Chapter 4. Techniques of Circuit Analysis - KNTU …wp.kntu.ac.ir/faradji/EC1/EC1_Ch4.pdf ·...

Chapter 4. Techniques of Circuit Analysis

By: FARHAD FARADJI, Ph.D.Assistant Professor,

Electrical Engineering,K.N. Toosi University of Technology

http://wp.kntu.ac.ir/faradji/ElectricCircuits1.htm

Reference: ELECTRIC CIRCUITS, J.W. Nilsson, S.A. Riedel, 10th edition, 2015.

Chapter Contents4.1. Terminology4.2. Introduction to the Node-Voltage Method4.3. NVM and Dependent Sources4.4. NVM: Some Special Cases4.5. Introduction to the Mesh-Current Method4.6. MCM and Dependent Sources4.7. MCM: Some Special Cases4.8. NVM Versus MCM4.9. Source Transformations4.10. Thévenin and Norton Equivalents4.11. More on Deriving a Thévenin Equivalent4.12. Maximum Power Transfer4.13. Superposition

Electric Circuits 1 Chapter 4. Techniques of Circuit Analysis 2

4.1. Terminology4.2. Introduction to the Node-Voltage Method4.3. NVM and Dependent Sources4.4. NVM: Some Special Cases4.5. Introduction to the Mesh-Currrrrrrreeeeeeennnnnntttttt MMMMMMMMeeeeeeettttthhhod4.6. MCM and Dependenttt SSSSoouuurrrrcceeeess4.7. MCM: Somee SSSSppppppeeecccciaaaalll CCCCCCCaaaasssseeeessss4.8. NVM Versusss MMMMMMCCCCCCCMMMMM4.9. Source Transformaatttiiioooooonnnsssss4.10. Thévenin and Nortonnn EEEEqqquuuuiiivvvaaallleeennnttsss4.11. More on Deriving a Thévenin Equivalent4.12. Maximum Power Transfer4.13. Superposition


4.1. Terminologyo To discuss the more involved methods of circuit analysis,

we must define a few basic terms.

o Planar circuits can be drawnon a plane with no crossing branches.

o A circuit that is drawn with crossing branches still is considered planar ifit can be redrawn with no crossover branches.


o To discuss the more involved methods of circuit analysis,we must define a few basic terms.

o Planar circuits can be drawnon a plane with no crossingggg bbbbbbbbrrrrrraaaannnncccchhhhhheeeeeesssssss.....

o A circuit that is drawn with crossssiiinnnnnnnggggggggggggg bbbbbbbbbrrrrrraaaaannnncches still is considered planar ifit can be redrawn wwwwwwwiiiiiitttttthhhhhhhhhhh nnnnnnnnnnnoooooooooooo cccccccccccccrrrrrrrrrrooooooooosssssssssssssssooooooovvvvvvvvvvveeeeeeeeeerrrrrrrrr bbbbbbbbbbrrrrrrrraaaaaaaaannnnnnnnccccccccchhhhhhhhhhheeeeeeeeessssssss..


4.1. Terminologyo Nonplanar circuits cannot be redrawn in such a way that

all the node connections are maintained andno branches overlap.

o The node-voltage method is applicable toboth planar and nonplanar circuits.

o The mesh-current method is limited toplanar circuits.


o Nonplanar circuits cannot be redrawn in such a way thatall the node connections are maintained andno branches overlap.

o The node-voltage method is apppppppppppppllllliiiiiiccccaaaabbbblllleeee tttttttooooooboth planar and nonplanar cccciiiirrrrrrcccccccuuuuuuuiiiiiiittttttttssssss....

o The mesh-current mettthhhhhhhhhhhhooooooooooddddddddddd iiiiiiisssssssssss llllllllliiiiiiimmmmmmmmmmmmmmiiiiiiiitttttttteeeeeeeeeeddddddd tttttttttttoooooooooplanar circccuuuuuuuuuuuuuiiiiiiiitttsssssssssss.....


4.1. Terminology

Electric Circuits 1 Chapter 4. Techniques of Circuit Analysis 5Electric Circuits 1 Chapter 4. Techniques of Circuit Analysis 5

4.1. Terminology


4.1. TerminologySimultaneous Equations—How Many?

o The number of unknown currents in a circuit equalsthe number of branches, b,

where the current is not known.

o We must have b independent equationsto solve a circuit with b unknown currents.


Simultaneous Equations—How Many?

o The number of unknown currents inn aa circuit equalsthe number of branches, b,,,,

where the current is not kkkkkkkknnnnnnoooowwwwnnnn....

o We must have b independent eeqqquuuuuaaaaaaaaaattttttttiiiiiioooooooonnnnnssssto solve a circuit wiiittttttthhhhhhh bbbbbbbbbbbb uuuuuuuuuunnnnnnnnnnnkkkkkkkkkkkknnnnnnnnnnnooooooooowwwwwwwwwwnnnnnnn cccccccccccuuuuuuuuuurrrrrrrrrrrrrrrrrrrrrrreeeeeeeeeennnnnnnnntttttttssssssss......



o If n is the number of nodes in the circuit,we can derive n-1 independent equations

by applying KCL to any set of n-1 nodes.

o Application of KCL to the nth nodedoes not generate an independent equation, because

this equation can be derived from the previous n-1 equations.

o Becausewe need b equations to describe a given circuit andwe can obtain n-1 of these equations from KCL,

we must apply KVL to loops or meshesto obtain the remaining b-(n-1) equations.



o If n is the number of nodes in the cirrcuuit,we can derive n-1 independdddeeeeennnnnnntttttttt eeeeeeqqqqqqquuuuuuaaaaattttions

by applying KCL to any seeeettttttt oooooofffff nnnn---11111 nnnnnnnoooooodes.

