Electric Motor Control
10th Edition
ISBN: 9781305177611
Author: Herman
Publisher: Cengage
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 39, Problem 1SQ
To determine
Explain the reason why both rotor and stator should possess equal number of poles.
Expert Solution & Answer
Answer to Problem 1SQ
The reason why both rotor and stator should possess equal number of poles is to synchronize with each other.
Explanation of Solution
An independent field is established around the rotor, which is energized by a direct current through slip rings mounted on the shaft. The rotor has the same number of coils as the stator, so that each and every rotor pole (north and south) can have an alternative stator poles (north and south), and synchronize with each other. Therefore, the motor’s stator and rotor are in relative speed due to magnetism.
Conclusion:
Thus, the reason why both rotor and stator should possess equal number of poles is to synchronize each other.
Want to see more full solutions like this?
Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Section 15-3 reversing motors using magnetic starters.Section 15-3Use data sheet B on page 383 to draw the wiring diagram. Note: use only the number of contacts required.
First 1. Wire the motor to operate in forward and reverse at 115 VAC.
Please solution
Use data sheet B on page 383 to draw the wiring diagram. Note: use only the number of contacts required.
First 1. Wire the motor to operate in forward and reverse at 115 VAC.
Chapter 39 Solutions
Electric Motor Control
Ch. 39 - Prob. 1SQCh. 39 - What is the effect of the starting winding of the...Ch. 39 - What are typical applications of synchronous...Ch. 39 - Prob. 4SQCh. 39 - A loaded synchronous motor cannot operate...Ch. 39 - Why must a discharge resistor be connected in the...Ch. 39 - Depending on their power factor ratings, what is...Ch. 39 - At what power factor do incandescent lights...Ch. 39 - Prob. 9SQCh. 39 - The speed of a synchronous motor is fixed by the...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.Similar questions
- B:A 20 MVA transformer which may be called upon to operate at 30% overload, feeds 11 KV busbars through a circuit breaker: other circuit breakers supply outgoing feeders. The transformer circuit breaker is equipped with 1000/5 A CTS and the feeder circuit breakers with 400/5 A CTS and all sets of CTs feed induction type over current relays. The relays on the feeder circuits breakers have a 125% plug setting, and 0.3 time setting. If 3 ph fault current of 5000 A flows from the transformer to one of the feeders, find the operating time of the feeder relay, the minimum plug setting of the transformer relay and its time setting assuming a discrimination time margin of 0.5 sec. Relays having the following characteristics for TMS=1 PSM T in sec. 2 3.6 5 10 15 20 10 6 3.9 2.8 2.2 2.1arrow_forward10.34 Determine the power readings of the two wattmetersshown in the circuit of Fig. P10.34 given that ZY = (15− j5) Warrow_forward10.29 A 208-V (rms) balanced three-phase source supports twoloads connected in parallel. Each load is itself a balanced threephaseload. Determine the line current, given that load 1 is 12 kVAat pf 1 = 0.7 leading and load 2 is 18 kVA at pf 2 = 0.9 lagging.arrow_forward
- 10.31 A 240-V (rms), 60-Hz Y-source is connected to a balancedthree-phase Y-load by four wires, one of which is the neutral wire.If the load is 400 kVA at pf old = 0.6 lagging, what size capacitorsshould be added to change the power factor to pf new = 0.95lagging?arrow_forwardCable A Cable A is a coaxial cable of constant cross section. The metal regions are shaded in grey and are made of copper. The solid central wire has radius a = 5mm, the outer tube inner radius b = 20mm and thickness t = 5mm. The dielectric spacer is Teflon, of relative permittivity &r = 2.1 and breakdown strength 350kV/cm. A potential difference of 1kV is applied across the conductors, with centre conductor positive and outer conductor earthed. Before undertaking any COMSOL simulations we'll first perform some theoretical analysis of Cable A based on the EN2076 lectures, to make sense of the simulations. Calculate the radial electric field of cable A at radial positions r b. Also calculate the maximum operating voltage of cable A, assuming a safety margin of ×2, and indicate where on the cable's cross section dielectric breakdown is most likely to occur.arrow_forward: For the gravity concrete dam shown in the figure, the following data are available: The factor of safety against sliding (F.S sliding)=1.2 Unit weight of concrete (Yconc)=24 KN/m³ - Neglect( Wave pressure, silt pressure, ice force and earth quake force) μ=0.65, (Ywater) = 9.81 KN/m³ Find factor of safety against overturning (F.S overturning) 6m3 80m Smarrow_forward
- I need help checking if its correct -E1 + VR1 + VR4 – E2 + VR3 = 0 -------> Loop 1 (a) R1(I1) + R4(I1 – I2) + R3(I1) = E1 + E2 ------> Loop 1 (b) R1(I1) + R4(I1) - R4(I2) + R3(I1) = E1 + E2 ------> Loop 1 (c) (R1 + R3 + R4) (I1) - R4(I2) = E1 + E2 ------> Loop 1 (d) Now that we have loop 1 equation will procced on finding the equation of I2 current loop. However, a reminder that because we are going in a clockwise direction, it goes against the direction of the current. As such we will get an equation for the matrix that will be: E2 – VR4 – VR2 + E3 = 0 ------> Loop 2 (a) -R4(I2 – I1) -R2(I2) = -E2 – E3 ------> Loop 2 (b) -R4(I2) + R4(I1) - R2(I2) = -E2 – E3 -----> Loop 2 (c) R4(I1) – (R4 + R2)(I2) = -E2 – E3 -----> Loop 2 (d) These two equations will be implemented to the matrix formula I = inv(A) * b R11 R12 (R1 + R3 + R4) -R4 -R4 R4 + R2arrow_forward10.2 For each of the following groups of sources, determineif the three sources constitute a balanced source, and if it is,determine if it has a positive or negative phase sequence.(a) va(t) = 169.7cos(377t +15◦) Vvb(t) = 169.7cos(377t −105◦) Vvc(t) = 169.7sin(377t −135◦) V(b) va(t) = 311cos(wt −12◦) Vvb(t) = 311cos(wt +108◦) Vvc(t) = 311cos(wt +228◦) V(c) V1 = 140 −140◦ VV2 = 114 −20◦ VV3 = 124 100◦ Varrow_forwardApply single-phase equivalency to determine the linecurrents in the Y-D network shown in Fig. P10.13. The loadimpedances are Zab = Zbc = Zca = (25+ j5) Warrow_forward
- 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
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
