In a particular circuit, if one of the circuit elements is
variable, then depending upon its value, the circuit characteristics varies. As
the value of the variable element is changed, the circuit parameters like
current, power factor, power losses etc. also change. The locus of the
extremity of the current phasor, obtained for various values of a variable
element is called a locus diagram.

From the equivalent
circuit of an induction motor, the motor can be treated as series R-L circuit
where the element resistance of the circuit is variable which varies as slip s.
Thus for variable load conditions, the resistance changes and hence the current
drawn by the motor also changes. The locus diagram of such a current phasor is
circular in nature and hence called circle diagram of a three phase induction
motor. Using this diagram, all the performance characteristics of an induction
motor like power factor, efficiency, stator losses, rotor losses, maximum
output, maximum torque etc. can be predicted. Thus, a circle diagram is a
graphical approach of predetermining the operation characteristics of an
induction motor.

Let us prove that the
locus diagram obtained for a current phasor is a circle, for a series R-L
circuit with an element R as variable.

#
**By using the
data obtained from the no load test and the blocked rotor test, the circle
diagram can be drawn using the following steps :**

**Step 1**: Take reference phasor V as vertical (Y-axis).

**Step 2**: Select suitable current scale such that diameter of circle is about 20 to 30 cm.

**Step3**: From no load test, I

_{o}and are Φ

_{o}obtained. Draw vector I

_{o}, lagging V by angle Φ

_{o}. This is the line OO

**'**as shown in the Fig. 1.

**Step 4**: Draw horizontal line through extremity of I

_{o}i.e. O

**'**, parallel to horizontal axis.

**Step 5**: Draw the current I

_{SN}calculated from I

_{sc}with the same scale, lagging V by angle Φ

_{sc}, from the origin O. This is phasor OA as shown in the Fig. 1.

**Step 6**: Join O

**'**A is called output line.

**Step 7**: Draw a perpendicular bisector of O

**'**A. Extend it to meet line O

**'**B at point C. This is the centre of the circle.

**Step 8**: Draw the circle, with C as a center and radius equal to O

**'**C. This meets the horizontal line drawn from O

**'**at B as shown in the Fig. 1.

**Step 9**: Draw the perpendicular from point A on the horizontal axis, to meet O

**'**B line at F and meet horizontal axis at D.

**Step 10**: Torque line.

The torque line separates
stator and rotor copper losses.

Note that as voltage axis is
vertical, all the vertical distances are proportional to active components of
currents or power inputs, if measured at appropriate scale.

Thus the vertical
distance AD represents power input at short circuit i.e. W

_{SN}, now which consists of core loss and stator, rotor copper losses.
Now FD = O

**'**G
= Fixed loss

Where O

**'**G is drawn perpendicular from O**'**on horizontal axis. This represents power input on no load i.e. fixed loss.
Hence AF α Sum
of stator and rotor copper losses

Then point E can be
located as,

AE/EF = Rotor copper loss /
Stator copper loss

The line O

**'**E under this condition is called torque line.**Power scale**: As AD represents W

_{SN}i.e. power input on short circuit at normal voltage, the power scale can be obtained as,

Power scale = W

_{SN}/l(AD) W/cm
where l(AD)
= Distance AD in cm

**Location of Point E**: In a slip ring induction motor, the stator resistance per phase R

_{1}and rotor resistance per phase R

_{2}can be easily measured. Similarly by introducing ammeters in stator and rotor circuit, the currents I

_{1}and I

_{2}also can be measured.

