The** Circle Diagram** of an Induction motor is very useful to study its performance under all operating conditions. The construction of the circle diagram is based on the approximate equivalent circuit shown below. It is the diagrammatic representation of the performance of the induction motor. The circle diagram provides information about the power output, losses, and the efficiency of the induction motor.

**Contents:**

- Construction of the Circle Diagram
- Result Obtained from the Circle Diagram
- Significance of lines on the Circle Diagram

Applying KCL (Kirchhoff’s Current Law)

Let the phase voltage V_{1} be taken along the vertical axis OY as shown in the figure below:

The **No-Load current I _{0} = OA** lags behind V

_{1}by an angle ϕ

_{0}. The no-load power factor angle ϕ

_{0}is of the order of 60 to 80 degrees because of the large magnetizing current needed to produce the required flux pole in a magnetic circuit containing the air gaps.

At the no-load condition, **s = 0** and **R _{2}/s** are

**infinite**, or we can say that the R

_{2}/s is an open circuit at no load.

Here, all the rotational losses are considered under R_{0} and the no-load loss is given by the equation shown below:

The rotor current referred to the stator is given by:

The current I’_{2} lags behind the voltage V_{1 }by the impedance angle ϕ as shown in the figure below:

Combining equation (1) and (2) we get

The above equation (3) is of the form** r = a sin φ** which represents a circle in polar form with the diameter a.

From the above figure (b) shown, the following points are illustrated.

- The locus of I’
_{2}is a circle of the diameter V_{1}/ X_{1}+ X’_{2} - The radius of the circle O’C = V
_{1}/ 2(X_{1}+ X’_{2}) - The center C has the coordinates [V
_{1}/ 2(X_{1}+ X’_{2}), 0]

The **resulting circle diagram** of the induction motor is shown below:

It is seen that the tip of the phasor I_{1} coincides with that of the phasor I’_{2}. Thus, the locus of both I_{1} and I’_{2} is the upper semicircle. I_{1} and I’_{2} radiate from the origin O and O’ respectively. When the motor is started **s = 1** with the rated voltage, the tip of I_{1} and I’_{2} will be at some point F of the circle.

As the motor accelerates, the tips of I_{1} and I’_{2} move around the circle in an anti-clockwise direction. This process continues until the output torque matches the load torque. If there is no shaft load, the motor accelerates to synchronous speed. At this point** I’ _{2} = 0** and

**I**

_{1}= 0.## Construction of the Circle Diagram

The following data are required for constructing the circle diagram:

- Stator phase voltage V
_{1}= V_{L}/√3 - No-load current I
_{0} - No-load power factor cosϕ
_{0} - Blocked rotor current and power factor
- Stator phase resistance R
_{1}.

**Steps to draw Circle Diagram of an Induction Motor:**

- Take the phasor voltage V
_{1}along the y-axis. - Choose a convenient current. With O as origin, draw a line OO’ = I
_{0}at an angle ϕ_{0 }with V_{1}. - Draw the line OKN perpendicular to V
_{1}. Similarly, draw a line O’D perpendicular to V_{1}. - From point O draw the line equal to the blocked rotor current I
_{1BR}to the same scale as I_{0}. This line lags behind V_{1}by the blocked rotor power factor angle ϕ_{1BR}. - Join O’F and measure its magnitude in amperes. The line O’F represents I’
_{2BR}. - From point F, draw a line FMN parallel to V
_{1}. This line is perpendicular to O’D and ON. - Calculate MS = I’
^{2}_{2BR}R_{1}/V_{1}and locate point S. Join O’S and extends it to meet the circle at J. - Draw the perpendicular bisector of the chord O’F. This bisector will pass through the center of the circle at point C. Now with the radius CD’ or CD draw the circle.

## Result Obtained from the Circle Diagram

Assume that the line current I_{1} is known. With the centre at O, draw an arc with the radius I_{1}. This arc intersects the circle at the operating point E. Draw the line EK and locates the point H, L, G.

The following results are obtained from the circle diagram shown above.

- Input power = 3V
_{1}KE - Rotational losses = 3V
_{1}KH - Stator copper loss = 3V
_{1}HL - Rotor copper loss = 3V
_{1}LG - Output power = 3V
_{1}GE - Output torque = 3V
_{1}LE/ω_{s} - Starting torque = 3V
_{1}SF/ω_{s} - Slip = LG/LE
- Speed = GE/LE X n
_{s} - Efficiency = GE/KE
- Power factor = KE/OE

### Significance of lines on the Circle Diagram

**Input line ON**

The vertical distance between any point on the circle and the line **ON** represents the input power. Therefore, line ON is called the** input line**.

**Output Line O’F**

The vertical distance between any point on the circle and the line **O’G** represents the **output power**. Hence, line **O’F** is called the **output line**.

**Air Gap Power Line O’J**

Line EL represents the air gap power P_{g}; line **O’L** is referred to as the **air gap power line**. Since, **Ʈ _{d} = P_{g}/ ω_{s}.** This line is also known as

**Torque line.**