2Voltage Current Power 1 V ≡ 1 J/C 1 J of work must be done to move a 1-C charge througha potential difference of 1V.CurrentPower
3Point 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 fieldE field at distance r from charge Q in free spaceForce on charge Q1 a distance r from QWork to move Q1 closer to Q
4Parallel Plate Capacitor Q=charge on plateA=area of plateε0=8.85x10-12 F/mElectric Field between PlatesCapacitancePotential difference (voltage) between the plates
5ResistivityExample: Find the resistance of a 2-m copper wire if the wire has a diameter of 2 mm.
6Resistor Power absorbed Energy dissipated Parallel Resistors Series Resistors
9KVL – Kirchhoff’s Voltage Law The sum of the voltage drops around a closed path is zero.KCL – Kirchhoff’s Current LawThe sum of the currents leaving a node is zero.Voltage DividerCurrent Divider
10Example: Find vx and vy and the power absorbed by the 6-Ω resistor.
11All nodes not connected to a voltage source are KCL equations 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
13Mesh Current Analysis Find the mesh currents i1 and i2 in the circuit Look 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-Ω resistor
14RL Circuit Put in some numbers R = 2 Ω L = 4 mH vs(t) = 12 V i(0) = 0 A
15RC CircuitPut in some numbersR = 2 ΩC = 2 mFvs(t) = 12 Vv(0) = 5V
20Phasors Only for: Sinusoidal sources Steady-state Impedance Admittance A 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 (Ω)Admittanceunits = Siemens (S)
21Phasors Converting from sinusoid to phasor Ohm’s Law for Phasors Current DividerOhm’s LawKVLKCLVoltage DividerMesh Current AnalysisNode Voltage Analysis
27No current flows through the impedance 50+j0.5 ΩNo current flows through the impedanceThévenin Equivalent
28AC Power Complex Power units = VA (volt-amperes) Average Power units = W (watts)a.k.a. “Active” or “Real” Powerunits = VAR(volt-ampere reactive)Reactive PowerPower Factorθ = impedance angleleading or laggingcurrent is leading the voltageθ<0current is lagging the voltageθ>0
29Complex Power Absorbed Example:v(t) = 2000 cos(100t) VCurrent, IComplex Power AbsorbedPower FactorAverage Power AbsorbedP=13793 W
30Power Factor Correction We want the power factor close to 1 to reduce the current.Correct the power factor to 0.85 laggingTotal Complex PowerAdd a capacitor in parallel with the load.
31S for an ideal capacitor v(t) = 2000 cos(100t) VWithout the capacitorCurrent after capacitor added
32RMS Current & Voltage a.k.a. “Effective” current or voltage RMS value of a sinusoid
33Balanced Three-Phase Systems Phase VoltagesY-connectedsourceLine CurrentsY-connectedloadLine Voltages
34Balanced Three-Phase Systems Line VoltagesΔ-connectedsourceΔ-connectedloadPhase CurrentsLine Currents
35Currents and Voltages are specified in RMS S for peakvoltage& currentS for RMSvoltage& currentS for 3-phasevoltage& currentAverage PowerComplex Powerfor Y-connected loadComplex Powerfor Δ-connected loadPower Factor
36Example:Find the total real power supplied by the source in the balanced wye-connected circuitGiven:P=1620 W
37With negative feedback Ideal Operational Amplifier (Op Amp)With negative feedbacki=0Δv=0Linear AmplifiermV or μV reading from sensor0-5 V output to A/D converter
38Magnetic Fields B – magnetic flux density (tesla) H – magnetic field strength (A/m)J - current densityNet magnetic flux througha closed surface is zero.Magnetic Flux φ 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 tothe current direction and the magnetic field direction
39Example: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 uniformlydistributed current in the opposite direction. Determine the approximate magneticenergy stored per unit length in this cable. Use μ0 for the permeablility of thematerial between the wire and conductor.
40Example:A cylindrical coil of wire has an air core and 1000 turns. It is 1 m long witha diameter of 2 mm so has a relatively uniform field. Find the currentnecessary to achieve a magnetic flux density of 2 T.