In one line
Back EMF is the voltage generated in the rotating armature of a DC motor that opposes the supply voltage; it automatically controls armature current according to the load.
Explained simply
A motor's armature is spinning inside a magnetic field, so it is also acting as a generator at the same time. The voltage it generates pushes back against the supply, hence the name. The faster it spins, the bigger the push-back, and the less current it lets in. This is the motor's built-in "cruise control".
Technically, by Faraday's law, armature conductors cutting the field induce \( E_b = \dfrac{P\phi NZ}{60A} \), whose direction (by Lenz's law) opposes the applied voltage. The net voltage driving armature current is \( V - E_b \).
How it works
- At start, \( N = 0 \) so \( E_b = 0 \), and \( I_a = V/R_a \) is very large because \( R_a \) is small. That is why a starter (series resistance) is used.
- As speed rises, \( E_b \) rises and \( I_a = (V - E_b)/R_a \) falls.
- When load increases, the motor slows a little; \( E_b \) falls; \( I_a \) rises; torque rises to meet the load. When load decreases, the reverse happens. The motor self-regulates.
- Power: multiplying \( V = E_b + I_aR_a \) by \( I_a \) gives \( VI_a = E_bI_a + I_a^2R_a \). \( E_bI_a \) is the power converted to mechanical form.
- Speed relation: \( N \propto E_b/\phi \).
Key formula
\( E_b = \dfrac{P\phi NZ}{60A}, \quad I_a = \dfrac{V - E_b}{R_a}, \quad P_m = E_bI_a \)
\( E_b \) = back EMF (V), \( P \) = number of poles, \( \phi \) = flux per pole (Wb), \( N \) = speed (rpm), \( Z \) = total armature conductors, \( A \) = number of parallel paths (2 for wave, \( P \) for lap winding), \( P_m \) = gross mechanical power developed (W).
Quick example
If a running permanent-magnet DC motor (for example a toy or cordless-drill motor) is stalled by hand, its armature current shoots up because back EMF has collapsed; if held for long, the winding overheats. This is why motor protection and starters are needed.
Common mistakes
- Saying back EMF is maximum at starting; it is zero at starting.
- Saying back EMF is a loss; \( E_bI_a \) is the useful converted power.
- Writing \( V = E_b - I_aR_a \); for a motor \( V = E_b + I_aR_a \) (the generator equation is \( E_g = V + I_aR_a \)).
How MU asks it
Numericals are not expected in this module (as per the MU syllabus), so questions are theory. Typical: "What is back EMF? Explain its significance" (3–5 marks), or "Why is a starter needed for a DC motor?". For full marks define back EMF, give its formula, write \( V = E_b + I_aR_a \), explain self-regulation with load, and link zero back EMF at start to the need for a starter.