_{phase}
\]
\[
V_{AB} = V_{phase} \left( \frac{3}{2} - j\frac{\sqrt{3}}{2} \right)
\]
4. **Magnitude of \( V_{AB} \)**:
To find the magnitude, we compute:
\[
|V_{AB}| = |V_{phase}| \sqrt{\left( \frac{3}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}
\]
\[
|V_{AB}| = V_{phase} \sqrt{\frac{9}{4} + \frac{3}{4}} = V_{phase} \sqrt{3}
\]
### General Result
Thus, the relationship between the line voltage \( V_{line} \) and phase voltage \( V_{phase} \) in a star connection can be expressed as:
\[
V_{line} = \sqrt{3} \cdot V_{phase}
\]
### Summary
In a star-connected three-phase system:
- The line voltage is the voltage measured between two of the three lines and is equal to the phase voltage multiplied by the square root of three (\(\sqrt{3}\)).
- This relationship is critical for understanding the operation and design of three-phase electrical systems, particularly in power distribution and motor applications.
### Definitions
1. **Phase Voltage (V_phase)**: This is the voltage across each individual load in the star connection. In a three-phase system, there are three phase voltages (V_A, V_B, V_C), corresponding to each of the three phases.
2. **Line Voltage (V_line)**: This is the voltage measured between any two of the three lines (conductors) in the system. In a three-phase system, there are three line voltages (V_AB, V_BC, V_CA).
### Star Connection Diagram
In a star connection:
- The three phases are connected at a common point (neutral).
- Each phase is connected to a load.
### Voltage Relationships
In a balanced three-phase system, the phase voltages and line voltages are related through geometry, specifically using vector relationships.
1. **Vector Representation**: The phase voltages can be represented as vectors in a plane:
- \( V_A \) is at 0 degrees.
- \( V_B \) is at 120 degrees.
- \( V_C \) is at 240 degrees (or -120 degrees).
2. **Line Voltage Calculation**:
To find the line voltage \( V_{AB} \), we consider the voltages:
\[
V_{AB} = V_A - V_B
\]
In complex notation, we can express:
\[
V_A = V_{phase}
\]
\[
V_B = V_{phase} \angle 120^\circ = V_{phase} \left( -\frac{1}{2} + j\frac{\sqrt{3}}{2} \right)
\]
3. **Calculating \( V_{AB} \)**:
Substituting the expressions:
\[
V_{AB} = V_{phase} - \left( V_{phase} \left( -\frac{1}{2} + j\frac{\sqrt{3}}{2} \right) \right)
\]
\[
V_{AB} = V_{phase} + \frac{1}{2} V_{phase} - j\frac{\sqrt{3}}{2} V_{phase}
\]
\[
V_{AB} = V_{phase} \left( \frac{3}{2} - j\frac{\sqrt{3}}{2} \right)
\]
4. **Magnitude of \( V_{AB} \)**:
To find the magnitude, we compute:
\[
|V_{AB}| = |V_{phase}| \sqrt{\left( \frac{3}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}
\]
\[
|V_{AB}| = V_{phase} \sqrt{\frac{9}{4} + \frac{3}{4}} = V_{phase} \sqrt{3}
\]
### General Result
Thus, the relationship between the line voltage \( V_{line} \) and phase voltage \( V_{phase} \) in a star connection can be expressed as:
\[
V_{line} = \sqrt{3} \cdot V_{phase}
\]
### Summary
In a star-connected three-phase system:
- The line voltage is the voltage measured between two of the three lines and is equal to the phase voltage multiplied by the square root of three (\(\sqrt{3}\)).
- This relationship is critical for understanding the operation and design of three-phase electrical systems, particularly in power distribution and motor applications.
In a star (or wye) connected three-phase system, the relationship between line voltage and phase voltage can be derived using basic principles of electrical engineering and trigonometry. Here’s a detailed derivation:
### Definitions and Basic Relationships
1. **Phase Voltage (\( V_{\text{ph}} \))**: The voltage measured across any single phase winding in the star connection. This is also the voltage between any one phase and the neutral point.
2. **Line Voltage (\( V_{\text{L}} \))**: The voltage measured between any two of the three phase conductors.
### Star Connection Setup
In a star connection:
- Three phase windings are connected in a star (Y) configuration.
- The neutral point is the common connection point of the three windings.
- Each phase winding is connected between one of the line conductors and the neutral point.
### Vector Representation
To understand the relationship between line voltage and phase voltage, it's helpful to use vector representation. Assume we have a balanced three-phase system. Let’s denote the phase voltages as \( V_{AN} \), \( V_{BN} \), and \( V_{CN} \), where \( N \) represents the neutral point. These can be represented as vectors in the complex plane:
- \( V_{AN} = V_{\text{ph}} \angle 0^\circ \)
- \( V_{BN} = V_{\text{ph}} \angle -120^\circ \)
- \( V_{CN} = V_{\text{ph}} \angle 120^\circ \)
Here, \( V_{\text{ph}} \) is the phase voltage.
