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T-08: Scott (T-T) Connection

Section: A | Priority: 🟡 MEDIUM | Exam Frequency: 3/7 years Sources: Theraja Ch-33 (Art. 33.10), VK Mehta Ch-7 (Art. 7.37), Slides L-11 S20–S24

Why This Topic Matters

The Scott connection appeared in 3/7 papers (2018, 2019, 2024). When asked, it carries 4-6 marks. The question is always: "Explain the Scott connection with a diagram" and sometimes includes a numerical.

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📝 Key Definitions

Scott (T-T) connection: "The Scott connection is a means of converting a 3-phase supply to a 2-phase supply, or vice versa, using two single-phase transformers. One transformer is called the 'main' (or 'T') transformer and the other is called the 'teaser' transformer. The teaser primary has $\sqrt{3}/2$ (= 86.6%) of the main transformer primary turns." — VK Mehta, Art. 7.37

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How the Scott Connection Works

Main transformer: Primary connected between phases A and B (full line voltage $V_{AB}$). Secondary provides one phase of the 2-phase output.

Teaser transformer: Primary connected from the midpoint (M) of the main transformer primary to phase C. The voltage from M to C:

$$V_{MC}=\frac{\sqrt{3}}{2}\times V_L$$

This is the height of the equilateral voltage triangle. The teaser primary needs $\frac{\sqrt{3}}{2}=86.6\%$ of the main transformer turns.

Why the outputs are 90° apart:

$V_{AB}$ is a line voltage. $V_{MC}$ is perpendicular to $V_{AB}$ in the phasor diagram (midpoint M bisects AB, so MC is the perpendicular bisector). The two secondary voltages inherit this 90° separation. Result: balanced 2-phase output.

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🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Explain the Scott connection with necessary diagrams.

Appeared: 2018 Q4(a), 2024 Q4(d) — 4 marks

Full Answer:

The Scott (T-T) connection converts 3-phase power to 2-phase power using two single-phase transformers.

Two transformers required:

(1) Main transformer: Standard transformer. Primary connected between two phases (e.g., A and B). Full line voltage $V_{AB}$ appears across its primary.

(2) Teaser transformer: Primary connected from the midpoint (M) of the main transformer primary to the third phase (C). Its primary has $\sqrt{3}/2=86.6\%$ of the main transformer's turns, because the voltage from midpoint M to C is $V_{MC}=(\sqrt{3}/2)V_L$ (the height of the equilateral voltage triangle).

Why 90° separation: In a balanced 3-phase system, the voltage from the midpoint of one line voltage to the third phase is always perpendicular to that line voltage. $V_{AB}$ is horizontal in the phasor diagram, $V_{MC}$ is vertical. The two secondaries inherit this 90° separation, producing a balanced 2-phase output.

Applications: Electric arc furnace power supplies, powering 2-phase induction motors, converting 2-phase to 3-phase (reverse operation is possible because transformers are reciprocal devices).

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🎯 Q2: Can you convert 3-phase to 2-phase or vice versa? If yes, explain.

Appeared: 2019 Q4(b) — 4 marks

Full Answer:

Yes. This is done using the Scott (T-T) connection.

Two single-phase transformers are needed:

(1) Main transformer: Primary connected between two phases of the 3-phase supply (e.g., lines A and B). Secondary provides one phase of the 2-phase output.

(2) Teaser transformer: Primary connected from the mid-point of the main transformer primary to the third line (C). The teaser primary has $\sqrt{3}/2$ (86.6%) of the main transformer's turns. Secondary provides the second phase of the 2-phase output, exactly 90° displaced from the first.

Why it works: The two primary voltages are 90° apart geometrically in the phasor diagram. The line-to-midpoint voltage ($V_{MC}$) is perpendicular to the line-to-line voltage ($V_{AB}$) in a balanced 3-phase system. This 90° separation transfers to the two secondary voltages, giving a balanced 2-phase output.

Reverse (2-phase to 3-phase): Connect the two-phase supply to the secondaries. The 3-phase supply comes from the primaries. The same transformation works in reverse because transformers are reciprocal devices.

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🎯 Q3: Two T-connected transformers supply 440V, 33 kVA balanced load from 3300V 3-phase. Find ratings.

Appeared: 2018 Q4(b) — 4 marks

Full Answer:

Given: Supply $V_L=3300$ V (3-phase). Load: 440V, 33 kVA (2-phase).

Secondary voltages (2-phase, equal): $V_2=440$ V per phase.

Secondary current per phase: $I_2=(S/2)/V_2=(33000/2)/440=\boxed{37.5\text{ A}}$

Main transformer primary: Connected across A-B: $V_{1,\text{main}}=V_L=3300$ V

$$I_{1,\text{main}}=\frac{S/2}{V_{1,\text{main}}}=\frac{16500}{3300}=\boxed{5.0\text{ A}}$$

Teaser transformer primary: Connected from midpoint of AB to C:

$$V_{1,\text{teaser}}=\frac{\sqrt{3}}{2}\times V_L=0.866\times 3300=\boxed{2858\text{ V}}$$

$$I_{1,\text{teaser}}=\frac{S/2}{V_{1,\text{teaser}}}=\frac{16500}{2858}=\boxed{5.77\text{ A}}$$

kVA ratings:

$$\text{Main}=3300\times 5.0=\boxed{16.5\text{ kVA}}$$

$$\text{Teaser}=2858\times 5.77=\boxed{16.5\text{ kVA}}$$

Both transformers have the same kVA rating. Total = 33 kVA ✓

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⚡ Exam Tips & Common Mistakes

1. Teaser has $\sqrt{3}/2$ = 86.6% turns, not 50%. This is the most common error.
2. Both transformers have the same kVA rating. The teaser has fewer turns but higher current.
3. The 90° comes from geometry, not from any special winding.

🔗 Related Topics

• T-07a: 3-Phase Connections — Standard 3-phase configurations
• T-10: Auto-Transformer — Another special transformer type

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