Question

If $N_s$ is the synchronous speed, $N$ is the rotor speed and 's' is the slip then relation is

• A. $N_s = (1 - s)N$
• B. $N = s N_s$
• C. $N = (s - 1)N_s$
• D. $N = N_s(1 - s)$

Hint:

At standstill (when the motor is just starting), $N = 0$, so slip $s = 1$. At synchronous speed (theoretical), $N = N_s$, so slip $s = 0$. In normal operation, slip is usually between 0.01 and 0.05.

Answer 1

The Correct Option is D

The concept of slip is fundamental to the operation of an induction motor, as it represents the difference between the speed of the magnetic field and the physical speed of the rotor. 1. Defining Synchronous Speed ($N_s$): The synchronous speed is the speed at which the stator's magnetic field rotates. It is determined by the supply frequency ($f$) and the number of poles ($P$) as $N_s = \frac{120f}{P}$.

2. Defining Slip ($s$): The rotor speed ($N$) is always slightly less than the synchronous speed in an induction motor. The relative difference between these two speeds, expressed as a fraction of the synchronous speed, is called slip: $$s = \frac{N_s - N}{N_s}$$

3. Deriving the Relation for Rotor Speed ($N$): By rearranging the slip formula, we can express the rotor speed in terms of synchronous speed and slip: $$s \cdot N_s = N_s - N$$ $$N = N_s - s \cdot N_s$$ $$N = N_s(1 - s)$$ This equation shows that the rotor speed is a fraction of the synchronous speed, specifically $(1 - s)$ times the synchronous speed. For example, if the slip is 5% (0.05), the rotor rotates at 95% of the synchronous speed.