
Concept explainers
Explain the process how 230/460-volt motor is used as a part winding motor.

Explanation of Solution
Part winding motor:
The part winding motor has two sets of identical windings which are connected in parallel with each other. The two identical windings are energized in a sequence to produce the reduced starting current and the reduced starting torque.
Dual voltage motor:
The dual voltage motor has two sets of windings (not identical) which are connected either in parallel or in series. When the 230/460-volt motor operates with the part winding starter, it is intended to connect the two sets of windings in parallel (voltage ratio is 1:2). An each set of windings is connected either in Y or delta connection.
The dual voltage motor can operate with different power supply voltages. When the dual voltage motor’s two sets of windings are connected in parallel, the motor can operate with low voltage connection. When the windings are connected in series, the motor can operate with high voltage.
230/460-volt motor as a part winding motor:
230/460-volt dual voltage motor can be used as a part winding motor with low voltage connection. At the starting of the motor, the low voltage supply is connected to the one set of winding. As soon as the motor reaches to the normal speed, apply the power to the other set of winding (the other winding from the parallel windings) in the running conditions for the normal running operation of the motor. The motor runs with reduced starting current and the reduced starting torque after both the windings are energized in a sequence.
The part winding starter is a reduced voltage starter and used in the starting process of the 230/460-volt part winding motor.
Refer to Figure 28.3 in textbook where the 230/460-volt motor can be used as a part winding motor with low voltage connection. One set of winding terminals
Figure 1 shows the low voltage connection of a part winding motor.
Conclusion:
Thus, the 230/460-volt motor can be used as a part winding motor.
Want to see more full solutions like this?
Chapter 28 Solutions
Electric Motor Control
- 10.8 In the network of Fig. P10.8, Za = Zb = Zc = (25+ j5) W.Determine the line currents.arrow_forwardUsing D flip-flops, design a synchronous counter. The counter counts in the sequence 1,3,5,7, 1,7,5,3,1,3,5,7,.... when its enable input x is equal to 1; otherwise, the counter count 0. Present state Next state x=0 Next state x=1 Output SO 52 S1 1 S1 54 53 3 52 53 S2 56 51 0 $5 5 54 S4 53 0 55 58 57 7 56 56 55 0 57 S10 59 1 58 58 S7 0 59 S12 S11 7 $10 $10 59 0 $11 $14 $13 5 $12 S12 $11 0 513 $15 SO 3 S14 $14 S13 0 $15 515 SO 0 Explain how to get the table step by step with drawing the state diagram and finding the Karnaugh map.arrow_forwardFor the oscillator resonance circuit shown in Fig. (5), derive the oscillation frequency Feedback and open-loop gains. L₁ 5 mH (a) ell +10 V R₁ ww R3 S C2 HH 1 με 1000 pF 100 pF R₂ 1 με RA H (b) +9 V R4 CA 470 pF C₁ R3 HH 1 με R₁ ww L₁ 000 1.5 mH R₂ ww Hi 1 μF L2 m 10 mHarrow_forward
- Expert handwritten solution onlyarrow_forwardB. For the oscillator circuit shown in frequency, feedback and open-loop gains. +10 V name the circuit, derive and find the oscillation P.Av +9 V -000 4₁ 5 mH w R₁ C₂ HH 1 με w 100 pF R₂ T R CA www. 470 pF w ww www 1000 pF HH 1μF C₁ HH 1μF Ra ww HI 4₁ 000 1.5 mH H 4 AF 000 10 mHarrow_forwardI want to check if the current that I have from using the mesh analysis is correct? I1 = 0.214mA I2 = -0.429mAarrow_forward
- I want to find the current by using mesh analysis pleasearrow_forwardI want to find the current by using mesh analysis pleasearrow_forwardR₁ W +10 V R3 +9 V C₂ R₁ CA C₁ 470 pF HH 1000 pF HH 1 με C4 1 μF 1 uF C₁ R₂ R4 100 pF Find Open-loop Jain L₁ 5 mH (a) Av=S,B={" H R₁₂ ✓ ww (b) R₁ L₁ 000 1.5 mH R₂ H 1 uF 12 10 mHarrow_forward
- A) Calculate the efficiency of the test transformer at the resistive loads (X-25%, 50%, 75%, 100%, 125% full load). B) From part (A) draw the plot (efficiency Vs power output) of the transformer. C) Discuss the plot of part (B).arrow_forwarda- Determine fH; and Ho b- Find fg and fr. c- Sketch the frequency response for the high-frequency region using a Bode plot and determine the cutoff frequency. Ans: 277.89 KHz; 2.73 MHz; 895.56 KHz; 107.47 MHz. 14V Cw=5pF Cwo-8pF Coc-12 pF 5.6kQ Ch. 40. pF C-8pF 68kQ 0.47µF Vo 0.82 kQ V₁ B=120 0.47µF www 3.3kQ 10kQ 1.2kQ =20µF Narrow_forwardUsing D flip-flops, design a synchronous counter. The counter counts in the sequence 1,3,5,7, 1,7,5,3,1,3,5,7,.... when its enable input x is equal to 1; otherwise, the counter. This counter is for individual settings only need the state diagram and need the state table to use 16 states from So to S15.arrow_forward
- Delmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage LearningElectricity for Refrigeration, Heating, and Air C...Mechanical EngineeringISBN:9781337399128Author:Russell E. SmithPublisher:Cengage Learning


