---
title: "DC shunt motor with open field circuit"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:shunt-motor-field-open"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-field-open
site: BCT Engineering
---

# DC shunt motor with open field circuit

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2078 Chaitra, 6 marks
- Video: [What will happen if field winding of DC shunt motor is open?](https://www.youtube.com/watch?v=2ngZLr-j0RE) — SKILL सीखो 
- 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-motor-field-open
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4

## Question

A dc shunt motor is operating at a constant speed supplied with rated voltage. Due to an accident, the field terminal of the motor gets disconnected. What will happen on the motor performance? Discuss on the implications.

## Short answer

Flux collapses to residual, back EMF drops nearly to zero, armature current surges to V/R_a. On light load the motor over-speeds dangerously (runaway); on heavy load it may stall while drawing high current. This risks mechanical damage from centrifugal forces, overheating, sparking, and potential flashover.

## Answer

If the field circuit of a running DC shunt motor opens, the flux falls suddenly to the small **residual flux**. Since $N \propto \dfrac{E_b}{\phi}$ and $T \propto \phi I_a$, both speed and current change drastically.

#### What happens
1. **Flux collapses:** $\phi$ drops to residual value, maybe 2-5% of normal.
2. **Back emf drops:** at that instant speed cannot change (inertia), so $E_b = k\phi N$ falls almost to zero.
3. **Armature current rises sharply:** $I_a = \dfrac{V - E_b}{R_a}$ approaches $V/R_a$, many times the rated value, like a start without a starter.
4. **Torque and speed:**
   - With a **light load**, the large $I_a$ times the small $\phi$ still gives enough torque, so the motor **accelerates rapidly**. To build enough back emf with tiny flux, speed must rise to a very high value. The motor **runs away** (over-speeds).
   - With a **heavy load**, the torque $k\phi_{res} I_a$ may be less than load torque. The motor slows or stalls while drawing a very large current.

#### Implications
- **Mechanical damage:** centrifugal forces at over-speed can throw out armature windings, damage the commutator and bearings, and burst the rotor.
- **Electrical damage:** very large armature current overheats the winding and insulation.
- **Heavy sparking and flashover** at the commutator due to large current and weak field (poor commutation).
- **Supply disturbance:** fuses/breakers trip; voltage dip affects other loads.
- **Safety hazard** to persons nearby and to the driven machine.
- **Inductive kick:** opening a highly inductive field can produce a high voltage arc at the break, damaging the field insulation.

#### Protection
- A **three-point starter** has its no-volt coil in series with the field, so a field break releases the handle and disconnects the motor.
- Field-failure relays, over-speed trips, and overload/over-current relays or circuit breakers.
- The field circuit should never contain a fuse or switch that can open it alone.

```text
 N, Ia
  |          ___ speed (runaway)
  |        _/
  |      _/   ___ armature current
  |    _/   _/
  |___/___/__________________ time
      ^ field opens
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. When the field of a running shunt motor opens, what happens to flux?
   A. It falls to the small residual value
   B. It doubles
   C. It reverses direction
   D. It stays unchanged
   Answer: A. Only residual magnetism remains. (Key point: Flux collapse)

2. Why does armature current surge?
   A. The supply voltage rises
   B. Eb falls almost to zero, so Ia approaches V/Ra
   C. Ra drops to zero
   D. The load torque becomes zero
   Answer: B. Speed cannot change instantly, so Eb = kφN drops with φ. (Key point: Armature current)

3. What happens at light load?
   A. The motor stops immediately
   B. Speed remains constant
   C. The motor accelerates to a dangerous speed (runaway)
   D. The motor reverses
   Answer: C. Eb must rebuild with a tiny flux, so speed must be very high. (Key point: Light load result)

4. Which protection responds to field failure?
   A. A fuse in the field circuit
   B. A switch in the field circuit
   C. A larger starting resistance
   D. A three-point starter's no-volt coil in series with the field releases the handle
   Answer: D. A field fuse or switch should never be used. (Key point: Protection)

## Related in Electrical Machines I

- [Shunt motor speed with 30 Ω field resistance added](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-220v-30ohm-field-torque) — Electrical Machines I, chapter 4, asked 5×
- [Shunt motor speed with 50 Ω field resistance (1200 rpm)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-250v-25a-half-load) — Electrical Machines I, chapter 4, asked 5×
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## Related in other subjects

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- [Shunt motor: field resistance to raise speed to 1100 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adcm-shunt-field-resistance-1100rpm) — Electrical Machines, chapter 4
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