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In the preceding section we shed light on the operation of a three-phase induction motor. It employs an ingenious scheme of placing three identical phase windings 120 -degr. electrical in space phase with respect to each other. When these windings are excited by a balanced three-phase power source, they set up a uniform magnetic field that revolves around the rotor periphery at synchronous speed. The flux-cutting action induces an electromotive force (emf) and thereby a current in the rotor conductors. The interaction of the rotor current and the revolving magnetic field causes the rotor to rotate at a speed somewhat lower than the synchronous speed of the motor. What happens when one of the three phases of the source is not connected to the motor? Is the rotor able to start from its standstill position and take up some mechanical load? In fact, we have already answered this question in Section 3. The motor behaves more like a two-phase motor and produces a revolving field that induces an emf and the current in the rotor conductors. The motor develops a starting torque and forces the rotor to rotate. Figure shows: Cut-away view of a single-phase, PSC, induction motor. Let us go one step further. What happens if only one of the three phases is excited? At this time you may be surprised to know that if the rotor is already rotating, it continues to rotate in that direction. However, if the motor is at rest and then one of the three phases of the motor is excited by a single-phase source, the rotor buzzes but does not rotate. Hence, a motor operating on a single phase is not capable of developing the starting torque. We will explain why this is so in the next section. A motor that operates on a single-phase source is called a single-phase induction motor. A single-phase induction motor requires only one single-phase winding to keep the motor running. However, such a motor is not self-starting. Therefore, we must provide some external means to start a single-phase induction motor. Most single-phase induction motors are built in the fractional-horsepower range and are used in heating, cooling, and ventilating systems. A properly designed direct-current (dc) series motor can be made to operate on both dc and alternating-current (ac) sources. For this reason, it’s appropriately called a universal motor. These motors operate at relatively high speeds and are an integral part of such units as vacuum cleaners, food blenders, and portable electric tools such as saws and routers. In this section we deal with many different types of single-phase induction motors. The operation of the universal motor is also explained. We have reserved another section for the study of specialty motors. We begin our study by explaining how a single-phase motor runs. Fgr1 A 2-pole, single-phase induction motor. Connecting wires.
In Section 3 we made it clear that in order to make the motor self-starting there must at least be two phase windings placed in space quadrature and excited by a two-phase source. This fact plays a major role in the design of all single-phase induction motors, as discussed later. A cross-sectional view of a 2-pole, single-phase induction motor with a squirrel-cage rotor is given in Fgr1. For clarity, we have placed the rotor conductors on the outer periphery of the rotor. In an actual motor, the rotor conductors are embedded in the rotor slots. Let us suppose that the supply voltage is increasing in the positive direction and causes a current in the stator (field) windings, as indicated in Fgr1, while the rotor is at rest. The current in each winding produces a magnetic field that is also increasing in the upward direction. Since two rotor conductors that are 180" electrical apart form a closed loop, the rotor conductors can then be paired as shown. Let us examine one of the closed loops, say the loop formed by conductors 2 and 2'. The flux passing through this loop induces an emf and thereby a current in this loop. The direction of the current in the loop is such that it produces a magnetic flux which tends to oppose the increase in the magnetic flux set up in the windings. For this to happen, the current must flow out of conductor 2 and into conductor 2', as shown by the dot and the cross, respectively. Ln the same way, we can determine the currents in the other conductors. Each current-carrying conductor must experience a force in accordance with the Lorentz force equation. The direction of the force acting on each conductor is also indicated in Fgr1. The forces experienced by conductors 1, 2, 3, and 4 in unison with conductors 1', 2', 3', and 4' tend to rotate the rotor in the counterclockwise direction. However, the rotation is opposed by the forces acting on the remaining conductors. The symmetric placement of the rotor conductors ensures that the motor develops equal torque in both directions and the net torque developed by it’s zero. Hence the rotor remains in its standstill position. As mentioned earlier, if the rotor is made to rotate in any direction while the single-phase winding is excited, the motor develops torque in that direction. Two theories have been put forth to explain this experimental fact: the double revolving-field theory and the cross-field theory. In this book, we confine our discussion to the double revolving-field theory.
