# 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

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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

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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,

Er = sE2

And, Xr = 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 = sm

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

After solving it,

R2 = sX

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

Er = 0.2 × 150 = 30 volt

For maximum torque,

R2 = smX2 = 0.1 Ω

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