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Slip of induction motor is obtained by

(Ns-N)/Ns

Ns-N

sNs

Correct Answer

⚙️ TE • Technical Concept & PrincipleElectricalMachine

Option A

(Ns-N)/Ns

Quick Summary:

The slip of an induction motor represents the relative difference between the synchronous speed of the rotating magnetic field and the actual speed of the rotor, expressed as a fraction of the synchronous speed. It is a dimensionless quantity that measures how much the rotor lags behind the stator field. A value of zero implies the rotor is moving at synchronous speed, while a value of one indicates a stationary rotor (standstill).

⚙️TETechnical SolutionConcept & Principle

💡 Explanation

🔢 Key Formulas

$s = \frac{N_s - N}{N_s}$ — Per-unit fractional slip of induction motor

$\%s = \frac{N_s - N}{N_s} \times 100$ — Percentage slip

$N_s = \frac{120f}{P}$ — Synchronous speed in RPM

$f_r = s \cdot f$ — Rotor induced frequency

$N = N_s(1 - s)$ — Actual rotor speed in RPM

⚙️ Working Principle

An induction motor operates on the principle of electromagnetic induction. The stator winding creates a magnetic field that rotates at synchronous speed $N_s$ . This moving field cuts the conductors of the stationary rotor, inducing an electromotive force (EMF) and current according to Faraday's Law. The resulting electromagnetic force causes the rotor to rotate in the direction of the stator field at speed $N$ , which must always be slightly less than $N_s$ to maintain relative motion and induced current.

📌 Key Points

▸
Slip at standstill ( $N = 0$ ) is equal to $1$ or $100\%$ .
▸
Under normal full-load operating conditions, slip typically ranges from $1\%$ to $5\%$ .
▸
If rotor speed reaches synchronous speed ( $N = N_s$ ), slip becomes $0$ , no torque is produced, and the motor stops accelerating.
▸
Rotor current frequency depends directly on slip ( $f_r = s f$ ).

✅ Advantages

▸
Simple parameter to analyze operating state (motoring, generating, or braking)
▸
Helps directly determine rotor current frequency and losses

❌ Disadvantages / Limitations

▸
Higher slip results in higher rotor copper losses ( $I_2°2 R_2$ ) and lower operating efficiency
▸
Large slip leads to poor speed regulation under varying loads

🛠️ Applications / Uses

▸
Determining motor efficiency and torque-slip performance characteristics
▸
Speed control of induction motors using slip power recovery schemes

📄 Additional Information

▸
Standard full-load slip values for industrial 3-phase induction motors are between 0.01 and 0.05.
▸
Option B ( $N_s - N$ ) represents the absolute slip speed in RPM, not the normalized slip parameter.
▸
Option C ( $s N_s$ ) is equal to slip speed ( $N_s - N$ ).
▸
Option D ( $N_s$ ) is the synchronous speed of the stator magnetic field.

📊 Diagram / Illustration

✅

A is correct — Slip is defined as the ratio of slip speed $(N_s - N)$ to synchronous speed $N_s$ , given by $s = \frac{N_s - N}{N_s}$ .

Core Concepts Used

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Synchronous Speed Slip and Slip Speed Rotor Frequency

💡 EXAM TIP

In competitive exams, pay close attention to whether the question asks for 'Slip' (dimensionless/per-unit ratio), 'Percentage Slip' (expressed in %), or 'Slip Speed' (expressed in RPM as $N_s - N$ ).

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