### Derivation of Line Voltage
Consider the line voltage \( V_{AB} \), which is the voltage between phase A and phase B. It can be computed as:
\[ V_{AB} = V_{AN} - V_{BN} \]
Using the vector representation:
- \( V_{AN} \) is a vector with magnitude \( V_{\text{ph}} \) at \( 0^\circ \)
- \( V_{BN} \) is a vector with magnitude \( V_{\text{ph}} \) at \( -120^\circ \)
Thus:
\[ V_{AB} = V_{\text{ph}} \angle 0^\circ - V_{\text{ph}} \angle -120^\circ \]
Expressing this in rectangular form:
- \( V_{AN} = V_{\text{ph}} + j0 \)
- \( V_{BN} = V_{\text{ph}} \left( \cos(-120^\circ) + j \sin(-120^\circ) \right) \)
- \( \cos(-120^\circ) = -\frac{1}{2} \)
- \( \sin(-120^\circ) = -\frac{\sqrt{3}}{2} \)
Thus:
\[ V_{BN} = V_{\text{ph}} \left( -\frac{1}{2} - j \frac{\sqrt{3}}{2} \right) \]
Now compute \( V_{AB} \):
\[ V_{AB} = V_{\text{ph}} - V_{\text{ph}} \left( -\frac{1}{2} - j \frac{\sqrt{3}}{2} \right) \]
\[ V_{AB} = V_{\text{ph}} \left( 1 + \frac{1}{2} + j \frac{\sqrt{3}}{2} \right) \]
\[ V_{AB} = V_{\text{ph}} \left( \frac{3}{2} + j \frac{\sqrt{3}}{2} \right) \]
The magnitude of \( V_{AB} \) is:
\[ |V_{AB}| = \sqrt{ \left( \frac{3}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 } \]
\[ |V_{AB}| = \sqrt{ \frac{9}{4} + \frac{3}{4} } \]
\[ |V_{AB}| = \sqrt{ \frac{12}{4} } \]
\[ |V_{AB}| = \sqrt{3} \cdot \frac{V_{\text{ph}}}{1} \]
\[ |V_{AB}| = \sqrt{3} \cdot V_{\text{ph}} \]
Therefore:
\[ V_{\text{L}} = \sqrt{3} \cdot V_{\text{ph}} \]
### Summary
In a star-connected three-phase system, the line voltage \( V_{\text{L}} \) is related to the phase voltage \( V_{\text{ph}} \) by the following expression:
\[ V_{\text{L}} = \sqrt{3} \cdot V_{\text{ph}} \]
This result is fundamental in three-phase power systems and reflects the geometric relationship between the voltages in a balanced three-phase system.
### Definitions and Basic Relationships
1. **Phase Voltage (\( V_{\text{ph}} \))**: The voltage measured across any single phase winding in the star connection. This is also the voltage between any one phase and the neutral point.
2. **Line Voltage (\( V_{\text{L}} \))**: The voltage measured between any two of the three phase conductors.
### Star Connection Setup
In a star connection:
- Three phase windings are connected in a star (Y) configuration.
- The neutral point is the common connection point of the three windings.
- Each phase winding is connected between one of the line conductors and the neutral point.
### Vector Representation
To understand the relationship between line voltage and phase voltage, it's helpful to use vector representation. Assume we have a balanced three-phase system. Let’s denote the phase voltages as \( V_{AN} \), \( V_{BN} \), and \( V_{CN} \), where \( N \) represents the neutral point. These can be represented as vectors in the complex plane:
- \( V_{AN} = V_{\text{ph}} \angle 0^\circ \)
- \( V_{BN} = V_{\text{ph}} \angle -120^\circ \)
- \( V_{CN} = V_{\text{ph}} \angle 120^\circ \)
Here, \( V_{\text{ph}} \) is the phase voltage.
### Derivation of Line Voltage
Consider the line voltage \( V_{AB} \), which is the voltage between phase A and phase B. It can be computed as:
\[ V_{AB} = V_{AN} - V_{BN} \]
Using the vector representation:
- \( V_{AN} \) is a vector with magnitude \( V_{\text{ph}} \) at \( 0^\circ \)
- \( V_{BN} \) is a vector with magnitude \( V_{\text{ph}} \) at \( -120^\circ \)
Thus:
\[ V_{AB} = V_{\text{ph}} \angle 0^\circ - V_{\text{ph}} \angle -120^\circ \]
Expressing this in rectangular form:
- \( V_{AN} = V_{\text{ph}} + j0 \)
- \( V_{BN} = V_{\text{ph}} \left( \cos(-120^\circ) + j \sin(-120^\circ) \right) \)
- \( \cos(-120^\circ) = -\frac{1}{2} \)
- \( \sin(-120^\circ) = -\frac{\sqrt{3}}{2} \)
Thus:
\[ V_{BN} = V_{\text{ph}} \left( -\frac{1}{2} - j \frac{\sqrt{3}}{2} \right) \]
Now compute \( V_{AB} \):
\[ V_{AB} = V_{\text{ph}} - V_{\text{ph}} \left( -\frac{1}{2} - j \frac{\sqrt{3}}{2} \right) \]
\[ V_{AB} = V_{\text{ph}} \left( 1 + \frac{1}{2} + j \frac{\sqrt{3}}{2} \right) \]
\[ V_{AB} = V_{\text{ph}} \left( \frac{3}{2} + j \frac{\sqrt{3}}{2} \right) \]
The magnitude of \( V_{AB} \) is:
\[ |V_{AB}| = \sqrt{ \left( \frac{3}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 } \]
\[ |V_{AB}| = \sqrt{ \frac{9}{4} + \frac{3}{4} } \]
\[ |V_{AB}| = \sqrt{ \frac{12}{4} } \]
\[ |V_{AB}| = \sqrt{3} \cdot \frac{V_{\text{ph}}}{1} \]
\[ |V_{AB}| = \sqrt{3} \cdot V_{\text{ph}} \]
Therefore:
\[ V_{\text{L}} = \sqrt{3} \cdot V_{\text{ph}} \]
### Summary
In a star-connected three-phase system, the line voltage \( V_{\text{L}} \) is related to the phase voltage \( V_{\text{ph}} \) by the following expression:
\[ V_{\text{L}} = \sqrt{3} \cdot V_{\text{ph}} \]
This result is fundamental in three-phase power systems and reflects the geometric relationship between the voltages in a balanced three-phase system.