Question

A balanced delta connected load is connected to a 415 V, 50 Hz supply. If the per phase impedance of the load is (8 + j12) ohm, then calculate the phase current of the load. Also, find out the line current and the power consumed by each phase.

Answer 1

To find the phase current, line current, and power consumed by each phase for a balanced delta-connected load, we can follow these steps:

### 1. **Calculate the Phase Current**

**Given Data:**
- Supply Voltage (Line-to-Line Voltage), \( V_{LL} = 415 \text{ V} \)
- Impedance per phase, \( Z = 8 + j12 \text{ ohms} \)

For a delta-connected load, the line-to-line voltage is the same as the voltage across each phase. So, the phase voltage \( V_{ph} \) is the same as the line-to-line voltage in this case.

**Phase Voltage \( V_{ph} \):**
\[ V_{ph} = V_{LL} = 415 \text{ V} \]

The phase current \( I_{ph} \) can be calculated using Ohm’s law:
\[ I_{ph} = \frac{V_{ph}}{Z} \]

First, we need to find the magnitude of the impedance \( Z \):
\[ |Z| = \sqrt{8^2 + 12^2} \]
\[ |Z| = \sqrt{64 + 144} \]
\[ |Z| = \sqrt{208} \]
\[ |Z| \approx 14.42 \text{ ohms} \]

Now, calculate the phase current:
\[ I_{ph} = \frac{415 \text{ V}}{14.42 \text{ ohms}} \]
\[ I_{ph} \approx 28.8 \text{ A} \]

### 2. **Calculate the Line Current**

For a delta-connected load, the line current \( I_{L} \) is related to the phase current \( I_{ph} \) by:
\[ I_{L} = \sqrt{3} \times I_{ph} \]

So,
\[ I_{L} = \sqrt{3} \times 28.8 \text{ A} \]
\[ I_{L} \approx 49.8 \text{ A} \]

### 3. **Calculate the Power Consumed by Each Phase**

The power consumed by each phase \( P_{ph} \) is given by:
\[ P_{ph} = I_{ph}^2 \times R \]

where \( R \) is the real part of the impedance (8 ohms).

Calculate the power:
\[ P_{ph} = (28.8)^2 \times 8 \]
\[ P_{ph} = 829.44 \times 8 \]
\[ P_{ph} \approx 6635.52 \text{ W} \]

So, the results are:
- **Phase Current:** \( \approx 28.8 \text{ A} \)
- **Line Current:** \( \approx 49.8 \text{ A} \)
- **Power Consumed by Each Phase:** \( \approx 6635.52 \text{ W} \)

Feel free to ask if you need any further clarification!

Answer 2

To solve this problem, we need to follow these steps:

1. **Calculate the Phase Voltage:**

   For a balanced delta-connected load, the line voltage \( V_L \) is related to the phase voltage \( V_{ph} \) by the following equation:
   \[
   V_{ph} = \frac{V_L}{\sqrt{3}}
   \]
   Here, \( V_L = 415 \text{ V} \). So:
   \[
   V_{ph} = \frac{415}{\sqrt{3}} \approx \frac{415}{1.732} \approx 239.6 \text{ V}
   \]

2. **Calculate the Phase Current:**

   The phase current \( I_{ph} \) can be found using Ohm’s law:
   \[
   I_{ph} = \frac{V_{ph}}{Z_{ph}}
   \]
   where \( Z_{ph} \) is the phase impedance, given as \( 8 + j12 \) ohms. First, calculate the magnitude of the phase impedance:
   \[
   |Z_{ph}| = \sqrt{8^2 + 12^2} = \sqrt{64 + 144} = \sqrt{208} \approx 14.42 \text{ ohms}
   \]
   Then:
   \[
   I_{ph} = \frac{239.6}{14.42} \approx 16.63 \text{ A}
   \]

3. **Calculate the Line Current:**

   For a delta-connected load, the line current \( I_L \) is related to the phase current \( I_{ph} \) by:
   \[
   I_L = I_{ph} \cdot \sqrt{3}
   \]
   Thus:
   \[
   I_L = 16.63 \times \sqrt{3} \approx 16.63 \times 1.732 \approx 28.8 \text{ A}
   \]

4. **Calculate the Power Consumed by Each Phase:**

   The power consumed by each phase \( P_{ph} \) can be calculated using:
   \[
   P_{ph} = I_{ph}^2 \cdot R_{ph}
   \]
   where \( R_{ph} \) is the real part of the impedance. For \( Z_{ph} = 8 + j12 \):
   \[
   R_{ph} = 8 \text{ ohms}
   \]
   So:
   \[
   P_{ph} = (16.63)^2 \times 8 \approx 276.9 \times 8 \approx 2215.2 \text{ W}
   \]

   Alternatively, you can calculate the power factor \( \text{pf} \) and use it to find the power:
   \[
   \text{pf} = \cos \theta = \frac{R_{ph}}{|Z_{ph}|} = \frac{8}{14.42} \approx 0.555
   \]
   The power per phase can also be computed using:
   \[
   P_{ph} = V_{ph} \times I_{ph} \times \text{pf}
   \]
   \[
   P_{ph} = 239.6 \times 16.63 \times 0.555 \approx 2215.2 \text{ W}
   \]

**Summary:**

- **Phase Voltage** \( V_{ph} \): 239.6 V
- **Phase Current** \( I_{ph} \): 16.63 A
- **Line Current** \( I_L \): 28.8 A
- **Power Consumed by Each Phase** \( P_{ph} \): 2215.2 W