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CT secondary open; auto transformer economy

Short answer

Opening a CT secondary removes opposing mmf, saturates the core, and produces dangerously high voltages. An autotransformer is economical when the ratio K is near unity because copper saving equals K times that of a two-winding transformer (e.g., 90% saving at K=0.9). Applications include motor starters and voltage regulators.

Why secondary of CT should not be left open? Show that auto transformer is economical when the transformation ratio is very close to unity mentioning its application.

Why the secondary of a CT must not be left open

The primary of a CT carries the line current, which is independent of the secondary. With the secondary closed, the mmf N2I2N_2I_2 nearly cancels N1I1N_1I_1, leaving only a small exciting mmf. If the secondary is open, I2=0I_2 = 0 and the whole primary mmf magnetises the core. The flux rises far into saturation, causing very high induced voltage in the many-turn secondary (insulation breakdown and shock hazard), excessive core heating, and residual magnetism that destroys accuracy.

Autotransformer is economical when the ratio is near unity

An autotransformer has a single winding, part of which is common to the primary and the secondary. Let the primary have N1N_1 turns (terminals A–C), the secondary N2N_2 turns (taps B–C), and take N2<N1N_2 < N_1. Define K=N2N1=V2V1=I1I2K = \dfrac{N_2}{N_1} = \dfrac{V_2}{V_1} = \dfrac{I_1}{I_2}.

      I1
 A o-->------+
             |   section AB: (N1-N2) turns, current I1
 B o---+-----+
       |  I2-I1  section BC: N2 turns, current (I2 - I1)
 C o---+

The amount of copper is proportional to (turns × current) of each winding portion.

Two-winding transformer (primary N1N_1 at I1I_1, secondary N2N_2 at I2I_2):

W2w∝N1I1+N2I2=2N1I1W_{2w} \propto N_1I_1 + N_2I_2 = 2N_1I_1

Autotransformer:

Wauto∝(N1−N2)I1+N2(I2−I1)=N1I1(1−K)+N1I1(1−K)(using N2I2=N1I1, N2I1=KN1I1)=2N1I1(1−K)\begin{aligned} W_{auto} &\propto (N_1 - N_2)I_1 + N_2(I_2 - I_1) \\ &= N_1I_1(1 - K) + N_1I_1(1 - K)\quad(\text{using } N_2I_2 = N_1I_1,\ N_2I_1 = KN_1I_1) \\ &= 2N_1I_1(1 - K) \end{aligned} WautoW2w=1−K,saving=W2w−Wauto=K W2w\frac{W_{auto}}{W_{2w}} = 1 - K,\qquad \text{saving} = W_{2w} - W_{auto} = K\,W_{2w}

The saving in copper is the fraction KK of the copper of a two-winding transformer. When KK is close to 1 (for example 0.9), the autotransformer needs only (1−K)(1 - K) = 10% of the copper, a saving of 90%. When KK is small (say 0.2), the saving is only 20%, so it is not economical. Also, only a small part of the power, (1−K)(1 - K), is transferred by induction, while the rest is passed directly by conduction, so the core, losses and cost fall in the same way.

Applications: starting of induction motors (autotransformer starter), voltage regulators (variacs), boosting or reducing line voltage by 10 to 20%, and interconnecting 220 kV and 132 kV systems.

✓ Answer checked by Claude Opus 5.5 (high effort)

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