Question

A three-phase transformer has 600 primary turns and 150 secondary turns. If the supply voltage is 1.5 kV determine the secondary line voltage on no-load when windings are connected in delta-star.

• A. 649.50 V
• B. 549.50 V
• C. 595.50 V
• D. 449.50 V

Hint:

In a delta-star transformer, primary line voltage equals phase voltage, while secondary line voltage is \(\sqrt{3}\) times the phase voltage.

Answer 1

The Correct Option is A

Step 1: Determine the turns ratio of the transformer.
\[ \text{Turns ratio} = \frac{N_1}{N_2} = \frac{600}{150} = 4 \]
Step 2: Identify the primary phase voltage.
Primary winding is connected in delta, therefore
\[ V_{\text{phase, primary}} = V_{\text{line, primary}} = 1.5 \text{ kV} \]
Step 3: Calculate the secondary phase voltage.
Using the turns ratio,
\[ V_{\text{phase, secondary}} = \frac{1500}{4} = 375 \text{ V} \]
Step 4: Convert secondary phase voltage to line voltage.
Secondary winding is connected in star, so
\[ V_{\text{line, secondary}} = \sqrt{3} \times V_{\text{phase, secondary}} \]
\[ V_{\text{line, secondary}} = \sqrt{3} \times 375 = 649.50 \text{ V} \]
Step 5: Conclusion.
The secondary line voltage on no-load is
\[ \boxed{649.50 \text{ V}} \]

Answer 2

For a delta-primary/star-secondary transformer, the overall relationship between line voltages combines the turns ratio with the \(\sqrt{3}\) factor introduced by the star connection, so we can compute a single combined line-to-line transformation factor and check each option against it.

Turns ratio: \( \dfrac{N_1}{N_2} = \dfrac{600}{150} = 4 \), so \( \dfrac{N_2}{N_1} = 0.25 \).

Since the primary is delta (line voltage = phase voltage) and the secondary is star (line voltage = \(\sqrt{3}\) times phase voltage), the combined line-to-line factor is:

\[ \frac{V_{\text{line,sec}}}{V_{\text{line,pri}}} = \sqrt{3} \times \frac{N_2}{N_1} = \sqrt{3} \times 0.25 \approx 0.4330 \]

Applying this to the given supply voltage of \(1.5\) kV: \( V_{\text{line,sec}} = 1500 \times 0.4330 \approx 649.50 \text{ V} \).

1. 649.50 V: Matches the combined line-to-line factor applied directly to the \(1500\) V supply.
2. 549.50 V: Would require a combined factor of about \(0.3663\), which does not correspond to \(\sqrt{3}/4\) for this winding ratio.
3. 595.50 V: Would require a factor of about \(0.3970\), again not matching \(\sqrt{3}\) times the actual turns-ratio reciprocal.
4. 449.50 V: Would require a factor of about \(0.2997\), noticeably lower than the \(0.4330\) that the actual turns ratio and star connection produce together.

Only the combined delta-star line factor applied to the supply voltage gives a value matching one of the options.

Therefore, the correct answer is 649.50 V.