Question

The rotor of a 6 pole, 3 phase, 60 Hz induction motor has per phase resistance and reactance of 0.1 Ω and 0.5 Ω respectively. The voltage induced per phase in the rotor at standstill condition is 150 V. When the motor develops maximum torque, the rotor current per phase will be
1. 150 A
2. 150 \(\sqrt{2}\) A
3. 750 A
4. 750 \(\sqrt{2}\) A

Answer 1

Correct Answer - Option 2 : 150 \(\sqrt{2}\) A

Concept:

Torque developed in Induction Motor:

• In the Induction motor, the torque is proportional to the product of flux per stator pole and rotor current.
• In addition to current and flux, the power factor also comes into existence.
   

 And we know that, T ∝ ϕ₂I₂cosϕ₂

Or, T = kϕ₂I₂cosϕ₂

Where I₂ is rotor current at standstill

Φ₂ is the angle between rotor EMF and rotor current

k is a constant

We have, E₂ ∝ ϕ₂

⇒ T = k₁E₂I₂cosϕ₂ …. (1)

Where E₂ is rotor EMF at a standstill and k₁ is another constant

If R₂ is rotor resistance per phase and X₂ is rotor reactance per phase at standstill,

Therefore rotor impedance at standstill per phase is given as,

\({Z_2} = \sqrt {R_2^2 + X_2^2} \)

\({I_2} = {\frac{{{E_2}}}{Z_2}} = \frac{{{E_2}}}{{\sqrt {R_2^2 + X_2^2} }}\)

\(cos{\phi _2} = \frac{{{R_2}}}{{{Z_2}}} = \frac{{{R_2}}}{{\sqrt {R_2^2 + X_2^2} }}\)

From equation (1)

\(T = {T_{st}} = {k_1}\frac{{E_2^2{R_2}}}{{R_2^2 + X_2^2}}\)

If the Induction motor is under the running condition and run with slip s,

E_r = sE₂

And, X_r = sX₂

Under this condition torque given as,

\(T = {T_r} = {k_1}\frac{{sE_2^2{R_2}}}{{{{\left( {{R_2}} \right)}^2} + {{\left( {s{X_2}} \right)}^2}}}\)

For developing maximum running torque,

\(\frac{{dT}}{{ds}} = 0\)

s = s_m

The torque at any load in an induction motor is the function of slip.

After solving it,

R₂ = s_m X

And, value of current at maximum starting torque is given as,
\({I_m} = \frac{{{E_r}}}{{\sqrt {R_2^2 + {{\left( {s_m{X_2}} \right)}^2}} }} = \frac{{E_r}}{{\sqrt { {{2R_2}^2}} }} \)

Calculation:

Given,

R₂ = 0.1 Ω

X₂ = 0.5 Ω

\(s_m = \frac{{{R_2}}}{{{X_2}}} = \frac{{0.1}}{{0.5}} = 0.2\)

E₂ = 150 volt

E_r = 0.2 × 150 = 30 volt

For maximum torque,

R₂ = s_mX₂ = 0.1 Ω

\({I_m} = \frac{{{E_r}}}{{\sqrt {R_2^2 + {{\left( {s_m{X_2}} \right)}^2}} }} = \frac{{E_r}}{{\sqrt { {{2R_2}^2}} }} = \frac{{30}}{{\sqrt { {{2(0.1)}^2}} }} =150\sqrt 2 \;A\)