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:
Show Hint
- 100 Hz
- 95 Hz
- 10 Hz
- 5 Hz
The Correct Option is B
Solution and Explanation
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: \[ \boxed{95 \; \text{Hz}} \]
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