Chapter 2 Resistive Circuits

Transcription

1 Chapter esistie Circuits

2 Goal. Sole circuits by combining resistances in Series and Parallel.. Apply the Voltage-Diision and Current-Diision Principles.. Sole circuits by the Node-Voltage Technique.. Sole circuits by the Mesh-Current Technique. 5. Find Théenin and Norton Equialents. 6. Apply the Superposition Principle. 7. Understand the Wheatstone Bridge.

3 esistance in Series Using KVL, Ohm s Law i i i ( )i eq

4 esistance in Parallel Using KCL, i i i i Ohm s Law i i i i (/ / / ) eq / / /

5 Example Series Parallel Series

6 Network Analysis using Series/Parallel Equialents. Begin by locating a combination of resistances that are in series or parallel. Often the place to start is farthest from the source.. edraw the circuit with the equialent resistance for the combination found in step.. epeat steps and until the circuit is reduced as far as possible. Often end with a single source and a single resistance.. Sole for the currents and oltages in the final equialent circuit. Transfer esults to Backward until Original Circuit Network Analysis : The process of determining the Current, Voltage & Power for each element gien in Circuit & Element Values

7 Step, & Step, Step

8 Step

9 Voltage Diision i i total total Of the total oltage, the fraction that appears across a gien resistance in a series circuit is the ratio of the gien resistance to total series resistance Voltage Diision Principle

10 Application of the Voltage-Diision Principle total V 5

11 Generalized Voltage Diider I V Z in total Z total Z Z V out I Z Vin Z Z Z

12 Current Diision i i i i total total For two resistance in parallel, the fraction of the total current flowing in a resistance is the ratio of the other resistance to sum of the two resistance (Two esistors Case) Current Diision Principle

13 Application of the Current-Diision Principle i eq 0 60 Ω eq i s eq 0A

14 Position Transducer Based on Voltage-Diision Principle o s Kθ Typical Knob in Electrical System based on Analog Signal such as Amplifier, Adjustable Light, Volume

15 Series Sources Ideal independent oltage sources in series add algebraically I V V V V n I - V Note : Parallel Voltage Sources are not esolable. WHY?

16 Parallel Sources Ideal independent current sources in parallel add algebraically I T V I n _ I I I Note : Series current sources are not resolable. WHY?

17 Example and are effectiely open circuited and therefore can be omitted 7 and 8 are short circuited, and can be omitted I and I are Algebraically added because they are in Series. V and V are Algebraically added because they are in Parallel and are in Series & 5 & 6 are in Parallel

18 Although they are ery important concepts, series/parallel equialents and the current/oltage diision principles are not sufficient to sole all circuits.

19 Wye - Delta Transformations - Need to find equialent resistance to determine current. - HOW? - (They are not in series, not in parallel) Use Y to transformation

20 Equating esistance's esistance between X - Y In a // ( b c ) In Y ) ( ) ( c a b c a b XZ b a c X Y Z X Y Z ) ( ) ( c b a c b a XY ) ( ) ( b a c b a c YZ Similarly Aboe Equations are Linearly Solable between,, & a, b, c

21 Y X X Z Y Z Y

22 Y Transformation : Not Enough for All Circuits With Voltage Source With Dependent Source

23 Node-Voltage Analysis. Select eference Node. Assign Node Voltages If the reference node is chosen at one end of an independent oltage source, one node oltage is known at the start, and fewer need to be computed.. Write Network Equations. -First, usekcl to write current equations for nodes and supernodes. (Write as many current equations as you can without using all of the nodes.) - If not hae enough equations because of oltage sources between nodes, (use KVL to write additional equations) - If the circuit contains dependent sources, Find expressions for the controlling ariables in terms of the node oltages. Substitute into the network equations Obtain equations haing only the node oltages as unknowns.. Sole for the Node Voltages. (Equation in standard form) 5. Calculate currents or oltages of interest. with the node oltages from Step

24 Select eference Node & Assign Node Voltage Node : Two More Circuit Elements are Joined Together eference Node : Node with Known & eferencing Voltage Ground Symbol : Zero Voltage - Typically eference Node is set to be Ground Symbol - Negatie Polarity to eference Node

25 Assign Element Unknown Voltages - Assign Node Voltage across, & - Use KVL In In y x In z

26 Writing KCL Equations with Node Voltages s Unknown Node Voltages Equations At Node At Node At Node KCL

27 Solution of Equations Soling KCL equation as Standard form # of Unknown Node Voltage # of Equation With Unknown Node Voltages g g Generally g g i i With Unknown Node Voltage g g g i g g g g g g i i g ij j i i Solution of Linear Simultaneous Equation gies Node Voltage Calculation of emaining Node Voltages & Currents

28 Tips on KCL Equation Writing - Writing KCL Node not Connected to Voltage Source (note : Supernode) - Direction of Current from Node : Positie (Node Voltage of Interest Node Voltage Surrounding ) / esistance Current Output from Node of Interest to Node Surrounding - Sum of All Currents from Surrounding Nodes At Node Node Node Node Ground Node Node 0

