EE-0001 PEEE Refresher Course. Week 1: Engineering Fundamentals

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EE-000 PEEE efresher Course Week : Engineering Fundamentals

Engineering Fundamentals Bentley Chapters & Camara Chapters,, & 3 Electrical Quantities Energy (work), power, charge, current Electrostatic pressure, resistance Circuit Elements esistor, Inductor, Capacitor Circuit Equations Series, Parallel, Wye-Delta Conversions Norton Equivalent Thevenin Equivalent Kirchhoff s Laws Loop and Node equations Complex Algebra Transients & esonance, C, C, LC

3 Fundamental Electrical Units Quantity Units Symbol Formula Equivalent Units energy (work) joule W W Pdt watt-second power watt P dw P dt volt-ampere or joule/second charge coulomb Q Q CV ampere-second current ampere I V I coulomb/second electrostatic pressure volt V V I joule/coulomb resistance ohm V volt/ampere I

4 Voltage Energy gained or lost when a charge is moved from one point to another. Potential difference EMF V ab E V W Q dw dq dw dq P I joule coulomb V E energy given up or generated potential difference between two terminals

5 Power Time rate of doing work BTU/second Joule/second watts horsepower P VI P avg V I [ watts] voltage current t t 0 vi dt [ watts]

6 Energy (work) W t t P dt [ watt -sec] where P watts - or - W Pt if power is constant

7 Ohm s Law I - V I I V V I V I V

8 esistance The resistance of a section of conductor of uniform cross section is: ρl A where Across-sectional area (square meters or circular mils) l length (meters or feet) resistance (ohms) ρ resistivity of material (ohm-meters) /conductivity A l

9 esistivity for Typical Conductive Materials Material esistivity, ohm-meters aluminum.6 x 0-8 7 copper.7 x 0-8 0 cast iron 9.7 x 0-8 58 lead x 0-8 3 silver.6 x 0-8 9.9 steel (-90) x 0-8 66-540 tin x 0-8 69 nichrome 00 x 0-8 60 esistivity, ohm-circular mills per foot Note: The area of a circle one mill (0.00 inch) in diameter is one circular mil; the area of any circle in circular mils equals the square of its diameter in mils.

0 Circuit Element Characteristics Element Schematic Symbol Current Through Voltage Drop Power Dissipation Stored Energy Units resistor inductor capacitor I I - I - V V V - V I φ I L L V dt dv I C dt V I di V L dt V or I zero zero W LI V I dt zero W CV C ohm Ω, KΩ, MΩ henry h, mh, µh farad f, µf, pf

Metric Prefixes Multiplier Prefix Symbol 0-8 atto a 0-5 femto f 0 - pico p 0-9 nano n 0-6 micro µ 0-3 milli m 0 - centi c 0 - deci d 0 0 ---- ---- 0 deka da 0 hecto h 0 3 kilo k 0 6 mega M 0 9 giga G 0 tera T 0 5 peta P 0 8 exa E Note the use of lower case letters for multipliers less than one and the use of upper case letters for multipliers greater than 0 6. Inconsistent usage on kilo. Commonly used multipliers shown in bold.

The inductance of a coil is: Inductance L KN where N number of turns L inductance in henries K a constant dependent upon geometry and material

3 Capacitance C ενε r A d DA Εd where 9 0 ε d ν farads/meter 8.84 pf/m 36π ε relative permittivity of dielectric (dielectric constant) r A common area (square meters) d plate spacing (meters) C capacitance (farads) D electric flux density (coulombs per square meter) Ε electric field intensity (volts per meter)

4 Coulomb s Law Often used as basis for developing theory of electrostatic fields. Defines force on a charge in an electric field or the force between two charges

5 Coulomb s Law In the field of a charge Q concentrated at a point, the total flux, φ, through a sphere of radius, r, is : φ 4πr D But flux is also equal to the charge, Q, so : Q and D, the flux density at distance 4πr Electric field intensity, Ε, at this radius is : Q Ε 4πεr If a charge, Q, is placed in the field of a charge, Q, the force, F, on Q, is : where Q 4πr F ΕQ Q charge in coulombs D QQ 4πεr F force in newtons r caused by point charge, Q. Q r Q ε permittivity of the medium 8.85 0 for air r distance between charged particles in meters - 6.8 0 8 electrons have the charge of one coulomb electron has a charge of.6 0-9 coulomb

6

7 Series and Parallel Combinations Circuit Element Series Parallel CE CE CE CE CE CE CE CE, Capacitor, Inductor, esistor C C L L or C C C C C C C C L L L C C C or or L L L L L L L L