According to this theory, a magnetic field that pulsates in time but is stationary in space can be resolved into two revolving magnetic fields that are equal in magnitude but revolve in opposite directions. Let us consider the standstill condition of the rotor again. The magnetic field produced by the motor pulsates up and down with time, and at any instant its magnitude may be given as B = B, cos of (eqn.1) … where B, is the maximum flux density in the motor. The flux density B can be resolved into two components B, and B, such that the magnitude of B, is equal to the magnitude of B,. Thus, B, = B, = 0.5 B. If we assume that B, rotates in the clockwise direction, the direction of rotation of B, is counterclockwise, as illustrated in Fgr2. Fgr2 Resolution of a pulsating vector into two equal and oppositely revolving vectors. Fgr3 An equivalent circuit of a single-phase induction motor at rest. Forward branch; Backward branch. We now have two revolving fields of constant but equal magnitude rotating synchronously in opposite directions. An emf is induced in the rotor circuit owing to each revolving field. The polarity of the induced emf in the rotor due to one revolving field is in opposition to the other. Thus, the rotor currents induced by the two revolving fields circulate in opposite directions. However, at standstill, the slip in either direction is the same (s = 1) and so is the rotor impedance. Therefore, the starting currents in the rotor conductors are equal and opposite. In other words, the starting torque developed by each revolving field is the same. Since the direction of the starting torque developed by one revolving field opposes the other, the net torque developed by the motor is zero. This is the same conclusion we arrived at before. However, we have gained some insight into the operation of a single-phase induction motor according to the double revolving-field theory. We can look upon a single-phase induction motor as if it consists of two motors with a common stator winding but with rotors revolving in opposite directions. At standstill the two rotors develop equal torques in opposite directions, and the net torque developed is zero. With that understanding we can develop an equivalent circuit of a single-phase induction motor at standstill, as shown in Fgr3. Note that the magnetization reactance and the rotor impedance at standstill have been resolved into two sections to highlight the effect of two coupled rotors. For the sake of simplicity we treat core loss as a part of the rotational loss. The core-loss resistance is, therefore, omitted from the equivalent circuit. If the core-loss resistance is known, it can be placed across the power source as we did for the approximate equivalent circuit of a transformer or a three-phase induction motor. One section of the rotor circuit is usually referred to as the forward branch, and the other is called the backward branch. When the motor rotates, say in the clockwise direction, the forward branch represents the effect of the revolving field in that direction. In this case, the backward branch corresponds to the rotor circuit associated with the counterclockwise revolving field. At standstill, both branches have the same impedance. The rotor circuit currents are also the same, and the same is true for the torques developed. Thus, when the rotor is at rest, the net torque developed by it’s zero. We usually speak of torque developed by a branch. What we really mean is the torque developed by the rotor resistance in that particular branch. Let us now assume that the rotor is rotating in the clockwise direction with a speed Nm. The magnetic field revolving in the clockwise direction has a synchronous speed of N, (N, = 12(?f/P). The synchronous speed of the revolving field in the counterclockwise direction is then -Ns. The per-unit slip in the forward (clockwise) direction is: The per-unit slip in the backward (counterclockwise) direction is ... Note that at standstill, N, = 0 and s = sb = 1. We can now incorporate the effect of slips in the forward and the backward rotor branches as we did for the three-phase induction motor. The modified equivalent circuit is given in Fgr4. In our discussion of three-phase induction motors we found that the torque developed is proportional to the effective resistance in the rotor branch. At standstill, s = sb = 1, the effective resistance in both branches of Fgr4 is the same. Thus, the torque developed by the two rotors is equal in magnitude but opposite in direction. That explains why there is no starting torque in a single-phase induction motor. On the other hand, let us now assume that the rotor is rotating with a slip s such that s < 1. The effective rotor resistance, RJs, in the forward branch is greater than that in the backward branch, RJ(2 -s). Thus, the torque developed by the forward branch is higher than that developed by the backward branch. The resultant torque is in the forward direction, and it tends to maintain the rotation in that direction. Thus, once a single-phase induction motor is made to rotate in any direction by applying an external torque, it continues rotating in that direction as long as the load torque is less than the maximum net torque developed by it. Fgr4 An equivalent circuit of a single-phase induction motor at any slip s. EXAMPLE 1 A 115-V, 60-Hz, 4-pole, single-phase induction motor is rotating in the clockwise direction at a speed of 1710 rpm. Determine its per-unit slip (a) in the direction of rotation and (b) in the opposite direction. If the rotor resistance at standstill is 12.5 a, determine the effective rotor resistance in each branch. (a) Slip in the forward direction is (b) Slip in the backward direction is The effective rotor resistances are - Forward branch
- Backward branch
Exercises: A 230-V, 50-Hz, 2-pole, single-phase induction motor is designed to operate at a slip of 3%. Determine the slip in the other direction. What is the speed of the motor in the direction of rotation? If the rotor resistance at standstill is 2.1 a, what is the effective resistance at the rated slip in either direction? The rotor resistance of a 208-V, SO-Hz, &pole, single-phase induction motor at standstill is 1.6 n. When the motor rotates at a slip of 5%, determine (a) its speed, (b) the effective rotor resistance in the forward branch, and (c) the effective rotor resistance in the backward branch.
