Complete and Approximate Equivalent Circuits of Induction Motor

---

---

Complete Equivalent Circuit of Induction Motor

In order to obtain the complete per-phase equivalent circuit of an induction motor, it is required to refer the rotor part to the stator side frequency and voltage level.

The transformation of the rotor circuit of the induction motor can be done by the means of effective turns ratio of the induction motor.

The figure shows the complete equivalent circuit of the induction motor.

Let suffixes "s" and "r" be used for stator and rotor quantities respectively.

Then,

a_eff = Effective Turns Ratio of Induction Motor

R'_r = Resistance of the Rotor Winding per Phase Referred to Stator

X'_r0 = Standstill Rotor Reactance per Phase Referred to Stator

Therefore, the rotor EMF referred to stator side is given by,

$$\mathrm{\frac{E_{r}}{N_{er}} \: = \: \frac{E'_r}{N_{es}} \:\: \dotso \: (1)}$$

$$\mathrm{\Rightarrow \: E'_{r} \: = \: \frac{N_{es}}{N_{er}}E_{r} \: = \: a_{eff} \: \cdot \: E_{r} \: = \: E_{s} \:\: \dotso \: (2)}$$

Similarly, the rotor current referred to stator side is,

$$\mathrm{I'_{r} \: = \: \frac{I_{r}}{a_{eff}} \:\: \dotso \: (3)}$$

Rotor impedance referred to stator side is given by,

$$\mathrm{Z'_{r0} \: = \: a_{eff}^{2}\left(\frac{Rr}{s} \:+\: jX_{r0}\right) \:\: \dotso \: (4)}$$

Where, s is the fractional slip of the rotor.

The standstill rotor reactance referred to stator side is given by,

$$\mathrm{X'_{r0} \: = \: a_{eff}^{2} X_{r0} \:\: \dotso \: (5)}$$

Approximate Equivalent Circuit of Induction Motor

The circuit shown in the figure below is known as the approximate equivalent circuit per phase of the induction motor. The approximate equivalent circuit of the induction motor is obtained by shifting the shunt branches R₀ and X₀ in the equivalent circuit. In the approximate equivalent circuit, the only component that depends upon the slip (s) is the resistance which is representing the developed mechanical power by the rotor. All other quantities are constant and the reactances correspond to those quantities at the fixed stator frequency (f_s).

This approximate equivalent circuit is used as the standard for all performance calculation of an induction motor.

By referring the approximate equivalent circuit of the induction motor, the following equations can be written down for one phase at a slip s.

The impedance beyond the terminals A and B is given by

$$\mathrm{Z_{AB} \: = \: \left(Rs \: + \: \frac{R'_r}{s}\right) \: + \: j(X_{s} \: + \: X'_r) \:\: \dotso \: (6)}$$

$$\mathrm{I'_r =\frac{V_s}{Z_{AB}}=\frac{V_s}{\left(Rs \: + \: \frac{R'_r}{s}\right) \: + \: j(X_s + X'_r)} \:\: \dotso \: (7)}$$

$$\mathrm{\therefore \: Magnitude \: of \:\: I'_r = |I'_r| =\frac{V_s}{\sqrt{\left(Rs +\frac{R'_r}{s}\right)^{2} \:+\: (X_s + X'_r)^{2}}} \:\: \dotso \: (8)}$$

Therefore,

$$\mathrm{I'_{r} \:=\: |I'_{r}| \: \angle - \varphi_{r} \: = \: I'_{r} \: \cos \varphi_{r} \: - \: jI'_{r} \: \sin\varphi_{r} \:\: \dotso \: (9)}$$

Where,

$$\mathrm{\varphi_{r} \:=\: tan^{-1} \left(\frac{X_{s} \:+\: X'_{r}}{R_{s} \:+\: \frac{R'_r}{s}}\right) \:\: \dotso \: (10)}$$

Then, the power factor from the approximate equivalent circuit is,

$$\mathrm{\cos \varphi_{r} \: = \: \frac{(R_{s} \: + \: \frac{R'_{r}}{s})}{|Z_{AB}|} \:\: \dotso \: (11)}$$

The no-load current is given by,

$$\mathrm{I_{0} \: = \: I_{w} \: + \: I_{m}}$$

$$\mathrm{\Rightarrow \: I_{0} \: =\: \frac{V_{s}}{R_{0}} \:+\: \frac{V_{s}}{jX_{0}} \:\: \dotso \: (12)}$$

Hence, the total stator current is given by the phasor sum of rotor current referred to stator and no-load current, i.e.,

$$\mathrm{I_{s} \: = \: I'_{r} \:+\: I_{0} \:\: \dotso \: (13)}$$

$$\mathrm{\text{Total core losses, } \: P_{c} \: = \: P_{h} \: +\: P_{e} \: = \: 3 \:V_{s}I_{0} \:\cos \varphi_{0} \:\: \dotso \: (14)}$$

The input power to the stator is given by,

$$\mathrm{P_{input} \:=\: 3\:V_{s}I_{s} \: \cos \varphi_{s} \: =\: 3 \: V_{s}I'_{r} \: \cos \varphi_{r} \: + \: P_{C} \: = \: 3 \: I'^{2}_{r}\left(Rs \:+\: \frac{R'_{r}}{s}\right) \:+\: P_{C} \:\: \dotso \: (15)}$$

The air-gap power per phase in the induction motor is given by,

$$\mathrm{P_{g} \: =\: V_{s}I'_r \:\cos \varphi_{r} \: = \: I'^{2}_{r} \: \times \: \left(\frac{R'_r}{s}\right)}$$

$$\mathrm{\Rightarrow \: P_{g} \: =\: \frac{V_{s}^{2}}{\left(Rs \:+\: \frac{R'_{r}}{s}\right)^{2} \:+\: (X_{s} \:+\: X'_{r})^{2}} \:\times \: \left(\frac{R'_{r}}{s}\right) \:\: \dotso \: (16)}$$

Therefore, the torque developed by the motor is given by,

$$\mathrm{\tau_{d} \: = \: \frac{P_{g}}{\omega_{S}} \:=\: \frac{V_{s}^{2}}{\omega_{S}\left[\left(Rs \:+\: \frac{R'_{r}}{s}\right)^{2}\:+\: (X_{s} \:+\: X'_{r})^{2}\right]}\:\times \: \left(\frac{R'_{r}}{s}\right) \:\: \dotso \: (17)}$$