Correct Answer - Option 1 : 900 rpm
Concept:
The torque equation of the three-phase induction motor is given as
\(T = \frac{3}{{2\pi {N_s}}}\frac{{sE_2^2}}{{R_2^2 + {{(s{X_2})}^2}}}\)
Rotor current \({I_2} = \frac{{s{E_2}}}{{{Z_2}}}\)
slip formula for induction motor \(s = \frac{{{N_s} - N}}{{{N_s}}}\)
Ns = synchronous speed of the motor
N = rotor speed
s = slip of the motor
R2 = rotor resistance
X2 = rotor reactance
E2 = rotor induced emf
Calculation:
Case 1
Rotor frequency = 5 Hz
Supply frequency f1 = 25 Hz
slip S1 = fr / fs = 5 / 25 = 0.2
Case 2
Supply frequency f2 = 50 Hz
From the above equation:
\(T ∝ \frac{{s{V^2}}}{f}\)
As torque is constant
So, s ∝ f
\(\frac{{{s_1}}}{{{s_2}}} = \frac{{{f_1}}}{{{f_2}}}\)
substituting all value from above
s2 = (0.2 × 50) / 25
= 0.4
Ns = (120 × 50) / 4
= 1500 rpm,
From slip relation
\(0.4 = \frac{{1500 - N}}{{1500}}\)
N = 900 rpm