From the equivalent circuit of a single-phase induction motor ( Fgr4), we obtain [...] as the effective impedance of the forward branch and [...] as the effective impedance of the backward branch. The simplified equivalent circuit -in terms of Zr and Z, is given in Fgr5. If Z, = R, + jX, is the impedance of the stator winding, the input impedance is [...] The stator winding current is ... The power input is ... ... where 8 is the power-factor angle by which the current il lags the applied voltage V, . The stator copper loss is ... When we subtract the stator copper loss from the total power input, we are left with the air-gap power. However, the air-gap power is distributed between ...the two air-gap powers: one due to the forward revolving field and the other due to the backward revolving field. In order to determine the air-gap power associated with each revolving field, we have to determine the rotor currents in both branches. If i2f is the rotor current in the forward branch, then Fgr5 Simplified equivalent circuit of a single-phase induction motor. Similarly, the rotor current in the backward branch f2b is ... Hence, the air-gap powers due to the forward and backward revolving fields are ... Since Rf and Rb are the equivalent resistances in the forward and backward branches of the rotor circuit, the power transferred to the rotor must also be consumed by these resistances. In other words, we can also compute the air-gap powers as ... ....where P, is the rotational loss of the motor. In this case, the rotational loss consists of the friction and windage loss, the core loss, and the stray-load loss. The load (shaft) torque of the motor is .... Fgr6 Speed-torque characteristic of a single-phase induction motor. Finally, the motor efficiency is the ratio of the power available at the shaft Po to the total power input Pi,,. We could also have computed the torque developed by the forward and the backward revolving fields as ... The net torque developed by the motor is .... The two expressions of torque developed, Eq. (16) and Eq. (10.21), are identical. We can use either expression to compute Td. The torques developed, Tdf and Tdb, are plotted in Fgr6. These curves are also extended into the region of negative speed. This is usually done to show .... ...the torque that must be overcome when the motor is driven in the backward direction by a prime mover. The following example illustrates how to use the above equations to determine the performance of a single-phase induction motor.
Each single-phase induction motor derives its name from the method used to make it self-starting. Some of the motors discussed in this section are split-phase motor, capacitor-start motor, capacitor-start capacitor-run motor, and permanent split-capacitor motor. Another induction motor discussed later in this section is called the shaded-pole motor. For an induction motor to be self-starting, it must have at least two phase windings in space quadrature and must be excited by a two-phase source, as detailed in Section 3. The currents in the two phase windings are 90-degree electrical out of phase with each other. The placement of the two phase windings in space quadrature in a single-phase motor is no problem. However, the artificial creation of a second phase requires some basic understanding of resistive, inductive, and capacitive networks. Let us now examine how the second phase is created in each induction motor.