**.**K = I

^{.}._{1}/I

_{2}= Transformation ratio

Now AF/EF = Rotor copper loss / Stator copper
loss = (I

_{2}^{2}R_{2})/(I_{1}^{2}R_{1}) = (R_{2}/R_{2})(I_{2}^{2}/I_{1}^{2}) = (R_{2}/R_{2}).(1/K^{2})
But R

_{2}**'**= R_{2}/K^{2}= Rotor resistance referred to stator**.**AE/EF = R

^{.}._{2}

**'**/R

_{1}

Thus point E can be
obtained by dividing line AF in the ratio R

_{2}**'**to R_{1}.
In

**a****squirrel cage****motor**, the stator resistance can be measured by conducting resistance tset.**.**Stator copper loss = 3I

^{.}._{SN}

^{2}R

_{1}where I

_{SN }is phase value.

Neglecting core loss, W

_{SN }= Stator Cu loss + Rotor Cu loss**.**Rotor copper loss = W

^{.}._{SN }- 3I

_{SN}

^{2}R

_{1}

**.**AE/EF = (W

^{.}._{SN }- 3I

_{SN}

^{2}R

_{1})/(3I

_{SN}

^{2}R

_{1})

Dividing line AF in
this ratio, the point E can be obtained and hence O

**'**E represents torque line.

__1.1 Predicting Performance From Circle Diagram__

Let motor is running by
taking a current OP as shown in the Fig. 1. The various performance parameters
can be obtained from the circle diagram at that load condition.

Draw perpendicular from
point P to meet output line at Q, torque line at R, the base line at S and
horizontal axis at T.

We know the power scale
as obtained earlier.

Using the power scale
and various distances, the values of the performance parameters can be obtained
as,

Total motor input = PT
x Power scale

Fixed loss = ST x power
scale

Stator copper loss = SR x
power scale

Rotor copper loss = QR
x power scale

Total loss = QT x power scale

Rotor output = PQ x power
scale

Rotor input = PQ + QR = PR x
power scale

Slip s = Rotor Cu loss =
QR/PR

Power factor cos = PT/OP

Motor efficiency = Output /
Input = PQ/PT

Rotor efficiency = Rotor
output / Rotor input = PQ/PR

Rotor output / Rotor input =
1 - s = N/N

_{s}= PQ/PR
The torque is the rotor input
in synchronous watts.

__1.2 Maximum Quantities__

The maximum values of
various parameters can also be obtained by using circle diagram.

**1. Maximum Output**: Draw a line parallel to O

**'**A and is also tangent to the circle at point M. The point M can also be obtained by extending the perpendicular drawn from C on O

**'**A to meet the circle at M. Then the maximum output is given by l(MN) at the power scale. This is shown in the Fig. 1.

**2. Maximum Input**: It occurs at the highest point on the circle i.e. at point L. At this point, tangent to the circle is horizontal. The maximum input given l(LL

**'**) at the power scale.

**3. Maximum Torque**: Draw a line parallel to the torque line and is also tangent to the circle at point J. The point J can also be obtained by drawing perpendicular from C on torque line and extending it to meet circle at point J. The l(JK) represents maximum torque in synchronous watts at the power scale. This torque is also called stalling torque or pull out torque.

**4. Maximum Power Factor**: Draw a line tangent to the circle from the origin O, meeting circle at point H. Draw a perpendicular from H on horizontal axis till it meets it at point I. Then angle OHI gives angle corresponding to maximum power factor angle.

**.**Maximum p.f. = cos

^{.}.**∟**{OHI}

= HI/OH

**5. Starting Torque**: The torque is proportional to the rotor input. At s = 1, rotor input is equal to rotor copper loss i.e. l(AE).

**.**T

^{.}._{start}= l(AE) x Power scale ...................in synchronous watts

__1.3 Full load Condition__

The full load motor output
is given on the name plates in watts or h.p. Calculates the distance
corresponding to the full load output using the power scale.

Then extend AD upwards
from A onwards, equal to the distance corresponding to full load output, say A

**'**. Draw parallel to the output line O**'**A from A**'**to meet the circle at point P**'**. This is the point corresponding to the full load condition, as shown in the Fig. 2.
Once point P

**'**is known, the other performance parameters can be obtained easily as discussed above.
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