Question

The frequency of the emf induced in the rotor circuit of a three phase. 50 Hz, 4 pole induction motor running at 1425 RPM is

• A. 0.25 Hz
• B. 0.025 Hz
• C. 2.5 Hz
• D. 25 Hz

Hint:

Rotor frequency is directly proportional to slip. At standstill (start), slip is 1, so $f_r = f$. As the motor speeds up and approaches $N_s$, the slip and rotor frequency both decrease toward zero.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The frequency of the induced rotor current ($f_r$) is not the same as the supply frequency ($f$). It depends on the "slip" of the motor—the difference between the synchronous speed and the actual rotor speed.

Step 2: Key Formula or Approach:
1. Synchronous speed: $N_s = \frac{120f}{P}$ 2. Slip: $s = \frac{N_s - N}{N_s}$ 3. Rotor frequency: $f_r = s \cdot f$

Step 3: Detailed Explanation:
Given: $f = 50\text{ Hz}$, $P = 4$, $N = 1425\text{ RPM}$. 1. Find $N_s$: \[ N_s = \frac{120 \times 50}{4} = \frac{6000}{4} = 1500\text{ RPM} \] 2. Find Slip ($s$): \[ s = \frac{1500 - 1425}{1500} = \frac{75}{1500} = 0.05 \] 3. Find Rotor Frequency ($f_r$): \[ f_r = 0.05 \times 50 = 2.5\text{ Hz} \]

Step 4: Final Answer:
The frequency of the emf induced in the rotor is 2.5 Hz.

Answer 2

This question requires finding the frequency of the induced rotor emf for a slipping induction motor. For a 4-pole motor connected to 50 Hz, the synchronous speed is a fixed standard value of 1500 RPM. Let's check each option against the physical relationship between slip speed and rotor frequency.

1. 0.25 Hz: This would correspond to an extremely small slip of about 0.5%, meaning the rotor speed would be around 1492-1493 RPM, much closer to synchronous speed than the 1425 RPM actually given.
2. 0.025 Hz: This is an almost negligible frequency corresponding to a slip of just 0.05%, far too small to be consistent with a rotor running 75 RPM below synchronous speed.
3. 2.5 Hz: The rotor runs at 1425 RPM against a synchronous speed of 1500 RPM, a shortfall of 75 RPM. As a fraction of synchronous speed, this deficit is \( \frac{75}{1500} = 0.05 \), a slip of 5%. Since the rotor frequency is this slip fraction applied to the supply frequency, \( f_r = 0.05 \times 50 = 2.5\text{ Hz} \), consistent with a rotor genuinely lagging by 75 RPM.
4. 25 Hz: This would require a slip of 50%, meaning the rotor would be running at only half its synchronous speed (750 RPM), far slower than the 1425 RPM stated in the problem.

Only a slip of 5%, consistent with the actual 75 RPM difference between synchronous and rotor speed, produces a rotor frequency of 2.5 Hz.

Therefore, the correct answer is 2.5 Hz.