--- title: "CT secondary open; auto transformer economy" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:ct-open-autotransformer-economical" source: https://www.bctengineer.com/q/semester_4/electrical_machines/ct-open-autotransformer-economical site: BCT Engineering --- # CT secondary open; auto transformer economy Electrical Machines · Chapter 2: Transformer - Asked in: 2079 Jestha, 2+6 marks - Video: [auto transformer in hindi | auto transformer working | auto transformer construction | animation](https://www.youtube.com/watch?v=wQ6QS7_jO4Y) — Mind of Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/ct-open-autotransformer-economical - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question 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. ## 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. ## Answer #### 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 $N_2I_2$ nearly cancels $N_1I_1$, leaving only a small exciting mmf. If the secondary is open, $I_2 = 0$ 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 $N_1$ turns (terminals A–C), the secondary $N_2$ turns (taps B–C), and take $N_2 < N_1$. Define $K = \dfrac{N_2}{N_1} = \dfrac{V_2}{V_1} = \dfrac{I_1}{I_2}$. ```text 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 $N_1$ at $I_1$, secondary $N_2$ at $I_2$): $$ W_{2w} \propto N_1I_1 + N_2I_2 = 2N_1I_1 $$ **Autotransformer:** $$ \begin{aligned} W_{auto} &\propto (N_1 - N_2)I_1 + N_2(I_2 - I_1) \\ &= N_1I_1(1 - K) + N_1I_1(1 - K)\quad(\text{using } N_2I_2 = N_1I_1,\ N_2I_1 = KN_1I_1) \\ &= 2N_1I_1(1 - K) \end{aligned} $$ $$ \frac{W_{auto}}{W_{2w}} = 1 - K,\qquad \text{saving} = W_{2w} - W_{auto} = K\,W_{2w} $$ The saving in copper is the fraction $K$ of the copper of a two-winding transformer. When $K$ is close to 1 (for example 0.9), the autotransformer needs only $(1 - K)$ = 10% of the copper, a saving of 90%. When $K$ is small (say 0.2), the saving is only 20%, so it is not economical. Also, only a small part of the power, $(1 - K)$, 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What happens to the core flux when a CT secondary is left open? A. I2 = 0 so the full primary mmf magnetises the core and it saturates B. The flux falls to zero C. The flux stays small because the primary current drops D. Only the leakage flux changes Answer: A. No opposing secondary mmf remains, so the flux rises far into saturation. (Key point: Open CT secondary) 2. In the autotransformer analysis, what current flows in the common section BC of N2 turns? A. I1 + I2 B. I2 - I1 C. I1 D. I2 Answer: B. The load current I2 and input current I1 partly share the common winding, so it carries the difference I2 - I1. (Key point: Copper in autotransformer) 3. What is W_auto / W_2w for an autotransformer with ratio K? A. K B. 1 + K C. 1 - K D. K^2 Answer: C. W_auto = 2N1I1(1-K) and W_2w = 2N1I1, so the ratio is 1 - K. (Key point: Copper ratio) 4. For K = 0.9, how much copper does the autotransformer save compared with a two-winding transformer? A. 10% B. 19% C. 81% D. 90% Answer: D. Saving = K times the two-winding copper, i.e. 90%. (Key point: Economy near unity) ## Related in Electrical Machines - [CT secondary should not be kept open](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Act-secondary-open-true-false) — Electrical Machines, chapter 2 - [Copper saving in auto transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aautotransformer-copper-saving) — Electrical Machines I, chapter 2 - [Auto transformer merits and Cu saving](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aautotransformer-merits-copper-saving) — Electrical Machines, chapter 2 - [Copper saving in auto-transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aautotransformer-copper-saving) — Electrical Machines, chapter 2 ## Related in other subjects - [CT secondary open and PT secondary short](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Act-open-pt-short) — Switchgear and Protection, chapter 2 - [True/false: CT secondary kept open](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atf-false-ct-secondary-open) — Electrical Machine, chapter 2 - [Current transformer vs conventional; CT open](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Acurrent-transformer-secondary-open) — Electrical Machine, chapter 2