This is one of the most widely used induction motors for mechanical applications in the fractional horsepower range. The motor employs two separate windings that are placed in space quadrature and are connected in parallel to a single-phase source. One winding, known as the main winding, has a low resistance and high inductance. This winding carries current and establishes the needed flux at the rated speed. The second winding, called the auxiliary winding, has a high resistance and low inductance. This winding is disconnected from the supply when the motor attains a speed of nearly 75% of its synchronous speed. A centrifugal switch is commonly used to disconnect the auxiliary winding from the source at a predetermined speed. The disconnection is necessary to avoid the excessive power loss in the auxiliary winding at full load. At the time of starting, the two windings draw currents from the supply. The main-winding current lags the applied voltage by almost 90-degr. owing to its high inductance (large number of turns) and low resistance (large wire size). The auxiliary-winding current is essentially in phase with the applied voltage owing to its high resistance (small wire size) and low inductance (few number of turns). As you may suspect, the main-winding current does not lag exactly by 90-degr., nor is auxiliary-winding current precisely in phase with the applied voltage. In addition, the two phase-winding currents may also not be equal in magnitude. In a well-designed split-phase motor, the phase difference between the two currents may be as high as 60-degr.. It’s from this phase-splitting action that the split-phase motor derives its name. Since the two phase-windings are wound in space quadrature and carry out-of-phase currents, they set up an unbalanced revolving field. It’s this revolving field, albeit unbalanced, that enables the motor to start. The starting torque developed by a split-phase motor is typically 150% to 200% of the full-load torque. The starting current is about 6 to 8 times the full-load current. The schematic representation of a split-phase motor is given in Fgr7 along with its typical speed-torque characteristic. Note the drop in torque at the time the auxiliary winding is disconnected from the supply. Fgr7 (a) Schematic representation and (b) speed-torque characteristic of a split-phase motor. Percent speed; Auxiliary winding ; Rotation; Percent speed; Both windings
In a capacitor-start motor a capacitor is included in series with the auxiliary winding. If the capacitor value is properly chosen, it’s possible to design a capacitor-start motor such that the main-winding current lags the auxiliary-winding current by exactly 90-degr.. Therefore, the starting torque developed by a capacitor motor can be as good as that of any polyphase motor. Once again, the auxiliary winding and the capacitor are disconnected at about 75% of the synchronous speed. Therefore, at the rated speed the capacitor-start motor operates only on the main winding like a split-phase induction motor. The need for an external capacitor makes the capacitor-start motor somewhat more expensive than a split-phase motor. However, a capacitor-start motor is used when the starting torque requirements are 4 to 5 times the rated torque. Such a high starting torque is not within the realm of a split-phase motor. Since the capacitor is used only during starting, its duty cycle is very intermittent. Thus, an inexpensive and relatively small ac electrolytic-type capacitor can be used for all capacitor-start motors. A schematic representation of a capacitor-start motor and its speed-torque characteristic are given in Fgr8. Fgr8 (a) Schematic representation and (b) speed-torque characteristic of a cap-start motor.
Although the split-phase and capacitor-start motors are designed to satisfy the rated load requirements, they have low power factor at the rated speed. The lower the power factor, the higher the power input for the same power output. Thus, the efficiency of a single-phase motor is lower than that of a polyphase induction motor of the same size. For example, the efficiency of a capacitor-start or a split-phase single-phase motor is usually 50% to 60% in the fractional horsepower range. On the other hand, for the same application, a three-phase induction motor may have an efficiency of 70%) to 80%. The efficiency of a single-phase induction motor can be improved by employing another capacitor when the motor runs at the rated speed. This led to the development of a capacitor-start capacitor-run (CSCR) motor. Since this motor requires two capacitors, it’s also known as the two-value capacitor motor. One capacitor is selected on the basis of starting torque requirements (the start capacitor), whereas the other capacitor is picked for the running performance (the run capacitor). The auxiliary winding stays in circuit at all times, but the centrifugal switch helps in switching from the start capacitor to the run capacitor at about 75% of the synchronous speed. The start capacitor is of the ac electrolytic type, whereas the run capacitor is of an ac oil type rated for continuous operation. Since both windings are active at the rated speed, the run capacitor can be selected to make the winding currents truly in quadrature with each other. Thus, a CSCR motor acts like a two-phase motor both at the time of starting and at its rated speed. Although the CSCR motor is more expensive because it uses two different capacitors, it has relatively high efficiency at full load compared with a split-phase or capacitor-start motor. A schematic representation of a CSCR motor and its speed-torque characteristic are given in Fgr9. Fgr9 (a) Schematic representation and (b) speed-torque characteristic of a cap-start cap-run motor. Auxiliary; Percent of speed. Fgr10 (a) Schematic representation and (b) speed-torque characteristic of a permanent-split capacitor motor.