29 0 s i 0 i s 5 KCL Equations with a Current Source - - Node Node Node

30 KCL Equations with Current Sources Node 0 i a Node 0 Node 0 5 b i

31 Different choice of eference Node Node Node Node V 7.7V 5.5V i x A Same Logic regardless of eference Node

32 KCL Equations with Voltage & Current Sources 0 Node V 6. 9V 0 Node Node : Not Writing KCL on, because of Voltage Source Same Logic for Current & Voltage Sources Containing Circuit

33 Circuits with Voltage Sources Supernode Supernode When a Voltage Source is connected between Nodes, KCL can not be written at Nodes because both Current & esistance are not defined in Voltage Source Aboe Circuit can not Write KCL, because All node are connected to Voltage Sources The Net Current flowing through any closed surface must equal to Zero Supernode : Combination of Node including Voltage Sources If Write KCL equations with All Node & Supernodes ( 5) ( 5) Supernode 0 ( 5) ( 5) Supernode 0 Dependant Eqns Use KVL at Supernode

34 Solution by Supernode Supernode at Supernode KVL KCL 0 ( 5) ( 5) 0 Independent Equation!! Solable 0 Write KVL at Supernode Write KCL for Super node

35 0 0 0 Supernode Example KVL at Supernode KCL around Supernode KCL at Node KCL at Ground Same Equation!

36 Node-Voltage Analysis with a Dependent Source First, write KCL equations at each node, including the current of the controlled source just as if it were an ordinary current source. i s i x 0 0 i x

37 Substitution yields i s 0 0 Next, find an expression for the controlling ariable i x in terms of the node oltages. i x Solable!

38 Example with Dependant Voltage Source Supernode (Node & ) Node Ground KVL KCL. 5x 0 x 0 is i s 0 Solution for,,

39 Mesh-Current Analysis. Define the Mesh Currents (usually a clockwise direction) If necessary, redraw the network without crossing conductors or elements.. Write Network equations ( # of equations # of mesh currents). - Use KVL for meshes with no current sources. - If current sources, write expressions for their currents in terms of the mesh currents. - If common current source to two meshes, Use KVL for Supermesh. - If dependent sources, Find expressions for the controlling ariables Substitute into the network equations, Obtain equations haing only the mesh currents as unknowns.. Sole for the mesh currents (Standard form equation). Calculate any other currents or oltages of interest.

40 Basic ule for Mesh Current Node-Voltage Analysis KVL for A, & i i KVL for B, & KCL for Node i i i i i A earrange B i i i ) ( ) ( i i i When Seeral Mesh Currents flow through one Elements, the Current in that Element to be Algebraic Sum of the Mesh Current A B

41 Choosing the Mesh Currents Only for Planar Network : Network without Crossing Element Mesh : Loop of Elements Mesh Current : Current flowing through Mesh Define Current Direction Clockwise for Consistency Example, Current flow in to Left i - i Current flow in to Up i - i

42 Writing Equations to Sole for Mesh Currents If Network with only resistances and independent oltage sources, write required equations by following each current around its mesh & applying KVL. Mesh ( i i ) ( i i ) 0 s A Mesh Mesh ( i i ) i 0 B ( i i ) i 0 B

43 Example of Mesh Current with Voltage Source i ( i i ) ( i i ) 0 A i ( i i ) ( i ) i ( i i ) ( i ) 0 7 i 6 8 i i ( i i ) ( i ) 0 8 i Solution of Network Equation & Find Voltage at Node

44 Example of Mesh-Current Analysis Mesh 0i 0(i i ) 50 0 Mesh 0(i i ) 5 i 00 0 In standard form 0i - 0(i i ) 50-0 i 5 i -00 i.a i -.08A Current through downward i i 6.59A

45 Mesh Currents in Circuits with Current Sources i A 0 KVL to Mesh 0 ( i i ) 5i 0 A common mistake made by beginning students is to assume that the oltages across current sources are zero.

46 Mesh Currents with Voltage & Current Sources KVL for Mesh & is impossible, because Voltage across 5A is unknown Combine meshes and into a Supermesh. Write KVL equation around the periphery of meshes & Mesh Current Source 5A i ( ii ) ( i ) 0 0 ( i i ) ( i ) 0 i i i i i 5

47 Example of Mesh-Current Analysis Supermesh i 5A ( i i ) 5i i i A 5 i 0i 0 0 0

48 Mesh-Current with Dependant Source - Write Equations exactly the same as Independent Source - Express Dependant Variable with Mesh Current - Substitute into Network Equation Supemesh Dependant Source 0 i 6i i x i i x i 0

49 Théenin Equialent Circuits b a a eal Complex Circuit a b Easy Understandable Equialent Circuit

50 Théenin Equialent Circuits V t oc t i oc sc

51 Example of Théenin Equialent Circuits To find Open Circuit Voltage s i 0. A oc i V eq Ω To find Short Circuit Current s 5 isc 0. 5A 00 oc 5 t. Ω i 0.5 sc t equals eq with Zero Voltage!

