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The frequency of induced electromotive force (EMF) in rotor of a three-phase 5 kW, 400 V, 5 Hz, 4 pole induction motor is 25 Hz. If the motor is supplied with 400 V, 50 Hz in alternating current source, what will be its speed of rotation?
1. 900 rpm
2. 1500 rpm
3. 1000 rpm
4. 1450 rpm

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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

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