Shunt generator that stalls at 12 volts
A 28 V, 100 A shunt-wound DC generator on a test bench, driven at its rated 3000 rpm, fails to build up. With the field disconnected the armature produces 1.8 V. The field winding measures 12 Ω at the slip rings, matching the nameplate, and insulation resistance is good. With the field reconnected through its own circuit the terminal voltage climbs to 12 V and then stops rising. A clamp meter on the field circuit reads 0.75 A at that stable point. Armature circuit resistance is 0.4 Ω. The technician's first proposal is to flash the field from a 24 V battery. The output load breaker is confirmed open.
1. Use the field current reading to identify the fault, and state whether the winding itself is serviceable.
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At the stable point, terminal voltage is 12 V and the field is fed from those terminals.
- Expected field current if only the 12 Ω winding were in circuit: 12/12 = 1.0 A
- Measured: 0.75 A → implied total field-circuit resistance = 12/0.75 = 16 Ω
- Therefore there is about 4 Ω of unwanted series resistance somewhere in the field path.
The winding measured 12 Ω at the slip rings, so the copper is sound; the 4 Ω appears only when the circuit is completed. Look for: brush seating and brush pressure, brush-box to terminal joints, a corroded or loose field terminal, a field rheostat or resistor that should be shorted out but is not, and crimped connections. A single bad joint of this size is enough to explain the symptom.
Why the machine stops at 12 V: build-up continues only while the field current produces more flux than the current operating point needs. It settles where the field-resistance line (16 Ω) crosses the magnetisation curve for 3000 rpm. Raising the field-circuit resistance lowers that intersection - the machine is sitting at a self-excited equilibrium well below rated output, not simply 'weak magnets'.
2. Evaluate the technician's proposal to flash the field.
Read the worked answer
Flashing is the wrong action here. Flashing restores or reverses residual magnetism; it does not reduce field-circuit resistance.
Evidence that residual magnetism is present and correct:
- 1.8 V is generated with the field open - that is residual voltage, which cannot exist without residual flux.
- The machine then builds up from 1.8 V to 12 V, i.e. the field is reinforcing the residual flux, not opposing it (a reversed field would drive the voltage down towards zero when connected).
Flashing now would at best be a no-op and at worst reverse the residual polarity, converting a simple high-resistance joint into a machine that will not build up at all. It also risks a dangerous spark and, if the machine were connected to a live bus, back-feeding the battery into the bus.
Flashing is justified only when: (a) field continuity, brush contact and polarity have all been proved; (b) speed is confirmed at rated value; (c) the field circuit resistance is confirmed below the critical value; and (d) there is essentially zero residual voltage. Then flash with the machine isolated, field disconnected, using the correct polarity, and re-check residual voltage before reconnecting.
3. Quantify what would happen if the output load breaker had been left closed with a 5 Ω load during the build-up.
Read the worked answer
Take the observed operating point: E ≈ 12 V + (0.75 A × 0.4 Ω) = 12.3 V at 0.75 A field.
With a 5 Ω load connected across the terminals:
- At 12 V terminal, load current = 12/5 = 2.4 A
- Armature current = 2.4 + 0.75 ≈ 3.15 A
- Internal drop = 3.15 × 0.4 = 1.26 V
- Terminal voltage would then be E − 1.26 = 11.04 V, not 12 V.
But a lower terminal voltage means a lower field current (11.04/16 = 0.69 A), which means less flux, which means lower E, which means lower terminal voltage again. The operating point collapses: the machine either settles at a very low voltage or fails to build up at all.
Judgement point: the load must be open during build-up and the field circuit must be the only significant current at that stage. This is also why a shunt generator with a load left on the bus appears 'dead' and is often wrongly diagnosed as a loss of residual magnetism - check the breaker before reaching for the battery.