In terms of slip ratio of rotor input to rotor copper loss is | Electrical MCQ | Examoogle

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In terms of slip ratio of rotor input to rotor copper loss is

$1/S$

$(1-S)/S$

$S/(1-S)$

$S$

Correct Answer

⚙️ TE • Technical Concept & PrincipleElectricalMachine

Option A

$1/S$

Quick Summary:

In a three-phase induction motor, the ratio of rotor input power ( $P_g$ , air-gap power) to rotor copper loss ( $P_{cu}$ ) is given by the inverse of the slip ( $S$ ). Mathematically, $P_g = \frac{P_{cu}}{S}$ , leading to the ratio $\frac{P_g}{P_{cu}} = \frac{1}{S}$ .

⚙️TETechnical SolutionConcept & Principle

💡 Explanation

🔢 Key Formulas

$P_{cu} = S \cdot P_g$ — Rotor copper loss in terms of air-gap power

$\frac{P_g}{P_{cu}} = \frac{1}{S}$ — Ratio of rotor input to rotor copper loss

⚙️ Working Principle

The electrical power transferred from the stator to the rotor via the air gap is known as air-gap power ( $P_g$ ). A fraction of this power, equal to $S \times P_g$ , is dissipated as heat in the rotor resistance (copper loss), while the remaining fraction $(1-S) \times P_g$ is converted into mechanical power. Since $P_{cu} = S \cdot P_g$ , the ratio of $P_g$ to $P_{cu}$ simplifies to $\frac{P_g}{S \cdot P_g} = \frac{1}{S}$ .

📌 Key Points

▸
Air-gap power ( $P_g$ ) is the total power transferred to the rotor.
▸
Rotor copper loss ( $P_{cu}$ ) is the power lost due to rotor resistance.
▸
Mechanical power developed ( $P_m$ ) is equal to $(1-S)P_g$ .
▸
The power flow sequence in an induction motor is Stator Input $\rightarrow$ Air-gap Power $\rightarrow$ Mechanical Power.

✅ Advantages

▸
Provides a simple method to calculate efficiency.
▸
Helps in understanding the speed-torque characteristics.

❌ Disadvantages / Limitations

▸
Applicable only under steady-state conditions.
▸
Assumes constant rotor resistance.

🛠️ Applications / Uses

▸
Efficiency calculation of induction motors.
▸
Predicting motor performance under variable load conditions.

📄 Additional Information

▸
The ratio of Power Input : Copper Loss : Mechanical Power is $1 : S : (1-S)$ .
▸
Option B, $(1-S)/S$ , represents the ratio of Mechanical Power to Rotor Copper Loss.

📊 Diagram / Illustration

✅

A is correct — The ratio of rotor input power to rotor copper loss is derived as $1/S$ .

Core Concepts Used

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Induction Motor Power Flow Slip Definition Rotor Copper Loss

💡 EXAM TIP

Remember the sequence $1 : S : (1-S)$ for Air-gap power : Copper loss : Mechanical power; it is the most efficient way to solve almost all power flow problems in induction motors.

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