--- title: "Starting shunt generator on no load and voltage build-up" subject: "Electrical Machines I" chapter: 3 key: "semester_4/electrical_machines_i:shunt-generator-no-load-buildup" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-generator-no-load-buildup site: BCT Engineering --- # Starting shunt generator on no load and voltage build-up Electrical Machines I · Chapter 3: DC Generator - Asked in: 2082 Kartik (new course), 2+4 marks - Video: [VOLTAGE BUILD UP PROCESS IN DC SHUNT GENERATOR(OCC)-(LECTURE-6)-ELECTRICAL MACHINE](https://www.youtube.com/watch?v=paAFKmfuXjw) — BEING ELECTRIFY - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-generator-no-load-buildup - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3 ## Question Why DC shunt generator shall not be started at load? Explain the voltage build-up process of a DC shunt generator. ## Short answer On load at start, residual EMF sends current to the load instead of the field, preventing voltage rise. Build-up: residual flux induces small EMF; field current (if aiding) strengthens flux, cumulative rise occurs until the field resistance line intersects the OCC. Conditions: residual magnetism, aiding field mmf, field resistance below critical. ## Answer #### Why a shunt generator is not started on load A shunt generator builds up from **residual magnetism**, and its field current is taken from its own terminal voltage ($I_f = V/R_f$). - If a load (low resistance) is connected at starting, the small residual EMF drives most of its current through the load, and the armature drop $I_aR_a$ becomes large compared with the tiny EMF. - The terminal voltage applied to the field therefore stays very low, so the field current, and hence the flux, cannot grow. - In terms of the characteristic: with the load connected, the field gets only part of the armature current, and the effective resistance line becomes $E = I_f\left[R_f + R_a\left(1 + \dfrac{R_f}{R_L}\right)\right]$. For a low load resistance $R_L$ (below the critical load resistance) this line is steeper than the critical-resistance line, so it cuts the OCC only near the residual voltage and the generator **fails to build up**. Therefore the load switch is kept open until the generator reaches rated voltage, and only then is the load connected. #### Voltage build-up process ```text +---------+------o/ o----+ | | switch | ( G ) [Rsh] (open) load | | | +---------+--------------+ ``` 1. **Residual flux:** the field poles keep a small residual magnetism. Driving the armature at rated speed induces a small EMF $E_r$ (2-5% of rated). 2. **Field current starts:** $E_r$ drives a small current $I_{f1} = E_r/R_f$ through the shunt field. 3. **Flux increases:** if the field winding is connected so that its MMF aids residual flux, flux rises and the EMF rises to $E_1$ (read from the OCC at $I_{f1}$). 4. **Cumulative rise:** $E_1$ drives a larger $I_{f2}$, giving a higher $E_2$, and so on, step by step. 5. **Stable point:** the process stops where the **OCC** meets the **field resistance line** $V = I_fR_f$. Iron saturation bends the OCC, so the intersection is definite. ```text E ^ Rf line | / | / _____-- OCC | /.-' | /' <- final voltage | .'/ | .' / | .' / | .' / Er +-----/----------------> If ``` **Conditions for build-up:** residual magnetism present, field MMF aiding residual flux, field resistance below critical resistance, speed above critical speed, and no (or light) load at starting. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why should a shunt generator not be started with load connected? A. Low-resistance load takes the small EMF's current, so terminal voltage and field current cannot grow B. The load would reverse residual flux C. Field winding would burn D. Speed would fall to zero Answer: A. The tiny EMF is dropped across Ia Ra with the load, leaving too little voltage to raise the field current, so it fails to build up. (Key point: Why not start on load) 2. A shunt generator's field current is obtained from: A. Its own terminal voltage (If = V/Rf) B. A battery C. The series winding D. Interpoles Answer: A. The field is across the armature, so If = V/Rf, which relies on the terminal voltage rising. (Key point: Field current source) 3. During build-up the voltage rises step by step because: A. Higher EMF gives more field current, which gives more flux and more EMF B. Speed increases C. Load decreases D. Residual flux increases itself Answer: A. This cumulative process continues until the OCC meets the field resistance line. (Key point: Cumulative rise) 4. For build-up, the field MMF must: A. Aid the residual flux B. Oppose the residual flux C. Be zero D. Be at 90 degrees to it Answer: A. If it opposes, residual flux is wiped out and the generator cannot build up. 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