Figure 9.2 R2 b 4. Referring to Figure 9.2, if the generator phase voltage is 230 volts and the load is balanced with each leg at 12 Q2, determine the line voltage, line current, generator phase current, load current, load voltage and total load power. E1 E2 E3 Z R1 R3 m ww X

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Referring to Figure 9.2, if the generator phase voltage is 230 volts and the load is balanced with each leg
at 12 Ω, determine the line voltage, line current, generator phase current, load current, load voltage and
total load power.

### Figure 9.2: Electrical Circuit Diagram

**Description:**
The circuit diagram shown in **Figure 9.2** consists of three alternating current (AC) voltage sources labeled \( E1 \), \( E2 \), and \( E3 \). These sources are connected in a three-phase configuration. Below is a breakdown of the elements and connections in the circuit:

- **Voltage Sources:**
  - \( E1 \) is connected between points **a** and **c**.
  - \( E3 \) is connected between points **c** and **z**.
  - \( E2 \) is connected between points **b** and **c**.
  
- **Resistive Loads:**
  - \( R1 \) is connected between points **x** and **z**.
  - \( R2 \) is connected between points **y** and a grounded point.
  - \( R3 \) is connected directly across points **z** and **y**.

**Given Information:**
- The generator phase voltage is 230 volts.
- The load is balanced with each leg having a resistance of 12 ohms.

**Tasks:**
Determine the following:
1. Line voltage
2. Line current
3. Generator phase current
4. Load current
5. Load voltage
6. Total load power

**Task 4: Calculation Instructions**

Let's assume each element is represented by its technical name and calculate the requested values based on the given data.

**a. Line Voltage:**
The line voltage \( V_L \) in a balanced three-phase system equals the phase voltage \( V_p \) multiplied by \( \sqrt{3} \).

\[ V_L = V_p \cdot \sqrt{3} \]

Given \( V_p = 230 \text{ volts} \):

\[ V_L = 230 \cdot \sqrt{3} \approx 230 \cdot 1.732 \approx 398 \text{ volts} \]

**b. Line Current:**
The line current \( I_L \) in a balanced three-phase system is given by:

\[ I_L = \frac{V_p}{Z} \]

Where \( Z \) is the impedance of each phase. For purely resistive loads:

\[ Z = R = 12 \Omega \]

Hence:

\[ I_L = \frac{230}{12} \approx 19.
Transcribed Image Text:### Figure 9.2: Electrical Circuit Diagram **Description:** The circuit diagram shown in **Figure 9.2** consists of three alternating current (AC) voltage sources labeled \( E1 \), \( E2 \), and \( E3 \). These sources are connected in a three-phase configuration. Below is a breakdown of the elements and connections in the circuit: - **Voltage Sources:** - \( E1 \) is connected between points **a** and **c**. - \( E3 \) is connected between points **c** and **z**. - \( E2 \) is connected between points **b** and **c**. - **Resistive Loads:** - \( R1 \) is connected between points **x** and **z**. - \( R2 \) is connected between points **y** and a grounded point. - \( R3 \) is connected directly across points **z** and **y**. **Given Information:** - The generator phase voltage is 230 volts. - The load is balanced with each leg having a resistance of 12 ohms. **Tasks:** Determine the following: 1. Line voltage 2. Line current 3. Generator phase current 4. Load current 5. Load voltage 6. Total load power **Task 4: Calculation Instructions** Let's assume each element is represented by its technical name and calculate the requested values based on the given data. **a. Line Voltage:** The line voltage \( V_L \) in a balanced three-phase system equals the phase voltage \( V_p \) multiplied by \( \sqrt{3} \). \[ V_L = V_p \cdot \sqrt{3} \] Given \( V_p = 230 \text{ volts} \): \[ V_L = 230 \cdot \sqrt{3} \approx 230 \cdot 1.732 \approx 398 \text{ volts} \] **b. Line Current:** The line current \( I_L \) in a balanced three-phase system is given by: \[ I_L = \frac{V_p}{Z} \] Where \( Z \) is the impedance of each phase. For purely resistive loads: \[ Z = R = 12 \Omega \] Hence: \[ I_L = \frac{230}{12} \approx 19.
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