Test: Three-phase Induction Motor Equivalent Circuit - Electrical Engineering (EE) MCQ

Concept:

• Slip at maximum torque is calculated with the help of the maximum power transfer theorem

           Z_L = Z_th

• Power developed from an induction motor is given by

• Torque of an induction motor

           here P is the power and ω is the speed in rad/sec

Solution:

To find the slip apply the MPT theorem and open circuit the current source, you will equate the load impedance with the source impedance.

To find the Torque we have to calculate the power consumed by the load and to know that we must calculate the value of the load current,

the load current is calculated by applying the current division rule.

∴ The maximum torque of the machine is given by

Concept:

per phase circuit diagram of 3 phase induction motor is shown below.

here

r₁ = Primary resistance

r₂ = Secondary resistance 

x₁ = Primary reactance

x₂ = Secondary reactance

X_m = Magnetising reactance

Solution:

To transfer maximum power to the load. Load resistance must be equal to source impedance, therefore, find the total impedance from the load side by open circuiting the independent current source.

Power flow in induction motor:

 

From the above diagram, the air gap power is the algebraic sum of rotor copper losses and total mechanical power developed.

P_g = P_c + P_m

Also, P_g : Pc : P_m = 1 : s : 1 - s

where, Pg = Air gap power

Pc = Rotor copper loss

Pm = Mechanical power developed

s = Slip

Reason for Induction Motor Lagging Power Factor:

• Induction motors are widely used in commercial and as well as residential applications they always operate on lagging power factors.
• Power factor is given by cosine angle of voltage and current.
• If the angle between voltage and current is zero degrees or both are in phase with each other then it is a purely resistive circuit.
• If current lags voltage by 90 degrees then it is a purely inductive circuit
• If current leads voltage by 90 degrees then it is a purely capacitive circuit.
• In the Induction motor, air gap exists between the rotor and stator. Due to the presence of an air gap reluctance is high.

We know that, 

Where ϕ is flux
S is reluctance and NI is MMF

And, 

• Since more amount of magnetizing current is required to transfer power from stator circuit to rotor circuit because of the presence of airgap.
• As air-gap has a relative permeability of 1 then the value of reluctance is low and hence the value of magnetic flux is increased.
• The magnetic flux is the function of inductance  hence inductance increased with an increase in the value of flux.
• Hence, the induction motor acts as an inductive load and it operates at a lagging power factor.

Induction motor on no-load:

Circuit diagram for induction motor on no-load:

Where,

V₁ is the applied voltage

I₁ is the no-load current

I_c is the working component of current through Rc (Magnetizing resistance)

I_m  is the magnetizing component of current through Xm (magnetizing reactance)

• The current drawn by the induction motor when it is not coupled to the driven equipment is called the no-load current of the motor.
• The no-load current produces the magnetic field in the motor.
• No-load current is 30 to 50 % of the full load current of the induction machine

Therefore when the full load current is 20 A, then the no-load current is 6 to 10 A

Important Points

There are two parts of no-load current.

1. Magnetizing component of the current
2. Loss components( Hysteresis and eddy current loss)​

​

Phasor Diagram when Induction motor on no load:

Where,

E₁ is the stator induced EMF

V₁ is the stator terminal voltage

E₂ is the rotor induced EMF

The power factor of the machine is poor when operating at no-load

Magnetic reactance (X_m) is depends on airgap flux and the flux is depends on V/f.

X_m ∝ ϕ ∝ V/f

Hence, magnetizing reactance gets affected by reducing the rms value of the supply at the rated frequency.

Additional Information

The total resistance at rotor is represented as  where s is slip. Now, if we create an equivalent transformer circuit for induction motor, the secondary resistance will be r₂. Thus the load resistance will be .

Induction motor modelled as a transformer

When all the rotor parameters are shifted to stator side induction motor circuit is given by

So is the resistance which shows the power which is converted to mechanical power output or useful power.

Given that, at 50 Hz, 400 V supply

R₁ = 5.39 Ω

R₂ = 5.72 Ω

X₁ = X₂ = 8.22 Ω

At, 100 V, 10 Hz supply,

R₁ and R₂ will remains same

As X is dependent on frequency, X₁ and X₂ will change.

I_st (ph) = 8.63 A

⇒ I_L = √3 I_ph = √3 × 8.63 = 14.94 A

Concept:

The current is given as for star connected motor,

Calculation:

Given that,

R₁ = 0.3 Ω, R₂ = 0.3 Ω, X₁ = 0.41 Ω, X₂ = 0.41 Ω

Frequency (f) = 60 Hz

For star connected induction motor,

When supply is charged to 80 V (line), 20 Hz, reactance offered by stator & rotor will be changed because of change in frequency {∴ x ∝ f}

At 20 Hz, equivalent circuit can be drawn as,

I_st = 70.05 A

Induction motor modelled as a transformer

When all the rotor parameters are shifted to stator side induction motor circuit is given by

So,  is the resistance which shows the power which is converted to mechanical power output or useful power.

The core losses and magnetizing current in a three-phase induction motor are primarily associated with the Magnetizing Reactance (Xm), which is represented by Option C.