**Subject :**Electrical Machines II (AC Machines)

**Unit :**Induction machines

## Construction of the Circle Diagram

**Construction of the Circle Diagram:**

*Fig:1*

Circle diagram of an induction motor can be drawn by using the data obtained from (1) no-load (2) short-circuit test and (3) stator resistance test, as shown below.

**Step No. 1**

From no-load test, I0 and φ0 can be calculated. Hence, as shown in Fig. 1, vector for I_{0} can be laid off lagging φ_{0} behind the applied voltage V.

**Step No. 2**

Next, from blocked rotor test or short-circuit test, short circuit current I_{SN} corresponding to normal voltage and φ_{S} are found. The vector OA represents I_{SN} = (I_{S}V/V_{S} ) in magnitude and phase. Vector O′A represents rotor current I_{2}′ as referred to stator. Clearly, the two points O′ and A lie on the required circle. For finding the centre C of this circle, chord O′A is bisected at right angles–its bisector giving point C.

The diameter O′D is drawn perpendicular to the voltage vector. As a matter of practical contingency, it is recommended that the scale of current vectors should be so chosen that the diameter is more than 25 cm, in order that the performance data of the motor may be read with reasonable accuracy from the circle diagram. With centre C and radius = CO′, the circle can be drawn. The line O′A is known as out-put line. It should be noted that as the voltage vector is drawn vertically, all vertical distances represent the active or power or energy components of the currents.

For example, the vertical component O′P of no-load current OO′ represents the no-load input, which supplies core loss, friction and windage loss and a negligibly small amount of stator I^{2}R loss. Similarly, the vertical component AG of short-circuit current OA is proportional to the motor input on short circuit or if measured to a proper scale, may be said to equal power input.

**Step No. 3**

*Torque line:* This is the line which separates the stator and the rotor copper losses. When the rotor is locked, then all the power supplied to the motor goes to meet core losses and Cu losses in the stator and rotor windings. The power input is proportional to AG. Out of this, FG (= O′P) represents fixed losses i.e. stator core loss and friction and windage losses. AF is proportional to the sum of the stator and rotor Cu losses. The point E is such that

line O′E is known as *torque line.*