ELECTENG 101 Chap.8 DC Motors and Electromechanical Response
DC Motors and Electromechanical Response
A permanent-magnet DC motor links armature voltage and current to torque and speed. Rotation generates back emf that opposes the supply. Load changes alter speed, back emf, current and developed torque as a causal loop. Power calculations distinguish electrical input, copper loss, converted power and mechanical shaft output. At standstill the back emf is nearly zero, so current can be large if armature resistance is small.
As speed rises, generated voltage reduces the current permitted by the armature balance. A greater mechanical load initially slows the rotor, lowering back emf and allowing current and electromagnetic torque to increase until a new operating point is reached. Torque and back-emf constants connect the electrical and mechanical domains, while shaft torque can be lower than developed torque because mechanical losses intervene.
What this chapter covers
- 01
Magnetic force and torque
- 02
Commutation
- 03
Armature equivalent circuit
- 04
Back emf and speed
- 05
Torque and current
- 06
Mechanical output power
- 07
Motor efficiency
Worked example · free
Reconcile a motor operating point
- +2Back emf is 24 − 3×1 = 21 V.
- +3Angular speed is 125.66 rad/s and shaft power is 15.08 W.
- +2Input is 72 W, so efficiency is 20.9%.
Key terms
- Back emf
- The rotation-generated voltage opposing the applied armature voltage.
- Commutator
- The switching structure that reverses conductor current to maintain one torque direction.
- Shaft power
- Useful mechanical output calculated as torque multiplied by angular speed.
DC Motors and Electromechanical Response FAQ
Why is motor starting current high?
At standstill back emf is near zero, so armature current is limited mainly by the comparatively small armature resistance. That current can exceed the safe continuous value, which is why starting methods, supply limits or control electronics matter. As the rotor accelerates, rising back emf reduces current under the same applied voltage.
What happens when mechanical load increases?
Speed initially falls, reducing back emf. Current and developed torque then rise until a new operating point is reached. The response is a causal sequence rather than an instantaneous claim that load directly changes current. If the required torque exceeds the motor and supply capability, no stable speed may be reached and the motor can stall.
How do I check motor efficiency?
Use useful shaft output divided by electrical input with consistent watt units. A value above 100% indicates a missing loss or unit error. Convert rotational speed to radians per second before multiplying by shaft torque, and distinguish shaft output from internally converted mechanical power when copper and rotational losses are given separately.
What does back emf do in the armature equation?
Back emf is generated by rotation and opposes the applied supply under normal motoring operation. The balance is applied voltage equals back emf plus armature resistance drop. It is not an optional loss term: it links speed to current and explains why a running motor draws less current than the same motor at standstill.
How are current and torque related?
For a permanent-magnet DC motor in the linear model, developed electromagnetic torque is proportional to armature current through the torque constant. Shaft torque may be smaller because friction and other mechanical losses consume part of the developed torque. State which torque is used when calculating mechanical output power.
Why convert revolutions per minute before finding power?
The rotational power relation uses angular speed in radians per second, so revolutions per minute must be multiplied by two pi and divided by sixty. Using rpm directly with newton metres produces a number with the wrong scale. Writing the unit conversion explicitly makes the common factor error easy to detect.
Exam move
Draw the armature voltage balance and the energy-flow path separately. Use the voltage equation to find current or back emf, then connect current to developed torque and speed to mechanical power. Check that input exceeds output and account for copper loss. Practise verbal load changes so the causal loop becomes automatic. Build a four-state comparison for standstill, no-load running, rated load and stall risk.
At each state, rank speed, back emf, armature current and developed torque without assigning numbers. Then solve numerical cases in a fixed order: back emf, current, torque, angular speed, power and efficiency. Maintain separate rows for electrical input, copper loss, converted power, mechanical loss and shaft output so no quantity is counted twice.
Finish with a perturbation exercise in which load torque rises, supply voltage falls or armature resistance changes; narrate the first effect and each subsequent feedback step before calculating the new equilibrium.
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