We have already seen that the induction motor can be treated as generalized transformer. Transformer works on the principle of electromagnetic induction. The induction motor also works on the same principle. The energy transfer from stator to rotor of the induction motor takes place entirely with the help of a flux mutually linking the two. Thus stator acts as a primary while the rotor acts as a rotating secondary when induction motor is treated as a tranformer.

If E

_{1}= Induced voltage in stator per phase E

_{2 }= Rotor induced e.m.f. per phase on standstill k = Rotor turns / Stator turns

then k = E

_{2}/ E_{1} Thus if V

_{1}is the supply voltage per phase to stator, it produces the flux which links with both stator and rotor. Due to self induction E_{1}, is the induced e.m.f. in stator per phase while E_{2}is the induced e.m.f. in rotor due to mutual induction, at standstill. In running condition the induced e.m.f. in rotor becomes E_{2r }which is s E_{2}.Now E

_{2r }= Rotor induced e.m.f. in running condition per phase R

_{2 }= Rotor resistance per phase X

_{2r }= Rotor reactance per phase in running condition R

_{1 }= Stator resistance per phase X

_{1 }= Stator reactance per phase So induction motor can be represented as a transformer as shown in the Fig. 1.

Fig. 1 Induction motor as a transformer |

When induction motor is on no load, it draws a current from the supply to produce the flux in air gap and to supply iron losses.

1. I

_{c }= Active component which supplies no load losses2. I

_{m }= Magnetizing component which sets up flux in core and air gap These two currents give us the elements of an exciting branch as,

R

_{o }= Representing no load losses = V_{1 }/I_{c }and X

_{o }= Representing flux set up = V_{1}/I_{m }Thus, Ī

_{o}= Ī_{c}+ Ī_{m} The equivalent circuit of induction motor thus can be represented as shown in the Fig. 2.

Fig. 2 Basic equivalent circuit |

The stator and rotor sides are shown separated by an air gap.

I

= E

_{2r }= Rotor current in running condition= E

_{2r }/Z_{2r }= (s E_{2})/√(R_{2}^{2}+(s X_{2})^{2}) It is important to note that as load on the motor changes, the motor speed changes. Thus slip changes. As slip changes the reactance X

_{2r }changes. Hence X_{2r }= sX_{2 }is shown variable.**Representing of rotor impedance :**

It is shown that, I

_{2r }= (sE_{2})/√(R_{2}^{2}+(s X_{2})^{2}) = E_{2 }/√((R_{2}/s)^{2}+ X_{2}^{2}) So it can be assumed that equivalent rotor circuit in the running condition has fixed reactance X

_{2}, fixed voltage E_{2 }but a variable resistance R_{2}/s, as indicated in the above equation. So the variable rotor resistance R

_{2}/s has two parts.1. Rotor resistance R

_{2}itself which represents copper loss.2. R

_{2}(1 - s)/s which represents load resistance R_{L}. So it is electrical equivalent of mechanical load on the motor.**Key**

**Point**: Thus the mechanical load on the motor is represented by the pure resistance of value R

_{2}(1 -s)/s.

Now let us obtain equivalent circuit referred to stator side.

Equivalent circuit referred to stator :

Transfer all the rotor parameters to stator,

k = E

_{2}/E_{1}= Transformation ratio E

_{2}**'**= E_{2}/ k The rotor current has its reflected component on the stator side which is I

_{2r}**'**. I

_{2r}**'**= k I_{2r}= (k s E_{2})/√(R_{2}^{2}+(s X_{2})^{2}) X

_{2}**'**= X_{2}/K^{2}= Reflected rotor reactance R

_{2}**'**= R_{2}/K^{2}= Reflected rotor resistance R

_{L}**'**= R_{L}/K^{2}= (R_{2}/K^{2})(1-s / s) = R

_{2}**'**(1-s / s) Thus R

_{L}**'**is reflected mechanical load on stator. So equivalent circuit referred to stator can be shown as in the Fig. 4

.

Fig. 4 Equivalent circuit referred to stator |

The resistance R

R

X

R

and X

While Ī

and Ī

Thus the equivalent circuit can be shown in the Fig.6.

where V

I

cos Φ = Power factor of stator

Stator core loss = I

Stator copper loss = 3 I

where R

P

P

Thus P

P

T = Torque developed

where N = Speed of motor

But N = N

and I

where R

I

See part 2

_{2}**'**(1 -s)/ s = R_{L}**'**is fictitious resistance representing the mechanical load on the motor.__1.1 Approximate Equivalent Circuit__

Similar to the transformer the equivalent circuit can be modified by shifting the exciting current (R

Now the resistance R_{o}and X_{o}) purely across the supply, to the left of R_{1}and X_{1}. Due to this, we are neglecting the drop across R_{1}and X_{1 }due to I_{o}, which is very small. Hence the circuit is called approximate equivalent circuit. The circuit is shown in the Fig.5.Fig. 5 Approximate equivalent circuit |

_{1}and R_{2}**'**while reactance X_{1}and X_{2}**'**can be combined. So we get,R

_{1e}= Equivalent resistance referred to stator = R_{1}+ R_{2}**'**X

_{1e}= Equivalent reactance referred to stator = X_{1}+**X**_{2}**'**R

_{1e}= R_{1 }+ (R_{2}/K^{2})and X

_{1e}= X_{1}+ (X_{2}/K^{2})While Ī

_{1}= Ī_{o}+ Ī_{2r}**'**.........phasor diagramand Ī

_{o}= Ī_{c}+ Ī_{m}Thus the equivalent circuit can be shown in the Fig.6.

Fig. 6 |

**Power Equations from Equivalent Circuit** With reference to approximate equivalent circuit shown in the Fig. 6, we can write various power equations as,

P_{in }= input power = 3 V_{1 }I_{1 }cos Φwhere V

_{1 }= Stator voltage per phaseI

_{1 }= Current drawn by stator per phasecos Φ = Power factor of stator

Stator core loss = I

_{m}^{2}_{ }R_{o }Stator copper loss = 3 I

_{1}^{2}R_{o }_{ }where R

_{1 }= Stator resistance per phaseP

_{2 }= Rotor input = (3 I_{2r}**'**^{2}_{ }R_{2}**'**)/sP

_{c }= Rotor copper loss = 3 I_{2r}**'**^{2}_{ }R_{2}**'**Thus P

_{c }= s P_{2 }P

_{m }= Gross mechanical power developedT = Torque developed

where N = Speed of motor

But N = N

_{s }(1-s) =, so substituting in aboveand I

_{2r}**'**= V_{1 }/(( R_{1e}+ R_{L}**'**) + j X_{1e})where R

_{L}**'**= R_{2}**'**(1-s)/sI

_{2r}**'**= V_{1}/√**(**( R_{1e}+ R_{L}**'**)^{2}_{ }+ X_{1e}^{2}_{ }**)****Key Point**: Remember that in all the above formula all the values per phase values.See part 2

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