--- title: "Shunt generator started with output open" subject: "Electrical Machine" chapter: 3 key: "semester_4/electrical_machine:dc-shunt-generator-start-open-circuit" source: https://www.bctengineer.com/q/semester_4/electrical_machine/dc-shunt-generator-start-open-circuit site: BCT Engineering --- # Shunt generator started with output open Electrical Machine · Chapter 3: DC Generator - Asked in: 2070 Bhadra, 4 marks - Video: [BUILD UP VOLTAGE IN A DC SHUNT GENERATOR](https://www.youtube.com/watch?v=XsPwA6noaMA) — St.Thomas ITI Kanhirapuzha - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machine/dc-shunt-generator-start-open-circuit - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-3 ## Question DC shunt generator shall be started keeping its output terminal open. Justify the statement. ## Short answer A shunt generator builds up voltage from residual magnetism. If loaded at start, the small initial emf supplies load current instead of field current, preventing flux growth. Starting with open terminals ensures all initial current flows into the shunt field, strengthening the flux until rated voltage is reached, after which load can be applied. ## Answer A shunt generator is self-excited: its field winding is fed from its own armature voltage. It must therefore build up voltage by itself, and this is possible only when the machine is not loaded at the start. #### Voltage build-up process ```text residual flux -> small emf (E_r) -> small field current If ^ | +-- more flux, more emf <--------------+ ``` 1. The poles keep a small **residual magnetism**, which gives a small emf E_r when the armature rotates. 2. This emf drives a small current I_f through the shunt field (R_sh), which strengthens the flux (if the field is connected so that it aids the residual flux). 3. More flux gives more emf and more field current, and so on, until the field resistance line meets the magnetisation curve (OCC). That point is the final no-load voltage. #### Why the output terminals must be open 1. **Build-up needs the field current to grow.** The emf from residual flux is only a few volts. If a load is connected, most of the small armature current flows into the load instead of the field, and the I_aR_a drop and armature reaction lower the terminal voltage V = E − I_aR_a further. The field current (V/R_sh) stays too small for the flux to grow, so the voltage may never build up. (Build-up on load is possible only if the load resistance is higher than the critical load resistance.) 2. A heavy load would draw current from a very small initial emf, and the armature reaction of that current weakens the small residual flux, so the voltage fails to rise. 3. Starting on no load keeps the prime mover load light, and avoids a sudden overload on the armature and prime mover. 4. After the voltage is built up to the rated value, the load can be applied gradually. Hence the output terminals are kept open (no load) while starting, and the load is switched on only after the voltage has built up. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What starts the voltage build-up in a shunt generator? A. The load current B. An external battery connected permanently C. The series field D. Residual magnetism in the poles giving a small emf Answer: D. Residual flux gives small E_r, which drives a small field current that strengthens the flux. (Key point: Voltage build-up source) 2. At what point does the no-load voltage settle? A. Where the field resistance line meets the magnetisation curve (OCC) B. Where the OCC crosses the zero axis C. At the highest point of the OCC D. Where the load line meets the field line Answer: A. Build-up continues until the field resistance line intersects the OCC. (Key point: Final voltage) 3. Why must a shunt generator be started on open circuit? A. Load would reverse the polarity of the poles at once B. With load the terminal voltage drops, field current becomes small and voltage may never build up C. Open circuit increases the field resistance D. The shunt field needs the load current to excite Answer: B. Load current causes drop I_aR_a so V and field current V/R_sh stay small, preventing build-up. 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