Intro Electrical Engineering
Intro Electrical Engineering EE 221
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This 36 page Class Notes was uploaded by Maudie Quitzon on Saturday September 12, 2015. The Class Notes belongs to EE 221 at West Virginia University taught by Staff in Fall. Since its upload, it has received 21 views. For similar materials see /class/202857/ee-221-west-virginia-university in Electrical Engineering at West Virginia University.
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Date Created: 09/12/15
in Nodal and Mesh Analysis in Superposition in Source transformation it Thevenin and Norton equivalent it Operational Amplifier A 959 Redraw circuit to emphasize nodes 2 Assign reference node and voltages N nodes result in N1 unknown voltages Use KCL to find N1 equations Relate dependent sources to node voltages Form supernode to enclose voltage sources and apply KCL Add voltage equations Reference node a KCL requires that all currents flowing into the region must sum to zero or we would pile up or run out of electrons At node 1 KCL 8 3 2 m u 3 At the supernodez KCL V2 V1 V3 V1 V3 325 V 2 3 4 5 1 At the supernode KVL 22 v3 v2 Al node 2 l U1 L 2 1 3 a 4 14 I 05 2 0 while at the 3 4 supemode 15 v m U4 v1 2 1 25 lquot We next relate the source we ages to the node voltages 05m Us 139 0213 021 02114 L I Finally we express the depen ent current source in terms of the assigned variables 05121 2 051393 131 Independent voltage source supernode containing reference V1 12 1 N 959 Redraw planar circuit to emphasize meshes Assign clockwise mesh currents M meshes result in M unknown currents Apply KVL around each mesh Relate dependent sources to mesh currents Use supermesh for current source shared between two meshes Add current equation 0 a 1i239i12i23i239i3 0 1 2 0 Finally we relate the currents in meshes 1 and 3 i1 i3 7 a Rearranging i1 4i24i3 1 2 is 3 Solving I1 612 313 i1 7 0 7 i19Ai2 25 A and i3 2A Creating a supermesh from meshes 1 and 3 quot71i139i2 3i339i21i3 Around mesh 2 1 2 3 T No voltage drop across terminals 0 V I but current can ow a No current ows but a volta B can 0 A v g r appeal across the terminals I b a A voltage source set to zero acts like a short circuit b A current source set to zero acts like an open circuit 3 Linear circuits allow superposition 10quot 9 0 O 25 b Keep only one independent source at a time activate a c Always keep dependent sources A general R L A general practical voltage practlcal source current source 39 l a connected to a quot5 9 RL connected A load resistor RL to a load reSIstor RL a a Convert between the two Sources are related by Rs RP and VsRSIS RplS Useful when asked for Maximum terminal voltage vs andor current is Maximum power transferred PL vL iL when RL RS Complex network RL O RL a b C quotDeadquot network to find equivalent source resistance RTH and RN Open loop voltage to determine VTH any method Short circuit current determines N Network A Network B Network A Network A a b c Source transformation is used here Network A Network A d e Open loop voltage to determine VTH and short circuit current determines lN Find equivalent source resistance RTH and RN use quotDeadquot network use RTH RN VTHI IN the only way in case of dependent sources Offset null V a Electrical symbol Input Output Offset null V Neglected a 1 Output voltage saturation b Minimum 2 Inputoutput resistance 0p amp39 3 Limited open loop gain Ideal 4 Input bias current 1 No Input current b 5 Input offset voltage 2 No voltage difference between input terminals Op amp connected as an Inverting amplifier Vout 39 Rf R1 Vin 60 40 Z Oul 20 Vin 0 g n 75 03 a A v 7quot Qy 7 I 20 40 60 Output characteristics 1st step Determine voltae at input terminals 2nd step Determine current i 3rd step Find output voltae vout a An op amp used to construct a noninverting amplifier circuit b Circuit with currents and voltages labeled c Output characteristics Vout Rf I R1 Vin Your choice Nodal analysis andor superposition o Vout 39Rbch V1 RdRaRb I RaRcRd V2 2 Basic components 2 Electric circuits ii Voltage and current laws Base quantity Name Symbol length meter m mass kilogram kg time second 5 electric current ampere A temperature kelvin K Factor Name Symbol Factor Name Symbol 103918 atto a 103915 femto f 103912 pico p 1012 tera T 10399 nano n 109 giga G 10396 micro p 106 mega M 10393 milli m 103 kilo k 10392 centi c it Two types of charge I positive proton E39 r negative electron 16 10490 mm 2 Continuoust transferring charge total amount of charge never changed neither created nor destroyed conservation i Defined in terms of ampere qt qtu gulf 2 Measured in coulomb C As r N H 539 rt 7 L C iiili Fill l Cross section i Charge in motion transfer of energy 03173335 dc related to charge i alt V Individual charges 3 Representing current numerical value unit eg 135 A direction gt unit is the ampere A Symbol for an I I independent current represented by l I It source Ix in Example W a b 6 ab Incomplete improper and incorrect definitions of a current c the correct de nition of Mt its General simple Circuit element two terminals cannot be decomposed further completely characterized by its voltagecurrent relationship 3 Pushing Charge general twoterminal expenditure of energy C39rcu39t element electrical voltage potential difference voltage quotacrossquot the element k Voltage measures work required to move charge k Representing voltage numerical value unit eg 25 V grg geg gggfermm39 direction sense V left terminal is V volts positive with i respect to the right terminal W Y T unit is volt V J C a b c represented by V v vt Symbols a DC voltage source b battery c ac voltage source d Power is the rate of energy expenditure Voltage Current Voltage defined in terms of energy Current is rate at which charge moves a Representing power I A general twoterminal numerical value unit eg 56 W circuit element quotdirectionquot by Passive Sign Convention PSC Current entering element through positive terminal unitiswattWVAJCAJAsAJs represented by P p pt b Is a choice we make convention i The current arrow is directed into the quotquot marked terminal in The power absorbed by the element is given by the product p v i Agenererrwerermrner circuit element p vi 3 A negative value indicates that power represents the power is actually generated embed in Or The power generated by the element is given by the product p vi W Resistance of conducting element 3 Ohm39s law v R i linear directly proportional Igt f H Passive element R 3 Power p v i i2 R v2 R lamps 3 Representing resistance 2 l numerical value unit eg 3 Q 3 i unitisohmQVA 3 i represented byR 1 1 t g All g t 35 HO more 45 H H H W wNwa 3 Branches Q 3 W Paths Hoops miiiquotie7fee I iwoNf el Eids v rfi mt b 3 Kirchhoffs current law e Conservation of charge The algebraic sum oflhc currents entering any node is zero 1A1B1C1D0 3 Kirch hoff39s voltage law hi Conservation of energy I39h ulgdw nu anI nt39lh ul nulmvl an a lmml Mill 39 v1vZv3 7 72 o 01 I 712 73 1 11 12L3 a b a Series connected voltage sources can be replaced by a single source b Parallel current sources can be replaced by a single source Examples of circuits with multiple sources some of which are illegal as they violate Kirchhoff s laws R o a a 1 IA 5A 3A R 1A 6 d a Series combination of N resistors b Electrically equivalent circuit ReqR1R2RN a Parallel combination of N resistors b Electrically equivalent circuit i 11Req 1IR11IR2 1IRN A special case worth remembering is is 1 Req RIHRZ an R1 R2 R1R2 R eq R1R2 An illustration of voltage division For a parallel combination of N resistors the current through Rk equals 1 An illustration of i Rk i current dIVIsIon k 1 1 1 R1 R2 RN What do we count as positive Direction of summation 1st Choice determines polarity v vzvhvc 2nd choice vzvzvhvc