52 Example of Théenin Equialent Circuits i oc oc i i sc To find Open Circuit Voltage i oc 5A oc i V eq Ω To find Short Circuit Current i sc 5A A oc 50 t 50Ω i sc t equals eq with Zero Current!

53 Finding the Théenin esistance Directly If No Dependant Source, When zeroing a Voltage source, it becomes a Short circuits When Zeroing a Current Source, it becomes a Open Circuit In Zeroing Independent Source, eplace a Voltage Source with Short Circuit eplace a Current Source with Open Circuit. Short Circuit When the Source is Zeroed, the resistance seen from the Circuit terminals is equal to the Théenin esistance

54 Direct Finding the Théenin esistance b) Zeroing Source eq Ω 5 0 c) Short Circuit s 0 i i i i A sc i 5 SC 6A i 0 Vt tisc 6 V

55 Find Théenin esistance Zero Zero eq 0 Ω 5 0 t 0Ω eq Zero 6 0Ω 5 0 Zero eq 0Ω eq 5 Ω 0 0 Exercise.

56 Théenin esistance with Dependant Source Can not use Zeroing Source Technique due to dependant source Open Circuit : Node-Voltage Tech KCL at Node KVL Loop oc 8. 57V oc i x ix 0 5i 0 x oc Short Circuit 0 i x A A isc ix 6A 5 t oc 8.57V. Ω i 6A sc

57 Norton-Equialent Circuit Analysis Norton Equialent Circuit : Independent Current Source Théenin esistance I n is i sc in Théenin Equialent Analysis

58 Step-by-step Théenin/Norton-Equialent- Circuit Analysis. Perform two of these: a. Determine the open-circuit oltage V t oc. b. Determine the short-circuit current I n i sc. c. Zero the sources and find the Théenin resistance t looking back into the terminals.. Use the equation V t t I n to compute the remaining alue.. Théenin equialent : a oltage source V t in series with t.. Norton equialent : a current source I n in parallel with t.

59 Example of Norton-Equialent-Circuit Analysis KCL at Node KVL at Loop 5 x oc oc x 0. 5 oc oc. 6V 0 oc Current along & is Zero : x 0 5 s isc t oc oc Ω i 0.75 sc

60 Maximum Power Transfer The load resistance that absorbs the maximum power from a two-terminal circuit is equal to the Théenin resistance. L t t L V i L L i p L ( ) L t L t L V p ( ) ( ) ( ) 0 L t L t L t L t t L L V V d dp At Power Maximum t L t t Lmax V p

61 SUPEPOSITION PINCIPLE In Circuit with esistance, Linear Dependant Source, Independent Source Superposition Principle The total response is the sum of the responses to each of the Independent Sources acting indiidually with Other Independent Source Zeroed. When Zeroed, Current Source become Open and Voltage Source become short. esponse : Voltage or Current In equation form, this is r r r... r T n

62 Example of Superposition Principle T K s If i s is set to Zero, esponse due to s : If s is set to Zero, esponse to i s : T T is Interesting KCL at Top T s T i x K i s T Ki s K x s K In General, dependant sources do not contribute a separate term to the total response, and we must zero dependant source in applying superposition i i s

63 WHEATSTONE BIDGE The Wheatstone bridge is used by mechanical and ciil engineers to measure the resistances of strain gauges in experimental stress studies of machines and buildings. x

64 Wheatstone Bridge V V V V b a ( ) ( ) V V V Voltage Zero V a V b V V a a V b b DC Wheatstone bridge V : Bridge Supply Voltage D : Voltage Detector

65 G Th Th G V I Th ( )( ) V V TH Bridge Circuits(Galanometer) Galanometer Bridge V : Bridge Supply Voltage G : Voltage Detector G

66 Lead Compensation Lead Compensation for emote Sensor Contact esistance of each esistor V V ( ) ( ) L L V V ( ( ) ( ) L L ) L Lead Problem - L makes V change - Enironmental Effect Temperature, Stress,Vapor - Transient effect

67 Lead Compensated Circuit V V ( ) ( ) L g L L Lead Compensation by Power Lead Extension Connection () : g c No Change on V Connection () & () : L, L Same Condition : & are identically changed Finally No Change in Bridge

68 Current Balance Bridge To make null bridge circuit ( ) 5 >> >> 5 & ( ) V V b V V 5 I5 5 V ( ) 5 I5 5. Splitting 5. Current on 5 << to make V null by adjusting magnitude and polarity of current I manually or electronically

69 a x C V V V b a x b C V V V V V V 0 V V V x ( ) I V V V x Potential Measurement Make V zero (null condition) Or current balance combination I 5 0 V x

70 Linearity of Bridge Circuit V V ( ) ( ) Bridge off-null oltage V (a) nonlinear for large-scale changes in resistance (b) nearly linear at small ranges of resistance change