A less expensive version of a CSCR motor is called a permanent split-capacitor (PSC) motor. A PSC motor uses the same capacitor for both starting and full load. Since the auxiliary winding and the capacitor stay in the circuit as long as the motor operates, there is no need for a centrifugal switch. For this reason, the motor length is smaller than for the other types discussed above. The capacitor is usually selected to obtain high efficiency at the rated load. Since the capacitor is not properly matched to develop optimal starting torque, the starting torque of a PSC motor is lower than that of a CSCR motor. PSC motors are, therefore, suitable for blower applications with minimal starting torque requirements. These motors are also good candidates for applications that require frequent starts. Other types of motors discussed above tend to overheat when started frequently, and this may badly affect the reliability of the entire system. With fewer rotating parts, a PSC motor is usually quieter and has a high efficiency at full load. The schematic representation of a PSC motor and its speed-torque characteristic are given in Fgr10.
Fgr11 Equivalent circuit of a PSC motor. We already know how to determine the performance of a single-phase motor running on the main winding only. This analysis can be used to determine the performance at full load of a split-phase or a capacitor-start induction motor. However, the analysis of a single-phase motor is not complete without knowledge of its starting torque. The questions that still remain unanswered are the following: What is the starting torque developed by a split-phase or a capacitor-start motor? What is the voltage drop across the capacitor at starting? Is it within the maximum allowable limits? What is the current in-rush at the time of starting the motor? Does it cause severe fluctuations on the line? How can we determine the performance of a motor that uses both windings at all times? How does the switching of a capacitor from one value to another affect the performance of the motor prior to and immediately thereafter? It must be clear by now that we cannot answer these questions based upon the information obtained from the single-winding analysis. Therefore, our analysis must include both windings. Before proceeding further, some assumptions that are commonly accepted in this area are as follows: The main and the auxiliary windings are in space quadrature with each other. This assumption implies that the flux produced by one winding does not induce an emf in the other winding. In other words, no transformer action exists between the two windings. If we define the a-ratio as the ratio of the effective number of turns in the auxiliary winding to the effective number of turns in the main winding, then the leakage reactance, the magnetization reactance, and the rotor resistance for the auxiliary winding can be defined in terms of the main-winding parameters and the a-ratio. When both main and the auxiliary windings are excited, they produce their own forward and backward revolving fields. Consequently, there are four revolving fields in a two-winding single-phase motor. Each winding can be represented by an equivalent circuit with two parallel branches, one for the forward branch and the other for the backward branch. A revolving field induces emf in both windings. It does not really matter which winding sets up that revolving field. In other words, the forward and the backward revolving fields of the auxiliary winding induce emfs in the main winding and vice versa. It’s common to refer to these induced emfs as the speed voltages. We assume that the main winding is displaced forward in space by 90-degr. electrical with respect to the auxiliary winding. The forward field created by the auxiliary winding induces an emf in the main winding that lags by 90-degr. the emf induced by the same field in the auxiliary winding. This is a very important concept and must be clearly understood in order to properly account for the induced emfs in one winding by the revolving fields of the other winding. The above assumptions allow us to represent a two-winding single-phase motor by an equivalent circuit, as given in Fgr11, where ... R, = resistance of the main winding X, = leakage reactance of the main winding a = ratio of effective turns in the auxiliary winding to effective turns in R, = rotor resistance as referred to the main winding at standstill X, = rotor leakage reactance as referred to the main winding X,,, = the magnetization reactance of the motor as referred to the main winding R, = resistance of the auxiliary winding El = induced emf in the forward branch of the main winding by the forward revolving field of the auxiliary winding - E, = induced emf in the backward branch of the main winding by the backward revolving field of the auxiliary winding - E, = induced emf in the forward branch of the auxiliary winding by the forward revolving field of the main winding - E, = induced emf in the backward branch of the auxiliary winding by the main winding backward revolving field of the main winding The other parameters of the auxiliary winding have been defined in terms of the a-ratio. The equivalent circuit shown applies strictly to a PSC motor. We replace the capacitor impedance, -jXc, by a short circuit if we want to analyze a split-phase motor. For a CSCR motor, -jX, has two different values, one for