Question

A 10-pole, 50 Hz, 240 V, single phase induction motor runs at 540 RPM while driving rated load. The frequency of induced rotor currents due to backward field is:

• A. 100 Hz
• B. 95 Hz
• C. 10 Hz
• D. 5 Hz

Hint:

For backward field in single-phase induction motors, slip is calculated as \(\tfrac{N+N_s}{N_s}\).

Answer 1

The Correct Option is B

Step 1: Synchronous speed.
\[ N_s = \frac{120f}{P} = \frac{120 \times 50}{10} = 600 \; \text{RPM} \]

Step 2: Slip w.r.t forward field.
\[ s_f = \frac{N_s - N}{N_s} = \frac{600 - 540}{600} = \frac{60}{600} = 0.1 \]

Step 3: Rotor frequency due to forward field.
\[ f_{r,f} = s_f f = 0.1 \times 50 = 5 \; \text{Hz} \]

Step 4: Slip w.r.t backward field.
Backward field rotates at \(-N_s\). Relative speed of rotor w.r.t backward field: \[ N + N_s = 540 + 600 = 1140 \; \text{RPM} \] So, \[ s_b = \frac{N+N_s}{N_s} = \frac{1140}{600} = 1.9 \]

Step 5: Rotor frequency due to backward field.
\[ f_{r,b} = s_b f = 1.9 \times 50 = 95 \; \text{Hz} \]

Final Answer: \[ \boxed{95 \; \text{Hz}} \]