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Back emf in DC motor and its role

Short answer

Back emf is the voltage induced in the rotating armature opposing the supply, E_b = KΦN. It limits starting current, enables power conversion (E_b I_a), and self-regulates: when load increases, speed falls, E_b drops, armature current rises, developing more torque to match the load.

What is back emf in a DC motor? Explain the significance of back emf and how it helps the motor develop the required torque according to the load on the shaft.

Back emf

When the armature of a DC motor rotates in the magnetic field, its conductors cut the flux, and by Faraday's law an emf is induced in them. By Lenz's law this emf opposes the cause producing it, which is the supply voltage. It is called back emf (counter emf).

E_b = PΦZN / (60A) = KΦN

The motor equation is: V = E_b + I_aR_a, so I_a = (V − E_b) / R_a

Significance

  1. Acts as a governor: E_b depends on speed, so it automatically regulates the armature current drawn.
  2. Limits the armature current: R_a is small, and at start (E_b = 0) the current V/R_a is very large. As the motor speeds up, E_b rises and current falls to the required value.
  3. Power conversion: V I_a = E_b I_a + I_a²R_a. The term E_b I_a is the gross mechanical power developed. If E_b were zero, there would be no conversion of electrical to mechanical energy.
  4. Gives the motor its self-regulating nature.

How it matches the torque to the load

 Load torque increases
   -> speed N falls
   -> E_b = KΦN falls
   -> I_a = (V - E_b)/Ra rises
   -> torque T = KΦ Ia rises
   -> torque equals load torque; speed becomes steady
  • With a lighter load, the reverse happens: speed rises, E_b rises, I_a falls, and torque falls to match the smaller load.
  • Hence a DC motor draws only the current that the load demands, because the back emf adjusts the armature current.

✓ Answer checked by Claude Opus 5.5 (high effort)

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