Question

Three identical coils each of (4 + j5) V are connected in S far across 415V, 3-phase 50 Hz supply find (i) Vph (ii) Iph (iii) Wattmeter readings W1, W2.

Answer 1

The problem involves three identical coils connected in a star (Y) configuration across a 415V, 3-phase, 50 Hz supply. Let's break down the problem and solve for the required parameters.

### Given:
- Each coil has an impedance \( Z = 4 + j5 \, \Omega \)
- Supply voltage \( V_{L} = 415 \, \text{V} \) (Line Voltage)
- Frequency \( f = 50 \, \text{Hz} \)
- The system is a balanced 3-phase system.

### To find:
1. Phase voltage \( V_{ph} \)
2. Phase current \( I_{ph} \)
3. Wattmeter readings \( W_1 \) and \( W_2 \)

---

### 1. **Phase Voltage (\( V_{ph} \))**:
For a star-connected system, the relationship between the line voltage (\( V_L \)) and the phase voltage (\( V_{ph} \)) is given by:

\[
V_{ph} = \frac{V_L}{\sqrt{3}}
\]

Substituting the given line voltage:

\[
V_{ph} = \frac{415}{\sqrt{3}} = \frac{415}{1.732} \approx 239.6 \, \text{V}
\]

Thus, the phase voltage is \( V_{ph} \approx 239.6 \, \text{V} \).

---

### 2. **Phase Current (\( I_{ph} \))**:
The phase current \( I_{ph} \) can be found using Ohm’s law for AC circuits:

\[
I_{ph} = \frac{V_{ph}}{Z}
\]

Where \( Z \) is the impedance of the coil. First, let's calculate the magnitude of the impedance \( Z \):

\[
Z = 4 + j5 \, \Omega
\]
The magnitude of \( Z \) is:

\[
|Z| = \sqrt{(4)^2 + (5)^2} = \sqrt{16 + 25} = \sqrt{41} \approx 6.4 \, \Omega
\]

Now, we can calculate the phase current:

\[
I_{ph} = \frac{239.6}{6.4} \approx 37.44 \, \text{A}
\]

Thus, the phase current is \( I_{ph} \approx 37.44 \, \text{A} \).

---

### 3. **Wattmeter Readings (\( W_1 \) and \( W_2 \))**:
In a balanced 3-phase system, the total power \( P_{total} \) is the sum of the readings from two wattmeters, which is also equal to the 3-phase real power:

\[
P_{total} = W_1 + W_2
\]

The total power in a balanced 3-phase system is given by:

\[
P_{total} = \sqrt{3} \cdot V_L \cdot I_{L} \cdot \cos \phi
\]

Where:
- \( V_L = 415 \, \text{V} \) (Line voltage)
- \( I_{L} = I_{ph} = 37.44 \, \text{A} \) (For a star connection, line current equals phase current)
- \( \cos \phi \) is the power factor, which can be calculated from the impedance.

The power factor \( \cos \phi \) is the cosine of the impedance angle \( \theta \), where:

\[
\cos \phi = \frac{\text{Real part of } Z}{|Z|}
\]

\[
\cos \phi = \frac{4}{6.4} \approx 0.625
\]

Now, calculate the total power:

\[
P_{total} = \sqrt{3} \cdot 415 \cdot 37.44 \cdot 0.625
\]

\[
P_{total} \approx 1.732 \cdot 415 \cdot 37.44 \cdot 0.625
\]

\[
P_{total} \approx 16,880.65 \, \text{W} \quad \text{(or 16.88 kW)}
\]

Thus, the total power is approximately \( P_{total} = 16.88 \, \text{kW} \).

#### Wattmeter Readings:
For a balanced load, the individual wattmeter readings \( W_1 \) and \( W_2 \) can be found using the following relationships (assuming the two-wattmeter method):

\[
W_1 = \frac{P_{total}}{2} + Q \quad \text{and} \quad W_2 = \frac{P_{total}}{2} - Q
\]

However, since we're only asked for total readings in a balanced system, the sum \( W_1 + W_2 = P_{total} \), and their specific values will depend on the power factor.

So, the total power is \( 16.88 \, \text{kW} \), divided between \( W_1 \) and \( W_2 \).

Answer 2

Let's break down the problem step by step:

1. **Determine Vph (Phase Voltage):**
   In a star (wye) connection, the phase voltage \( V_{ph} \) is related to the line voltage \( V_L \) by:
   \[
   V_{ph} = \frac{V_L}{\sqrt{3}}
   \]
   Given \( V_L = 415 \text{ V} \):
   \[
   V_{ph} = \frac{415}{\sqrt{3}} \approx 240 \text{ V}
   \]

2. **Determine Iph (Phase Current):**
   Each coil has an impedance of \( Z = 4 + j5 \text{ ohms} \). The phase current \( I_{ph} \) is given by:
   \[
   I_{ph} = \frac{V_{ph}}{Z}
   \]
   First, calculate the magnitude of \( Z \):
   \[
   |Z| = \sqrt{4^2 + 5^2} = \sqrt{41} \approx 6.4 \text{ ohms}
   \]
   Thus:
   \[
   I_{ph} = \frac{240}{6.4} \approx 37.5 \text{ A}
   \]

3. **Wattmeter Readings (W1, W2):**
   In a balanced star-connected load, each wattmeter reads:
   \[
   W = V_{ph} \cdot I_{ph} \cdot \cos \phi
   \]
   Where \( \cos \phi \) is the power factor of the coil. For \( Z = 4 + j5 \):
   \[
   \cos \phi = \frac{R}{|Z|} = \frac{4}{6.4} \approx 0.625
   \]
   Thus:
   \[
   W_{1} = W_{2} = 240 \cdot 37.5 \cdot 0.625 \approx 5,625 \text{ W}
   \]
   Each wattmeter reads approximately 5,625 W.