In an induction motor for a value slip(s) = 0, the torque (T) is
zero
The question asks about the torque produced by an induction motor when the slip ($s$) is equal to zero. Let's break down what slip means and how it affects the torque.
Slip ($s$) in an induction motor is a measure of the difference between the synchronous speed ($N_s$) of the rotating magnetic field and the actual rotor speed ($N_r$). It is usually expressed as a fraction or percentage:
$$s = \frac{N_s - N_r}{N_s}$$
In a working induction motor, the rotor must always rotate slower than the synchronous speed ($N_r < N_s$) to produce torque. This difference in speed is crucial for inducing voltage and current in the rotor.
Torque in an induction motor is generated due to the interaction between the rotor's magnetic field and the stator's rotating magnetic field. The process is as follows:
When the slip $s = 0$, it implies:
$$0 = \frac{N_s - N_r}{N_s}$$
This means $N_s - N_r = 0$, or $N_s = N_r$. This is the condition where the rotor speed is exactly equal to the synchronous speed.
At synchronous speed ($N_r = N_s$), there is no relative motion between the stator's rotating magnetic field and the rotor conductors. Because there is no relative motion:
Therefore, when the slip $s = 0$, the torque ($T$) developed by the induction motor is zero.
| Condition | Rotor Speed ($N_r$) | Relative Speed | Induced Rotor Voltage | Rotor Current | Torque ($T$) |
|---|---|---|---|---|---|
| Slip $s=0$ | $N_r = N_s$ | 0 | 0 | 0 | Zero |
| Slip $s=1$ (Standstill) | $N_r = 0$ | $N_s$ | Maximum (approx.) | Maximum (approx.) | Starting Torque |
| Slip $0 < s < 1$ | $0 < N_r < N_s$ | $N_s - N_r$ | Induced | Induced | Produces Torque |
The condition $s = 0$ represents the motor running at its synchronous speed, a theoretical state where no torque is generated because the fundamental requirement for induction (relative motion) is absent.
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