- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

A 1-phase supply produces a pulsating magnetic field which does not rotate in the space. Therefore, a 1-phase supply cannot produce rotation in a stationary rotor. Although, like a 3-phase supply, the 2-phase supply can also produce a rotating magnetic field of constant magnitude. Therefore, all the single-phase induction motors, except shaded pole induction motor, are started as 2-phase motor. Once so started, the motor will continue to run on the 1-phase supply.

Consider a 2-phase, 2 pole motor, where the phases A and B are fed by a balanced 2-phase supply and the currents in these phases are I_{A} and I_{B} respectively as shown in the figure.

It is clear that the current I_{A} leads the current I_{B} by angle of 90°. Therefore, the fluxes produced by these currents can be written as,

$$\mathrm{φ_{𝐴} = φ_{𝑚}\:sin(\omega 𝑡 + 90°) = φ_{𝑚}\:cos\:\omega 𝑡 … (1)}$$

$$\mathrm{φ_{𝐵} = φ_{𝑚}\:sin\:\omega 𝑡 … (2)}$$

Where, *φ _{m}* is the maximum value of the flux produced by the current in either phase.

We shall now prove that this two-phase supply produces a magnetic flux of constant magnitude which rotates in the space.

Refer the waveform of the currents I_{A} and I_{B}. Here, at the instant-1, the current is zero in phase B and positive maximum in the phase A. Also, the instant 1 corresponds to ωt = 0°, hence, the fluxes are given by,

$$\mathrm{φ_{𝐴} = φ_{m}\:cos\:\omega 𝑡 = φ_{m}\:cos\:0 = φ_{m}}$$

And

$$\mathrm{φ_{𝐵} = φ_{m}\:sin\:\omega 𝑡 = φ_{m}\:sin\:0 = 0}$$

Therefore, the resultant flux at instant 1 is

$$\mathrm{φ_{𝑅} = \sqrt{{φ^{2}_{𝐴}}+{φ^{2}_{B}}}=\sqrt{(φ_{𝑚})^{2}+(0)^{2}}= φ_{𝑚}}$$

Hence, the magnitude of resultant flux is equal to φ_{m} and is directed towards right as shown in the figure instant-1

At this instant, the current in the phase A is still flowing in the same direction and an equal current is flowing in the phase B. This instant corresponds to ωt = 45°, so that,

$$\mathrm{φ_{𝐴} = φ_{m}\:cos\:\omega 𝑡 = φ_{m}\:cos\:45 =\frac{1}{\sqrt{2}}φ_{m}}$$

And

$$\mathrm{𝜑_{𝐵} = 𝜑_{𝑚}\:sin\:\omega 𝑡 = 𝜑_{𝑚} \:sin\:45 =\frac{1}{\sqrt{2}}φ_{m}}$$

Therefore, the resultant flux is given by,

$$\mathrm{𝜑_{𝑅} =\sqrt{{φ^{2}_{𝐴}}+{φ^{2}_{B}}}=\sqrt{(\frac{φ_{m}}{\sqrt{2}})^{2}+(\frac{φ_{m}}{\sqrt{2}})^{2}}=φ_{m}}$$

Hence, the resultant flux has same value but turned by an angle of 45° in the clockwise direction from Instant-1 (see the figure of Instant-2).

At Instant-3, the current in the phase A has reduced to 0 and the current in phase B has increased to the positive maximum value. This instant is corresponds to ωt = 90° in the waveform diagram, so that,

$$\mathrm{φ_{𝐴} = φ_{m}\:cos\:ωt = φ_{m}\:cos\:90 = 0}$$

And

$$\mathrm{𝜑_{𝐵} = φ_{m}\:sin\:ωt = φ_{m}\:sin\:90 = φ_{m}}$$

Therefore, the resultant flux is given by,

$$\mathrm{𝜑_{𝑅} =\sqrt{{φ^{2}_{𝐴}}+{φ^{2}_{B}}}=\sqrt{({0})^{2}+({φ_{m}})^{2}}=φ_{m}}$$

Hence, again the resultant flux has the same magnitude equal to **φ _{m}**, but has turned by an angle of 90° in the clockwise direction from the instant 1 (see the figure of Instant-3).

At Instant-4, the current in phase A is reversed and has the same value as that in the phase B (the current I_{B} is positive). This instant corresponds to ωt = 135°, hence the flux is given by,

$$\mathrm{𝜑_{𝐴} = 𝜑_{𝑚}\:cos\:ωt = 𝜑_{𝑚} \:cos\:135 =-\frac{1}{\sqrt{2}}𝜑_{𝑚}}$$

And

$$\mathrm{𝜑_{𝐵} = 𝜑_{𝑚}\:cos\:ωt = 𝜑_{𝑚} \:cos\:135 =\frac{1}{\sqrt{2}}𝜑_{𝑚}}$$

Therefore, the resultant flux is given by,

$$\mathrm{𝜑_{𝑅} =\sqrt{{φ^{2}_{𝐴}}+{φ^{2}_{B}}}=\sqrt{(-\frac{φ_{m}}{\sqrt{2}})^{2}+(\frac{φ_{m}}{\sqrt{2}})^{2}}=φ_{m}}$$

Again, the resultant flux has the same magnitude, but has turned by an angle of 135° from Instant-1 as shown in the figure of Instant-4. We shall continue to consider other instants to this fact.

From the above discussion, it is clear that a balanced 2-phase supply produces a rotating magnetic field of constant value equal to **φ _{m}**. The speed of the rotation of the magnetic field is called as

$$\mathrm{𝑁_{𝑆} =\frac{120𝑓}{𝑃}}$$

Where,

*P*is the number poles in the machine, and*f*is the supply frequency.

- Related Questions & Answers
- Rotating Magnetic Field in Three-Phase Induction Motor
- Rotating Magnetic Field in a 3-Phase Induction Motor
- Magnetic Field Intensity of an Electromagnet
- Magnetic Field around a Current Carrying Conductor
- Advantages of Stationary Armature–Rotating Field Alternator
- What kind of output is produced by UNIX_TIMESTAMP() function?
- How to embed fonts in PDFs produced by Matplotlib
- How to check android mobile supports MAGNETIC FIELD sensor?
- Two-Phase AC Servo Motor and Three-Phase AC Servo Motor
- Force on a Current Carrying Conductor in a Magnetic Field
- DC Power Supply Filter Types
- Double Revolving Field Theory of Single-Phase Induction Motors
- Difference between Magnetic Tape and Magnetic Disk
- Write a program in C++ to check if a string can be obtained by rotating another string by two places
- Explain about two phase locking (2PL) protocol(DBMS)

Advertisements