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## Elements of Electrical Engr II

by: Dr. Reina Hane

10

0

6

# Elements of Electrical Engr II ECE 207

Dr. Reina Hane
RHIT
GPA 3.79

Staff

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COURSE
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Staff
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Class Notes
PAGES
6
WORDS
KARMA
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## Popular in Electrical Engineering & Computer Science

This 6 page Class Notes was uploaded by Dr. Reina Hane on Monday October 19, 2015. The Class Notes belongs to ECE 207 at Rose-Hulman Institute of Technology taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/225088/ece-207-rose-hulman-institute-of-technology in Electrical Engineering & Computer Science at Rose-Hulman Institute of Technology.

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Date Created: 10/19/15
81 Magnetic circuits The source of magnetic flux is current Not just current but the product of current and the number of times the current is wrapped around a core NI The units for NI are ampereturns W This quantity magnetomotive force MMF is analogous to voltage in electric circuits 5N Magnetic ux 4 plays the role of current By convention the pole from which magnetic ux leaves is called the north pole and the one to which it enters is called the south pole The earth is a notable exception In magnetic circuits eluctance l2 plays the role that resistance does in electric circuits Like resistance reluctance is determined by material property and geometry The inductance of a coil is determined by the core s reluctance and the number of turns L N2l2 Example iii Determine the inductance ofthe coil i Find the current needed to produce a ux of5 me in the air gap ii What is the direction of the ux in the core based on the shown direction of the current 83 Relays and solenoids These are devices that produce mechanical movement from an electrical signal Its basic components are coil which when connected to a supply creates the MMF which is the source of magnetic flux in the core core which carries the flux from the coil to the airgap sometimes more than one airgap is used The core is usually fixed and has a low reluctance armature which carries the flux between the airgaps It is movable and like the core has a low reluctance The armature is usually connected to the mechanical device that has to be actuated eg shutoff valve spring which holds the armature in the nonactuated position Armature J L l moveable V l Spring WW UUUUK Core fixed In the ideal case the reluctance of the steel in the core and armature is negligible this means that all MMF is dropped across the two airgaps which are in series 222g The magnetic circuit is The energy stored by an inductance is ELF Using this to find the energy stored in each air gap Now taking the derivative with respect to the gap usually denoted as X or g the force per air gap is obtained 85 Example For the relay shown previously the spring exerts a force of 02 N The gap length is 5mm when fully open and 2mm when fully closed The coil has 5000 turns and is wound on a core of 1cm square cross section Predict the pickup and dropout currents Example A lifting magnet is shown on the left The core has a square crosssection of6 X 6 cm2 The coil has 300 turns and a resistance of6Q Determine the lifting force when the airgap is 5mm and a 120V DC supply is used Neglect the reluctance ofthe core and the mass being lifted V030 Mass to be lifted magnetic material Example Repeat with a 120 V 60 Hz AC supply instead of DC The reactance of the AC coil reduces the current which greatly reduces the lifting force Lifting magnets are nearly always DC

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