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This question is about a three-phase system with a delta (Δ) connected source and load. Here’s how to approach each part:

1. **E_line (Line Voltage)**: 
   Since the source is connected in delta and the line voltage is given as 480 V, this is the line-to-line voltage directly measured across each phase.

   \[
   E_{\text{line}} = 480 \, \text{V}
   \]

2. **I_line (Line Current)**:
   The line current in a delta configuration is related to the phase current by the formula:
   
   \[
   I_{\text{line}} = I_{\text{phase}} \times \sqrt{3}
   \]

   We’ll need to calculate the phase current from the load impedance and line voltage to find this value.

3. **E_phase (Source Phase Voltage)**:
   For a delta configuration, the phase voltage in the source is equal to the line voltage:

   \[
   E_{\text{phase(source)}} = E_{\text{line}} = 480 \, \text{V}
   \]

4. **I_phase (Source Phase Current)**:
   This can be calculated using Ohm’s law for each phase of the load:
   
   \[
   I_{\text{phase(load)}} = \frac{E_{\text{phase(load)}}}{Z_{\text{load}}}
   \]

   Here, \( Z_{\text{load}} = 30 \, \Omega \). So,

   \[
   I_{\text{phase(load)}} = \frac{480 \, \text{V}}{30 \, \Omega} = 16 \, \text{A}
   \]

5. **E_phase (Load Phase Voltage)**:
   In a delta load, the phase voltage across each load is equal to the line voltage:

   \[
   E_{\text{phase(load)}} = 480 \, \text{V}
   \]

6. **I_phase (Load Phase Current)**:
   This is the same as the source phase current calculated above:

   \[
   I_{\text{phase(load)}} = 16 \, \text{A}
   \]

7. **P_total (Total Power)**:
   For a three-phase system, total power \( P_{\text{total}} \) is given by:

   \[
   P_{\text{total}} = \sqrt{3} \times E_{\text{line}} \times I_{\text{line}}
   \]

   First, we need to find \( I_{\text{line}} \):

   \[
   I_{\text{line}} = I_{\text{phase}} \times \sqrt{3} = 16 \times \sqrt{3} \approx 27.71 \, \text{A}
   \]

   Then,

   \[
   P_{\text{total}} = \sqrt{3} \times 480 \times 27.71 \approx 23,038.65 \, \text{W}
   \]

### Summary of Answers
- **E_line** = 480 V
- **I_line** ≈ 27.71 A
- **E_phase(source)** = 480 V
- **I_phase(source)** = 16 A
- **E_phase(load)** = 480 V
- **I_phase(load)** = 16 A
- **P_total** ≈ 23,038.65 W

Would you like further explanation on any specific part, or do you have other questions?

#### Related Questions:
1. How does a delta configuration differ from a wye (Y) configuration in terms of voltage and current calculations?
2. What would change if the system were wye-connected rather than delta-connected?
3. How is total power affected by changes in load impedance in a delta connection?
4. Why is line current in a delta connection higher than the phase current by a factor of √3?
5. What would be the effect on total power if the load impedance were increased?

#### Tip:
In three-phase delta systems, remember that line voltage equals phase voltage, but line current is √3 times the phase current.

Calculate the values for E_line, I_line, E_phase(source), I_phase(source), E_phase(load), I_phase(load), and P_total in a three-phase delta-connected system with a 480 V line-to-line voltage and 30 Ω load impedance.

Explore the step-by-step calculations for E_line, I_line, E_phase, I_phase, and P_total in a three-phase delta system with a 480 V line voltage and 30 Ω impedance. This problem covers AC circuit analysis in delta configurations, essential for electrical engineering students.

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