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: Volts per turn on the primary (delta phase voltage \(1500\) V across \(600\) turns): \( \dfrac{1500}{600} = 2.5 \) V/turn.
Step 2: Secondary phase voltage (star winding, \(150\) turns) at the same volts-per-turn: \( 2.5 \times 150 = 375 \) V.
Step 3: Convert to secondary line voltage for a star connection: \( \sqrt{3} \times 375 \).
\[ V_{\text{line,sec}} = 1.732 \times 375 \approx \boxed{649.50 \text{ V}} \]

Answer 2

A third check works backward from each candidate secondary line voltage to see which one is consistent with a star-connected secondary phase voltage that in turn matches the correct turns ratio from the delta primary.

1. 649.50 V: Dividing by \(\sqrt{3}\) gives a secondary phase voltage of \(375\) V. Multiplying this by the turns ratio \(N_1/N_2 = 4\) gives back \(1500\) V, exactly the given primary supply voltage, so this option is fully consistent.
2. 549.50 V: Dividing by \(\sqrt{3}\) gives a secondary phase voltage of about \(317.3\) V; scaling up by the turns ratio of \(4\) gives about \(1269\) V, which does not match the \(1500\) V supply.
3. 595.50 V: Dividing by \(\sqrt{3}\) gives about \(343.9\) V; scaling by \(4\) gives about \(1375.6\) V, again not matching the \(1500\) V supply.
4. 449.50 V: Dividing by \(\sqrt{3}\) gives about \(259.6\) V; scaling by \(4\) gives about \(1038.2\) V, well short of the \(1500\) V actually supplied.

Only working back from \(649.50\) V reconstructs the actual \(1500\) V primary supply through the correct turns ratio and star geometry.

Therefore, the correct answer is 649.50 V.