Give step-by-step explanation and final answerA balanced star connected
3
-
phase load having an impedance of
(
6
+
j
9
)
Ω
on each phase is connected to a
3
-
phase,
2
3
0
V
5
0
Hz supply. Cal

Question

Give step-by-step explanation and final answerA balanced star connected 
3
-
phase load having an impedance of 
(
6
+
j
9
)
Ω
 ﻿on each phase is connected to a 
3
-
phase, 
2
3
0
V 
5
0
Hz supply. Calculate:
﻿i
)
 ﻿the line currents 
(
and verify with simulator
)
 ﻿
ii
)
 ﻿the phase currents 
(
and verify with simulator
)
 ﻿
iii
)
 ﻿the power factor iv
)
 ﻿total apparent power 
(
and verify with simulator
)
 ﻿
v
)
 ﻿total reactive power 
(
and verify with simulator
)
 ﻿
vi
)
 ﻿total active power 
(
and verify with simulator
can you work it out step by step stating what equations were used aswell as draw the diagram 
(
if possible can you draw the circuit on proteus to for me to see the diagram its no problem if you cant
)

VIPSolutions · Accepted Answer

Sure! Let's solve the problem step by step.

### **Given Data**
- **Star (Y) connected balanced 3-phase load**
- **Load Impedance per phase:** $ Z_{ph} = (6 + j9)\ \Omega $
- **Line-to-line Voltage (V$_{LL}$):** $ 230\ V $
- **Frequency:** $ 50\ Hz $

---

## **Step 1: Draw the Circuit Diagram**

Here’s a simple text-based representation:

```
      R         Y         B
     |         |         |
     |         |         |
   (Z)       (Z)       (Z)
     |         |         |
     ---------------------
             N
```

Where each (Z) is $ 6 + j9 $ Ω.

---

## **Step 2: Calculate Phase Voltage ($ V_{ph} $)**

For star connection:
$$
V_{ph} = \frac{V_{LL}}{\sqrt{3}}
$$

$$
V_{ph} = \frac{230}{\sqrt{3}} = \frac{230}{1.732} \approx 132.79\,V
$$

---

## **Step 3: Calculate Phase Current ($ I_{ph} $)**

$$
I_{ph} = \frac{V_{ph}}{Z_{ph}}
$$

First, calculate the magnitude of $ Z_{ph} $:

$$
|Z_{ph}| = \sqrt{6^2 + 9^2} = \sqrt{36 + 81} = \sqrt{117} \approx 10.816\, \Omega
$$

Now, calculate the phase current:

$$
I_{ph} = \frac{132.79}{10.816} \approx 12.28\,A
$$

**Angle of Z:**
$$
\theta_Z = \arctan\left(\frac{9}{6}\right) = \arctan(1.5) \approx 56.31^\circ
$$

So,
$$
I_{ph} = 12.28 \angle -56.31^\circ\,A
$$

---

## **Step 4: Calculate Line Current ($ I_{line} $)**

For star connection:
$$
I_{line} = I_{ph}
$$

So,
$$
\boxed{I_{line} = 12.28 \angle -56.31^\circ\,A}
$$

---

## **Step 5: Calculate Power Factor ($ pf $)**

$$
pf = \cos(\theta_Z) = \cos(56.31^\circ) \approx 0.552
$$
(Lagging, since the load is inductive.)

---

## **Step 6: Calculate Total Apparent Power ($ S $)**

$$
S = \sqrt{3} \, V_{LL} \, I_{line}
$$

$$
S = 1.732 \times 230 \times 12.28 \approx 4,899\, VA
$$

---

## **Step 7: Calculate Total Active Power ($ P $)**

$$
P = \sqrt{3} \, V_{LL} \, I_{line} \, pf
$$

$$
P = 1.732 \times 230 \times 12.28 \times 0.552 \approx 2,705\, W
$$

---

## **Step 8: Calculate Total Reactive Power ($ Q $)**

$$
Q = \sqrt{3} \, V_{LL} \, I_{line} \, \sin(\theta_Z)
$$
$$
Q = 1.732 \times 230 \times 12.28 \times \sin(56.31^\circ)
$$
$$
\sin(56.31^\circ) \approx 0.8321
$$
$$
Q = 1.732 \times 230 \times 12.28 \times 0.8321 \approx 4,079\, VAr
$$

---

## **Summary Table**

| Quantity                    | Value                 | Formula Used                                    |
|-----------------------------|-----------------------|-------------------------------------------------|
| **Phase Voltage**           | 132.79 V              | $ V_{ph} = V_{LL}/\sqrt{3} $                  |
| **Phase Current**           | 12.28 A @ -56.31°     | $ I_{ph} = V_{ph}/Z_{ph} $                    |
| **Line Current**            | 12.28 A @ -56.31°     | $ I_{line} = I_{ph} $ (Star connection)       |
| **Power Factor**            | 0.552 (Lagging)       | $ \cos(\theta_Z) $                            |
| **Apparent Power (S)**      | 4,899 VA              | $ \sqrt{3} V_{LL} I_{line} $                  |
| **Active Power (P)**        | 2,705 W               | $ \sqrt{3} V_{LL} I_{line} pf $               |
| **Reactive Power (Q)**      | 4,079 VAr             | $ \sqrt{3} V_{LL} I_{line} \sin(\theta_Z) $   |

---

## **Proteus/Simulator Diagram**

I can't draw directly here, but in **Proteus**:

- Place 3 AC voltage sources, connect in star (Y) configuration with neutral.
- Connect each phase to a series impedance of $ 6 + j9 $ Ω (series resistor and inductor).
- Measure phase and line currents, voltages, and verify calculated values.

---

## **Final Answers**

### **i) Line Currents**
$$
\boxed{I_{line} = 12.28\,A \ \text{at}\ -56.31^\circ}
$$

### **ii) Phase Currents**
$$
\boxed{I_{ph} = 12.28\,A \ \text{at}\ -56.31^\circ}
$$

### **iii) Power Factor**
$$
\boxed{0.552\ \text{(Lagging)}}
$$

### **iv) Total Apparent Power**
$$
\boxed{4,899\,VA}
$$

### **v) Total Reactive Power**
$$
\boxed{4,079\,VAr}
$$

### **vi) Total Active Power**
$$
\boxed{2,705\,W}
$$

---

**Equations Used:**
1. $ V_{ph} = V_{LL}/\sqrt{3} $
2. $ I_{ph} = V_{ph}/Z_{ph} $
3. For star, $ I_{line} = I_{ph} $
4. $ pf = \cos(\theta_Z) $
5. $ S = \sqrt{3} V_{LL} I_{line} $
6. $ P = S \cdot pf $
7. $ Q = \sqrt{3} V_{LL} I_{line} \sin(\theta_Z) $

---

**If you need the actual Proteus file, you’ll have to build it as described above (since I can't upload files).** Let me know if you need more help!

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