Electricity & MagnetismFE Review

VoltageCurrentPower1 V 1 J/C1 J of work must be done to move a 1-C charge througha potential difference of 1V.

Point Charge, QExperiences a force F=QE in the presence of electric field E(E is a vector with units volts/meter)

Work done in moving Q within the E field

E field at distance r from charge Q in free space

Force on charge Q1 a distance r from Q

Work to move Q1 closer to Q

Parallel Plate CapacitorElectric Field between PlatesQ=charge on plateA=area of plate0=8.85x10-12 F/mPotential difference (voltage) between the platesCapacitance

ResistivityExample: Find the resistance of a 2-m copper wire if the wire has a diameter of 2 mm.

ResistorPower absorbed

Energy dissipated

Parallel ResistorsSeries Resistors

CapacitorStores Energy

Parallel CapacitorsSeries Capacitors

InductorStores Energy

Parallel InductorsSeries Inductors

KVL Kirchhoffs Voltage LawThe sum of the voltage drops around a closed path is zero.KCL Kirchhoffs Current LawThe sum of the currents leaving a node is zero.Voltage DividerCurrent Divider

Example: Find vx and vy and the power absorbed by the 6- resistor.

Node Voltage AnalysisFind the node voltages, v1 and v2Look at voltage sources firstNode 1 is connected directly to ground by a voltage source v1 = 10 VAll nodes not connected to a voltage source are KCL equations Node 2 is a KCL equation

KCL at Node 2v1 = 10 Vv2 = 6 V

Mesh Current AnalysisFind the mesh currents i1 and i2 in the circuitLook at current sources firstMesh 2 has a current source in its outer branchAll meshes not containing current sources are KVL equationsKVL at Mesh 1Find the power absorbed by the 4- resistor

RL CircuitPut in some numbers

R = 2 L = 4 mHvs(t) = 12 Vi(0) = 0 A

RC CircuitPut in some numbers

R = 2 C = 2 mFvs(t) = 12 Vv(0) = 5V

DC Steady-State0Short Circuiti = constant0Open Circuitv = constant

Example:Find the DC steady-state voltage, v, in the following circuit.

Complex ArithmeticRectangularExponentialPolara+jbAejAPlot z=a+jb as an ordered pair on the real and imaginary axesEulers Identity

ej = cos + j sinComplex Conjugate(a+jb)* = a-jb(A)* = A-

Complex ArithmeticAdditionMultiplicationDivision

PhasorsA complex number representing a sinusoidal current or voltage.Only for:Sinusoidal sourcesSteady-stateImpedanceA complex number that is the ratio of the phasor voltage and current.units = ohms ()Admittanceunits = Siemens (S)

PhasorsConverting from sinusoid to phasorOhms Law for PhasorsOhms LawKVLKCLCurrent DividerVoltage DividerMesh Current AnalysisNode Voltage Analysis

Impedance

Example: Find the steady-state output, v(t).

Source TransformationsThvenin EquivalentNorton EquivalentTwo special cases

Example: Find the steady-state voltage, vout(t)-j1

j1 j0.5

50+j0.5 Thvenin EquivalentNo current flows through the impedance

AC PowerComplex Powerunits = VA (volt-amperes)Average Powera.k.a. Active or Real Powerunits = W (watts)Reactive Powerunits = VAR (volt-ampere reactive)Power Factor = impedance angleleading or laggingcurrent is leading the voltage0

v(t) = 2000 cos(100t) VExample:Current, IPower FactorComplex Power AbsorbedAverage Power AbsorbedP=13793 W

Power Factor CorrectionWe want the power factor close to 1 to reduce the current.Correct the power factor to 0.85 laggingAdd a capacitor in parallel with the load.Total Complex Power

S for an ideal capacitorv(t) = 2000 cos(100t) VCurrent after capacitor addedWithout the capacitor

RMS Current & Voltagea.k.a. Effective current or voltageRMS value of a sinusoid

Balanced Three-Phase SystemsY-connectedsourceY-connectedloadPhase VoltagesLine CurrentsLine Voltages

Balanced Three-Phase SystemsPhase CurrentsLine CurrentsLine Voltages-connectedsource-connectedload

Currents and Voltages are specified in RMSS for peakvoltage& currentS for RMSvoltage& currentS for 3-phasevoltage& currentComplex Powerfor Y-connected loadComplex Powerfor -connected loadAverage PowerPower Factor

Example:Find the total real power supplied by the source in the balanced wye-connected circuit Given:P=1620 W

Ideal Operational Amplifier (Op Amp)With negative feedbacki=0v=0Linear Amplifier0mV or V reading from sensor0-5 V output to A/D converter

Magnetic FieldsB magnetic flux density (tesla)H magnetic field strength (A/m)J - current densityNet magnetic flux through a closed surface is zero.Magnetic Flux passing through a surfaceEnergy stored in the magnetic fieldEnclosing a surface with N turnsof wire produces a voltage acrossthe terminalsMagnetic field produces a force perpendicular to the current direction and the magnetic field direction

Example:A coaxial cable with an inner wire of radius 1 mm carries 10-A current. The outercylindrical conductor has a diameter of 10 mm and carries a 10-A uniformly distributed current in the opposite direction. Determine the approximate magneticenergy stored per unit length in this cable. Use 0 for the permeablility of the material between the wire and conductor.

Example:A cylindrical coil of wire has an air core and 1000 turns. It is 1 m long with a diameter of 2 mm so has a relatively uniform field. Find the current necessary to achieve a magnetic flux density of 2 T.

Questions?