8 Wye-Delta Transformations 3 r r r 3 3 3 3 3 3 3 3 3 r r r r r r r r r r r r r r r r r r r r r 3 3 3 3 3 3 r r r

9 Turn Turn Turns Turns

0 Turn Turn Turns Turns

Turn Turn

Turn Turn Turns Turns

3 Turn Turn Turn Turn

4 Turn Turn Turns Turns

5 Complex Numbers Square root of (-) i Complex algebra is very much like vector algebra. Keep like parts together. Addition and subtraction: Add or subtract real and imaginary parts. Multiplication and division: Use cross multiplication for rectangular form. Polar form: Multiply/divide magnitude, Add/subtract phase angle Conversions: ectangular to polar: Magnitude from Pythagorean theorem, Phase angle is arctan( imaginary / real ) Polar to rectangular: eal magnitude * cos(angle) Imaginary magnitude * sin(angle)

Complex Algebra ectangular Form j-operator: j imaginary component, referred to as the j - operator 6 Addition: Multiplication: ( a jb) ( c jd ) ( a c) j( b d ) ( a jb)( c jd ) ( ac bd ) j( ad bc) Division: complex conjugate of a jb a jb a jb a c jb jd ( a jb)( c jd ) ( c jd )( c jd ) ( a jb)( c jd ) ( ac bd ) j( bc ad ) c d c d

7 Complex Algebra - Polar Form Notation: Multiplication: Division: A B Addition: A θ Acosθ jasinθ θ α δ arctan A B A θ A B α B A θ B α where C Ae jθ ( A θ )( B α ) AB ( θ α ) ( θ α ) ( θ α ) ( Acosθ B cosα ) j( Asinθ Bsinα ) ( real) ( imaginary) imaginary real C δ

8 Complex Algebra - aising to Powers ( ) m m A θ A ( mθ ) ( A θ ) where / m n / m θ πn A m m 0,,,...,( m )

9 Complex Notation ectangular form: Polar form: Z Z Exponential form: Z Z e Trigonometric form: Z ± jx X, ( θ ) whereθ arctan in degrees ± ± jθ whereθ is ( cosθ j sinθ ) Z Z ± expressed in radians

30 Complex Numbers - Examples Example : What is the sum of the complex numbers (4i7) and (6i9)? (A) 0i6 (B) ii5 (C) i6 (D) 0i7 Answer: (A) Example : What is the product of the complex numbers 4 i and 3e Solution: i i 3e iπ 4 has a magnitude of 8e 8e iπ 4 iπ 4 3 8e iπ iπ 4 4 o 8 and an associated angle of - 45, so 6 π i Alternatively, -i 3 45 8 45 o o 8 45 o 6

Circuit Element Equations 3 Need a convenient notational format. Use the j operator to indicate phase relation. Inductive circuit: current lags voltage by 90º V V Z j ji I L jx L [ ohms] Capacitive circuit: current leads voltage by 90º Z V ji V j I C jx C [ ohms] Now we can directly add the resulting impedances: Z jx L jx C [ ohms] Where X X L C ωl πfl ωc πfc [ ohms ]

eciprocal Elements 3 Sometimes it is more convenient to work with the reciprocal of impedance and its real and imaginary components. For example, when working with parallel circuits and current (rather than voltage) sources. Then we use the following: Complex term : admittance impedance > Y esistive term : conductance > G resistance - eactive term : susceptance > B reactance Thus: Z Y G jx jb [ ohms] [mhos or siemens] Z X [ mhos] or [ siemens] [ mhos] or [ siemens] [ mhos] or [ siemens]

33 Single Energy Transients Circuits with only one type of reactive element can be easily analyzed using a transient response formula of the form: f ( t) where f f ss 0 f ss ( f f ) steady state final value of initial value of 0 t / τ f ( t) τ circuit time constant ss e (i.e., at t ( C) [ seconds] This formula works for any voltage or current in the circuit, even with initial non-zero values of voltage or current, SO LONG AS ONLY A SINGLE TIME CONSTANT IS PESENT (ie, the equation is a first-order equation. L f ( t) or 0 (i.e., at t ) )

34 L Circuit Inductance is associated with the magnetic field of the current. di e L dt t / τ i I I I e I t0 where I ss ss ( ) Stored energy is: 0 V / τ L / ss [ seconds] V L W Li

Find the equivalent resistance of the network shown below: 35 0 40 4 8 4

36 Norton and Thevenin Equivalent Circuits Thevenin Norton

Wye-Delta Transform Problem 37 The circuit below is a Wheatstone bridge with a resistance in place of the galvanometer. Find the current supplied by the battery. a 50 V r r 4 50 50 r 80 3 c r 00 r 5 80

38 Work, Energy, and Power A hoist driven by a three-phase electric motor has a 4000- pound load capacity. a) How much energy in joules and power in KW is required to raise a full capacity load 00 feet in one minute? b) What motor capacity in HP is required to drive the hoist if the hoisting machinery is only 75% efficient?

Inductor Formulas 39

Inductor Formulas (continued) 40

Inductor Formulas (continued) 4

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