Question

Derive the expression of line current in terms of phase voltage in delta connection.

Answer 1

To derive the expression for line current in terms of phase voltage in a delta-connected system, let's start by understanding the basic configuration of a delta connection and the relationship between line and phase quantities.

### Delta Connection Overview

In a delta connection (Δ), three phases are connected in a closed loop. Each of the three loads (or impedances) is connected between two phase terminals. The key characteristics of a delta connection are:

1. **Phase Voltages (V_phase)**: The voltage across each load (Z1, Z2, Z3).
2. **Line Voltages (V_line)**: The voltages between any two lines (A, B, C) connected to the system.

In a delta connection, the line voltage is equal to the phase voltage:
\[
V_{line} = V_{phase}
\]

### Current Relationships

1. **Phase Currents (I_phase)**: The currents flowing through each load (I1, I2, I3).
2. **Line Currents (I_line)**: The currents flowing in each line (IA, IB, IC).

For a delta connection, the relationship between the phase currents and line currents can be understood by using Kirchhoff's Current Law (KCL).

### KCL in Delta Connection

When we apply KCL at each node in the delta connection, we have:

- For line A:
\[
I_A = I_1 + I_3
\]

- For line B:
\[
I_B = I_2 + I_1
\]

- For line C:
\[
I_C = I_3 + I_2
\]

### Current in Terms of Phase Voltage

The phase current can be expressed in terms of the phase voltage and the impedance of each phase:
\[
I_1 = \frac{V_{phase}}{Z_1}, \quad I_2 = \frac{V_{phase}}{Z_2}, \quad I_3 = \frac{V_{phase}}{Z_3}
\]

Here, \( Z_1, Z_2, Z_3 \) are the impedances of the respective phases.

### Finding Line Current

Let’s focus on finding the expression for line current \( I_A \):
\[
I_A = I_1 + I_3 = \frac{V_{phase}}{Z_1} + \frac{V_{phase}}{Z_3}
\]
Factoring out \( V_{phase} \):
\[
I_A = V_{phase} \left( \frac{1}{Z_1} + \frac{1}{Z_3} \right)
\]

For line B and line C, you can derive similar expressions:
\[
I_B = \frac{V_{phase}}{Z_2} + \frac{V_{phase}}{Z_1} = V_{phase} \left( \frac{1}{Z_2} + \frac{1}{Z_1} \right)
\]
\[
I_C = \frac{V_{phase}}{Z_3} + \frac{V_{phase}}{Z_2} = V_{phase} \left( \frac{1}{Z_3} + \frac{1}{Z_2} \right)
\]

### Summary Expression for Line Current

Thus, the line current in a delta-connected system can be expressed in terms of phase voltage and the impedances of the loads:
\[
I_A = V_{phase} \left( \frac{1}{Z_1} + \frac{1}{Z_3} \right)
\]
\[
I_B = V_{phase} \left( \frac{1}{Z_2} + \frac{1}{Z_1} \right)
\]
\[
I_C = V_{phase} \left( \frac{1}{Z_3} + \frac{1}{Z_2} \right)
\]

### Simplified Current Expressions

If the impedances are equal (say \( Z_1 = Z_2 = Z_3 = Z \)):
\[
I_A = \frac{V_{phase}}{Z} \cdot 2 \quad \text{(since each line current combines two phase currents)}
\]
This results in:
\[
I_A = \frac{2V_{phase}}{Z}
\]
And similarly for \( I_B \) and \( I_C \).

### Conclusion

In conclusion, the relationship between line currents and phase voltages in a delta-connected system allows us to derive the line current in terms of the phase voltage and the impedances of the connected loads. This understanding is crucial for analyzing and designing three-phase electrical systems.

Answer 2

To derive the expression for the line current in terms of the phase voltage in a delta (Δ) connection, let's start by understanding the basics of a three-phase delta connection.

### Delta Connection Overview

In a three-phase delta connection:
- The three-phase loads are connected in a triangular (delta) configuration.
- Each corner of the delta represents one phase of the load.
- The line voltage (\(V_L\)) is the voltage across any two of the three lines (phases).
- The phase voltage (\(V_{\phi}\)) is the voltage across each phase of the delta-connected load.

### Relationship Between Line Voltage and Phase Voltage

In a delta connection, the phase voltage (\(V_{\phi}\)) is the same as the line voltage (\(V_L\)) across each phase. However, the relationship between the line voltage and phase voltage can be given by:

\[ V_L = \sqrt{3} \cdot V_{\phi} \]

where \(V_L\) is the line voltage and \(V_{\phi}\) is the phase voltage.

### Derivation of Line Current

1. **Phase Voltage and Phase Current**:

   For a delta connection, the phase voltage (\(V_{\phi}\)) is the voltage across each load. If each load has an impedance \(Z_{\phi}\), the current through each phase (phase current, \(I_{\phi}\)) is:

   \[ I_{\phi} = \frac{V_{\phi}}{Z_{\phi}} \]

2. **Line Current in Terms of Phase Current**:

   In a delta connection, the line current (\(I_L\)) is the current flowing through one of the lines. This line current is the sum of the currents through the two phases that connect at that line.

   For a line connected between phase A and phase B:
   - The line current is the current flowing through line A and line B.
   - The line current \(I_L\) can be computed using vector addition of the two phase currents \(I_{A}\) and \(I_{B}\) that are connected to that line.

   Using the symmetrical property of a balanced three-phase system:
   \[ I_L = \sqrt{3} \cdot I_{\phi} \]

3. **Expressing Line Current in Terms of Phase Voltage**:

   Substitute \(I_{\phi}\) into the line current formula:

   \[ I_{\phi} = \frac{V_{\phi}}{Z_{\phi}} \]

   Therefore:

   \[ I_L = \sqrt{3} \cdot I_{\phi} = \sqrt{3} \cdot \frac{V_{\phi}}{Z_{\phi}} \]

   Hence, the line current in terms of the phase voltage is:

   \[ I_L = \frac{\sqrt{3} \cdot V_{\phi}}{Z_{\phi}} \]

### Summary

In a three-phase delta connection, the line current (\(I_L\)) is related to the phase voltage (\(V_{\phi}\)) and the phase impedance (\(Z_{\phi}\)) by:

\[ I_L = \frac{\sqrt{3} \cdot V_{\phi}}{Z_{\phi}} \]

This relationship shows that the line current is \(\sqrt{3}\) times the phase current and is inversely proportional to the phase impedance.