9. A 3-phase A connected generator feeds a A connected load consisting of 3 legs of 10 in series with j4 of inductive reactance, as shown in Figure 9.5. If the line voltage is 208 volts, find the voltage across each load leg, the current through the wires connecting the load to to the generator, and the apparent and real powers drawn by the load. Figure 9.5 R1 3PH) 5 L2 wwm R3 R2 L3
9. A 3-phase A connected generator feeds a A connected load consisting of 3 legs of 10 in series with j4 of inductive reactance, as shown in Figure 9.5. If the line voltage is 208 volts, find the voltage across each load leg, the current through the wires connecting the load to to the generator, and the apparent and real powers drawn by the load. Figure 9.5 R1 3PH) 5 L2 wwm R3 R2 L3
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Question
A 3-phase Δ connected generator feeds a Δ connected load consisting of 3 legs of 10 Ω in series
with j4 Ω of inductive reactance, as shown in Figure 9.5. If the line voltage is 208 volts, find the
voltage across each load leg, the current through the wires connecting the load to to the generator,
and the apparent and real powers drawn by the load.
![**Lesson Module: Understanding 3-Phase Delta Connected Generators and Loads**
**Problem Statement:**
A 3-phase Δ connected generator feeds a Δ connected load consisting of 3 legs of 10 Ω in series with j4 Ω of inductive reactance, as shown in Figure 9.5. If the line voltage is 208 volts, determine the following:
- The voltage across each load leg
- The current through the wires connecting the load to the generator
- The apparent and real powers drawn by the load
**Diagram:**
*Figure 9.5* illustrates a 3-phase Δ (Delta) connected system. It includes a generator connected to a load through three legs, labeled L1, L2, and L3. Each leg of the load consists of a resistor (R1, R2, R3) in series with an inductive reactance.
**Detailed Explanation:**
1. **Voltage Across Each Load Leg:**
Since it is a Δ connected system, the line voltage of 208 volts is the same as the voltage across each load leg.
Therefore, the voltage across each load leg is:
\[
V_{\text{line}} = V_{\text{load}} = 208 \text{ volts}
\]
2. **Current Through Each Load Leg:**
Each load leg consists of a series combination of resistance (R) and inductive reactance (X). The impedance (Z) of each leg is given by:
\[
Z = R + jX = 10 + j4 \;\Omega
\]
The magnitude of the impedance \( |Z| \) is calculated as:
\[
|Z| = \sqrt{R^2 + X^2} = \sqrt{10^2 + 4^2} = \sqrt{100 + 16} = \sqrt{116} \approx 10.77 \;\Omega
\]
The current through each load leg \( I_{\text{load}} \) is given by Ohm's Law:
\[
I_{\text{load}} = \frac{V_{\text{load}}}{|Z|} = \frac{208}{10.77} \approx 19.31 \text{ A}
\]
3. **Apparent and Real Powers:**
The apparent power (S](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5842debd-c7e0-436d-83d9-66a4971dddbf%2Fba3b713b-7f95-4654-864a-93eaaef2cc1e%2Fsg3guib_processed.png&w=3840&q=75)
Transcribed Image Text:**Lesson Module: Understanding 3-Phase Delta Connected Generators and Loads**
**Problem Statement:**
A 3-phase Δ connected generator feeds a Δ connected load consisting of 3 legs of 10 Ω in series with j4 Ω of inductive reactance, as shown in Figure 9.5. If the line voltage is 208 volts, determine the following:
- The voltage across each load leg
- The current through the wires connecting the load to the generator
- The apparent and real powers drawn by the load
**Diagram:**
*Figure 9.5* illustrates a 3-phase Δ (Delta) connected system. It includes a generator connected to a load through three legs, labeled L1, L2, and L3. Each leg of the load consists of a resistor (R1, R2, R3) in series with an inductive reactance.
**Detailed Explanation:**
1. **Voltage Across Each Load Leg:**
Since it is a Δ connected system, the line voltage of 208 volts is the same as the voltage across each load leg.
Therefore, the voltage across each load leg is:
\[
V_{\text{line}} = V_{\text{load}} = 208 \text{ volts}
\]
2. **Current Through Each Load Leg:**
Each load leg consists of a series combination of resistance (R) and inductive reactance (X). The impedance (Z) of each leg is given by:
\[
Z = R + jX = 10 + j4 \;\Omega
\]
The magnitude of the impedance \( |Z| \) is calculated as:
\[
|Z| = \sqrt{R^2 + X^2} = \sqrt{10^2 + 4^2} = \sqrt{100 + 16} = \sqrt{116} \approx 10.77 \;\Omega
\]
The current through each load leg \( I_{\text{load}} \) is given by Ohm's Law:
\[
I_{\text{load}} = \frac{V_{\text{load}}}{|Z|} = \frac{208}{10.77} \approx 19.31 \text{ A}
\]
3. **Apparent and Real Powers:**
The apparent power (S
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