--- title: "Back emf and its role in DC motor" subject: "Electrical Machines I" chapter: 4 key: "semester_4/electrical_machines_i:back-emf-dc-motor-role" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/back-emf-dc-motor-role site: BCT Engineering --- # Back emf and its role in DC motor Electrical Machines I · Chapter 4: DC Motor - Asked in: 2078 Chaitra, 8 marks; 2077 Chaitra, 8 marks; 2073 Magh, 2+6 marks; 2072 Asoj, 8 marks; 2071 Bhadra, 4 marks; 2070 Bhadra, 4 marks (asked 6 times) - Video: [What is Back EMF & what is its significance | DC Motor | TheElectricalGuy](https://www.youtube.com/watch?v=JgnPSOUoVJI) — Gaurav J - TheElectricalGuy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/back-emf-dc-motor-role - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4 ## Question What is back emf in dc motor? Explain the roles of back emf in dc motor. ## Short answer Back emf opposes applied voltage, limiting armature current and enabling self-regulation with load changes. It converts electrical power to mechanical (Eb Ia) and sets speed; without it, starting current would be excessive. ## Answer **Back emf** is the emf induced in the armature conductors of a running DC motor. When the armature rotates in the magnetic field, its conductors cut the flux and an emf is induced (generator action). By Lenz's law this emf **opposes the applied voltage**, so it is called back emf or counter emf, $E_b$. $$ E_b = \frac{P \phi Z N}{60 A}, \qquad V = E_b + I_a R_a $$ ```text +----Ra----+ | Ia -> | V (Eb) opposes V | | +----------+ ``` So the armature current is $$ I_a = \frac{V - E_b}{R_a} $$ #### Roles of back emf 1. **Limits armature current:** $R_a$ is very small (e.g. 0.5 Ω). Without $E_b$, a 220 V motor would draw $220/0.5 = 440$ A. With $E_b = 210$ V, $I_a = (220 - 210)/0.5 = 20$ A only. Back emf acts like a self-adjusting resistance. 2. **Makes the motor self-regulating (adjusts current to load):** - Load increases → speed falls slightly → $E_b$ falls → $I_a$ rises → torque rises to meet the load. - Load decreases → speed rises → $E_b$ rises → $I_a$ falls → torque falls. The motor automatically draws just the current needed by the load. 3. **Makes energy conversion possible:** multiplying $V = E_b + I_a R_a$ by $I_a$: $$ V I_a = E_b I_a + I_a^2 R_a $$ $E_b I_a$ is the **gross mechanical power developed**. Without back emf there would be no conversion; all input would be wasted as heat. Maximum mechanical power occurs when $E_b = V/2$. 4. **Determines speed:** $N \propto \dfrac{E_b}{\phi} = \dfrac{V - I_a R_a}{\phi}$. This is the basis of speed control by flux or armature voltage. 5. **Need for a starter:** at starting $N = 0$ so $E_b = 0$, and current would be very large. That is why a starter resistance is put in series until $E_b$ builds up. | Condition | Speed | $E_b$ | $I_a$ | |---|---|---|---| | Starting | 0 | 0 | Very large (limited by starter) | | Light load | High | Close to $V$ | Small | | Full load | Slightly lower | Lower | Rated | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the back emf oppose the applied voltage? A. It is a motor winding fault B. Generator action in the rotating armature induces an emf that opposes V by Lenz's law C. It comes from the supply D. It is the residual flux Answer: B. Rotating conductors cut flux and induce emf that opposes the cause, i.e. the supply. (Key point: Meaning of back emf) 2. What is the armature current expression? A. I_a = V/R_a B. I_a = (V - E_b)/R_a C. I_a = (V + E_b)/R_a D. I_a = E_b/R_a Answer: B. V = E_b + I_a R_a. (Key point: Armature current) 3. What happens when the load increases? A. Speed rises, E_b rises, I_a falls B. Speed falls, E_b falls, I_a rises and torque rises C. Speed is unchanged D. E_b rises and I_a rises Answer: B. Lower E_b allows more current and torque to meet the load. (Key point: Self-regulation) 4. What does E_b I_a represent? A. Input power B. Armature copper loss C. Gross mechanical power developed D. Iron loss Answer: C. Multiplying V = E_b + I_a R_a by I_a shows E_b I_a is the mechanical power. 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