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

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

A) $s$

B) $(1-s)/s$

C) $s/(1-s)$

D) $s$

Correct Answer

⚙️ TE • Technical Concept & PrincipleElectricalMachine

Option C

$s/(1-s)$

Quick Summary:

In a 3-phase induction motor, the total rotor copper loss ( $P_{rcu}$ ) and the mechanical power developed ( $P_{m}$ ) are related to the rotor input power ( $P_{g}$ ) by slip $s$ . The rotor copper loss is given by $P_{rcu} = s P_{g}$ and the rotor mechanical output is $P_{m} = (1 - s) P_{g}$ . Therefore, the ratio of rotor copper loss to rotor output is $\frac{P_{rcu}}{P_{m}} = \frac{s P_{g}}{(1 - s) P_{g}} = \frac{s}{1 - s}$ .

⚙️TETechnical SolutionConcept & Principle

💡 Explanation

🔢 Key Formulas

$P_{rcu} = s P_{g}$ — Rotor copper loss in terms of rotor input power

$P_{m} = (1 - s) P_{g}$ — Mechanical power output in terms of rotor input power

$\frac{P_{rcu}}{P_{m}} = \frac{s}{1 - s}$ — Ratio of rotor copper loss to rotor output

$P_{g} : P_{rcu} : P_{m} = 1 : s : (1 - s)$ — Power flow ratio in induction motor rotor

⚙️ Working Principle

The power transferred across the air gap from the stator to the rotor is the rotor input power $P_{g}$ . A fraction $s$ of this power is dissipated as rotor copper loss ( $s P_{g}$ ) due to rotor winding resistance, while the remaining fraction $(1 - s)$ is converted into mechanical power ( $P_{m} = (1 - s) P_{g}$ ) available at the shaft.

📌 Key Points

▸
Rotor efficiency is given by $\eta_{rotor} = \frac{P_m}{P_g} = 1 - s$ .
▸
Rotor copper loss increases directly with an increase in slip $s$ .
▸
The ratio of rotor copper loss to rotor mechanical power developed is $\frac{s}{1-s}$ .

📄 Additional Information

▸
Option A ( $s$ ) represents the ratio of rotor copper loss to rotor input power ( $\frac{P_{rcu}}{P_g}$ ).
▸
Option B ( $\frac{1-s}{s}$ ) represents the ratio of rotor mechanical output to rotor copper loss ( $\frac{P_m}{P_{rcu}}$ ).
▸
Option C ( $\frac{s}{1-s}$ ) correctly represents the ratio of rotor copper loss to rotor output.

📊 Diagram / Illustration

✅

C is correct — The ratio of rotor copper loss ( $s P_g$ ) to rotor mechanical output power ( $(1-s) P_g$ ) simplifies to $\frac{s}{1-s}$ .

Core Concepts Used

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Power flow diagram of 3-phase induction motor Rotor copper loss Slip ratio in induction machines

💡 EXAM TIP

Remember the fundamental power division ratio $P_g : P_{rcu} : P_m = 1 : s : (1-s)$ ; express any power ratio by dividing their respective proportional terms directly.

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