the start capacitor and the other for the run capacitor. In summary, for a split-phase motor … Z, = R, … and the auxiliary winding is in circuit for speeds below the operating speed of the centrifugal switch. Thereafter the motor operates on the main winding only. For a capacitor-start motor Z, = R, - jX where X, is the reactance of the start capacitor. The auxiliary winding is included in the analysis as long as the speed is less than the operating speed of the centrifugal switch. After that the motor operates only on the main winding. For a CSCR motor ...as long as the motor speed is below the operating speed of the centrifugal switch. Thereafter, ...where X, is the reactance of the run capacitor. (d) For a PSC motor ... where X, is the reactance of the capacitor in the auxiliary circuit. In this case, both windings are in circuit at all times. The forward and backward impedances of the main winding are ... A simplified equivalent circuit in terms of 2, and ib of a two-winding single The induced emfs in the main winding by its forward and backward revolving phase induction motor is given in Fgr12. fields are ... The induced emf in the auxiliary winding by its forward and backward revolving fields are .... Since the main winding is displaced 90-degr. electrical ahead of the auxiliary winding, the induced emf in the main winding by the forward revolving field of the auxiliary winding must lag by 90-degr. the induced emf in the auxiliary. In addition, the induced emf in the main winding must be l/a times the induced emf in the auxiliary. That is, ... Fgr12 Simplified version of Fgr1 1. By the same token, the induced emf in the main winding by the backward revolving field set up by the auxiliary winding must lead by 90-degr. the emf it induces in the auxiliary winding. Thus, ... Similarly, the induced emfs in the forward and backward branches of the auxiliary winding by the forward and backward revolving fields of the main winding are .... Since all the induced emfs are now known, the application of Kirchhoff's voltage law to the coupled circuit yields .... After substituting for the induced emfs, we can express the above equations in concise form as ... The currents in the main and the auxiliary windings are .... The line current is ... The power supplied to the motor is .... ...where 0 is the power-factor angle by which the line current lags the applied voltage. The stator copper losses for both the windings are ... By subtracting the stator copper losses from the power supplied to the motor, we obtain the air-gap power. The air-gap power is distributed among the four revolving fields in the motor. We can also write an expression for the air-gap power just as we did for the motor operating on the main winding only. However, we have to take into account the presence of speed voltages and the power associated with them. On this basis, the air-gap power developed by the forward revolving field of the main winding is .... Similarly, the air-gap power produced by the forward revolving field of the auxiliary winding is ... The net air-gap power due to both forward revolving fields is .... By the same token, the air-gap power developed by the backward revolving fields is .... Hence, the net air-gap power developed by the motor is .... At standstill (i.e., the blocked-rotor condition, or at the time of starting) the per-unit slip of the motor is unity. The rotor impedance in the forward and the backward branches is the same. The net air-gap power developed by the motor, from the above equation when Rf = Rb, is .... Note that the net power developed at the time of starting is proportional to the sine of the angle between the currents in the two windings. The power developed is maximum when the 8 is 90 -degr.. For split-phase motors, 8 may be as low as 30 -degr. and as high as 60 -degr.. For capacitor motors, 8 is usually close to 90 -degr.. This is why a capacitor motor can develop more starting torque than a split-phase motor of the same size. EXAMPLE 3: A 230-V, SO-Hz, 6-pole, single-phase PSC motor is rated at 940 rpm. The equivalent circuit parameters for the motor are R, = 34.14 R, X, = 35.9 a, R, = 149.78 a, X, = 29.32 R, X, = 248.59 R, R, = 23.25 a, a = 1.73, and C = 4 p,F, rated at 440 V. The core loss is 19.88 W, and the friction and windage loss is 1.9 W. Determine (a) the line current, (b) the power input, (c) the efficiency, (d) the shaft torque, (e) the voltage drop across the capacitor, and (f) the starting torque. Since the voltage drop across the capacitor is higher than its normal rating of 440 V, for continuous operation either the capacitor with a higher voltage rating must be used or the windings must be redesigned. (f) For the starting torque, s = 1 and Sb = 1. Using these values and following the above steps, we obtain ....
A 120-V, 60-Hz, C_pole, single-phase, capacitor-start motor has a main-winding impedance of 6 + j50 R and an auxiliary-winding impedance of 5 + j25 R. The rotor impedance at standstill is 8 + j5 R. The magnetization reactance is 150 R. If the starting capacitor is 100 pF, determine the starting torque developed by the motor. What is the torque developed by the motor in Exercise 5 when it operates at a full-load speed of 1575 rpm? Determine its efficiency if the rotational loss is 10 W. Note that the motor runs on the main winding only at its rated speed. |

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