Electric Motor Control
10th Edition
ISBN: 9781133702818
Author: Herman
Publisher: CENGAGE L
expand_more
expand_more
format_list_bulleted
Question
Chapter 40, Problem 9SQ
To determine
Explain the reason for the operation of field contacts (F) overlapped in Figure 40-1.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Calculate the current magnitude in the coils e1, e2 of theMagnetic circuit, if:ɸa = 3.00 x 10^-3 Wb, φb = 0.80 x 10^-3 Wb, ɸc = 2.20 x 10^-3 Wb
L ab = 0.10 m,A ab = 5.0 cm^2L afeb = L acdb = 0.40 mA afeb = A acdb = 20 cm^2
MATERIAL CHARACTERISTICSH (At/m) 240 350 530 1300 5000 9000B (T) 0.7 0.9 1.1 1.3 1.5 1.6
A toroid magnetic circuit is composed of three sections A, B and C, thesection C has an air gap, section A has an 850 round coil thatconsumes a current of 1.2 A. the physical and magnetic properties of each sectionare:
Section A: Length = 80 mm, Cross section = 120 mm^2, μr = 400
Section B: Length = 60 mm, Cross section = 40 mm^2, μr = 250
Section C: Length = 50 mm, Cross section = 200 mm^2, μr = 600
Gap: Length = 1 mm, Cross section = 40 mm^2, μr = 1
Calculate:The magnetic field density in each of the sections
3) Compute the input impedance of Fig. 3. (10 points)
Rin
R1
R₂
Figure 3
T
Vcc
Chapter 40 Solutions
Electric Motor Control
Ch. 40 - What are the two basic functions of synchronous...Ch. 40 - Prob. 2SQCh. 40 - If all of the control circuits are ACand DC is not...Ch. 40 - When the rotor and stator fields lock into step,...Ch. 40 - Prob. 5SQCh. 40 - Prob. 6SQCh. 40 - In Figure 402, approximately what is the...Ch. 40 - Why are the AC ammeter and DC meter necessary in...Ch. 40 - Prob. 9SQ
Knowledge Booster
Similar questions
- Show the workarrow_forward2) A bypass capacitor CE in parallel with RE is added to the above circuit. a) Draw the equivalent small-signal circuit. (10 points) b) Find the input resistance Rib looking into the base. (10 points) c) Find the output resistance looking into the collector, while the source is shorted, i.e. Vs 0 V and Rs = 0 2. (10 points) Vo Vs d) Find the voltage gain A₁ = ✓ using the above equivalent small signal circuit. (10 points)arrow_forwardhelp about this question in control systems?arrow_forward
- A conductor 300 mm long carries a current of 13A and is at right-angles to a magnetic fieldbetween two circular pole faces, each of diameter 80 mm. If the total flux between the polefaces is 0.75 mWb, calculate the force exerted on the conductor. [ANS = 0.582 N]arrow_forwarda) find Rthb) Find Vth in the circuit c)Draw the Thevenin Equivalent of the circuit to tge left of the a and b terminalsarrow_forwardAn electric car runs on batteries, but needs to make constant stops to re-charge. If a trailer is attached to the car that carries a generator, and the generator is turned by a belt attached to the wheels of the trailer, will the car be able to drive forever without stopping?arrow_forward
- A singl core cable of voltage 30 kv. The diameter of Conductor is 3 cm. The diameter of cable is 25 cm. This cable has Two layer of insulator having arelative permittivity 5-3 respectively of The ratio of maximum electric stress of maximum electric stress 8 First layer to the of second layer is 10 Find & 1- The thickness of each layers. 3- The voltage of each layers. §. Layers The saving in radius of cable if another ungrading cable has the Same maximum electric stress, Total village, Conductor diameter of grading cable.arrow_forward66 KV sing care Cable has a drameter of conductor of 3 cm. The radius of cable is 10 cm. This Cable house Two relative permmitivity of insulation 6 and 4 respectively. If The ratio of maximum electric stress of first layer to the maximum eledric streep & second layer is s 1- find the village & each layers. 2- Min- electric stress J Cable 3- Compare the voltage of ungrading Cable has the same distance and relectric stresses.arrow_forwardPrelab Information 1. Laboratory Preliminary Discussion First-order Low-pass RC Filter Analysis The first-order low-pass RC filter shown in figure 1 below represents all voltages and currents in the time domain. It is of course possible to solve for all circuit voltages using time domain differential equation techniques, but it is more efficient to convert the circuit to its s-domain equivalent as shown in figure 2 and apply Laplace transform techniques. vs(t) i₁(t) + R₁ ww V₁(t) 12(t) Lic(t) Vout(t) = V2(t) R₂ Vc(t) C Vc(t) VR2(t) = V2(t) + Vs(s) Figure 1: A first-order low-pass RC filter represented in the time domain. I₁(s) R1 W + V₁(s) V₂(s) 12(s) Ic(s) + Vout(S) == Vc(s) Vc(s) Zc(s) = = VR2(S) V2(s) Figure 2: A first-order low-pass RC filter represented in the s-domain.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Delmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage Learning
Electricity for Refrigeration, Heating, and Air C...Mechanical EngineeringISBN:9781337399128Author:Russell E. SmithPublisher:Cengage Learning
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning
Electricity for Refrigeration, Heating, and Air C...
Mechanical Engineering
ISBN:9781337399128
Author:Russell E. Smith
Publisher:Cengage Learning