o Application of KCL to the nth nooddddeeeedoes not generate aaaannnnnnn iiiiiiiiinnnnnnnnnnnndddddddddddeeeeeeeeeeepppppppppppppeeeeeeeeeeennnnnnnnnddddddddddeeeeeeennnnnnnnnnnttttttttt eeeeeeeeeeeeeqqqqqqqqqqqqquuuuuuuuuuuaaaaaaaaattttttiiiiiioooooooooooonnnnnnnn,,,,,,, bbbbecause

this equuuaaaaaaaaaaaaatttttttttttiioooooooooooonnnnnnnnnn ccccaaaaaaaaaaannnnnnnnnnn bbbbbbbbbbbbbbeeeeeeeeee dddddeeeeeeeeerrrrrrriiiiiiiivvvvvvvvveeeeeeeeeeeeeeeddddddddddd fffffffffrrrrrrrooooooooooommmmmmmmmmmmmm ttttthhhhhhhhhheeeeeeeeeee ppppppppppprrrrrreeeeeeeeeeeevvvvvvvvvviiiiiioooooooooooooouuuuuuuuuuuuusssssssss nnnnnnnnnnnnnn---11111111111 eeeeeqqquations.

o Becausewe need b equationssssssssss ttttttttttttoooooooooooooo dddddddddddddeeeeeeeeeessssssssssccccccccccrrrrrrrrrriiiiiiiibbbbbbbbbbbbbbeeee aaaaaaa gggggggggggggiiiiiiiivvvvvvvvveeeeeeeeeeeennnnnnnnnn cccccccccccciiiiiiiirrrrrrrrrrccccccccuuuit andwe can obtain n-1 of these equations from KCL,

we must apply KVL to loops or meshesto obtain the remaining b-(n-1) equations.



o Thus by countingnodes,meshes, andbranches with unknown currents,

we have establisheda systematic method for

writing the necessary number of equations to solve a circuit.

o We apply:KCL to n-1 nodes andKVL to b-(n-1) loops (or meshes).



o Thus by countingnodes,meshes, andbranches with unknown currrrrrrreeeeeeennnnnnnttttttsssssss,,,

we have establishhhhhhhheeeeeedddddda syssstttteeeeeeeeeeemmmmmmmmmmmaaaaaaattttiiccccccccc mmmmmmmeeeetttthhhhhhhhhhoooooooodddddddddd ffffooooooooorrrr

wriitttiing ttthhhhhhhheeeeeeeeeee nnnneeeeccccccccceeeeessssssssssaaaaaaaaaaarrrryyyyyyyyyy nnnnnnuuuuuuuuummmmmbbbbeeeeerrrr ooooooooofffffffff eeeeqqqqqqqqqqqqquuuuuuuaaaattttttttttiiiiiiooooonnnss tttto solve a circuit.

o We apply:KCL to n-1 nodes andKVL to b-(n-1) loops (or meshes).



o These observations also are valid in terms ofessential nodes andessential branches.

o Ifne is the number of essential nodes andbe is the number of essential branches with unknown currents,

we can apply:KCL at ne-1 nodes andKVL around be-(ne-1) loops or meshes.



o These observations also are valid in tterrms ofessential nodes andessential branches.

o Ifne is the number offf eeeeeeeessssssssssssssssssssseeeeeeeeeeeennnnnnnnnnnnttttttttttiiiiiiiaaaaaaaaaaaaalllllll nnnnnnnnnnooooddddddddeeeeeeeeeeeesssssssss aaaaaaaaaaaaannnnnnnnnnddddddddddbe is the nnuuuuuummmmmmmmmmmmmmmbbbbbbbbbbbeeeeeeeeeerrr ooooooooooooffffffffff eeeeeeeeeeesssssssssssssssssseeeeeeeeeeennnnnnnnnttttttttttttiiiiiiiaaaaaaaaaalllllllllll bbbbbbbbbbbbrrrrrrrrrrraaaaaaaaaannnnnnnnnnncccccccccccccchhhhhhhhhhheeeeesssssssssss wwwwwwwwwwwwiiiiiiitttttttttthhhhhhhhhhh uuuuuuuuuunnnnnnnnnnnnnnkkkkkkkkkkkknnnnnnnnnoooooooooowwwwwwwwwwwwwwnnnnnnnnnnn cccccuuuurrents,

we can apply:KCL at ne-1 nodes annnnddddddddddddKVL around be-(ne-1) loops or meshes.



o The number of essential nodes isless than or equal to the number of nodes.

o The number of essential branches isless than or equal to the number of branches.

o Thus it is often convenientto use essential nodes and essential branches,

because they produce fewer independent equations to solve.



o The number of essential nodes isless than or equal to the nuummmmmmbbbbbbbbeeeeeeerrrrr ooooooofffffff nnnnnoooodes.

o The number of essential branchhhhhhheeeeeessssss iiiissssssless than or equal to the nuummmbbbbbbbbbeeeeeeeeeerrrrrrrr oooooooofffff bbbbrranches.

o Thus it is often convenniieeeeeeeeeeennnnnnnnnnntttttttttttttto use esssseeeeeeennnnnnnnnnnnttttttttttttiiiiiiiiiaaaaaaaaaaalllllll nnnnoooooooooooodddddddddddddeeeeeeeeeeeeeessssssssssss aaaaaaaannnnnnnnnnndddddddddddd eeeeeeeeeeeessssssssssssssssssssssssssssseeeeeeeeeeennnnnnnnnnntttttttttttttiiiiiiiiiiaaaaaaaaaaaaaallllllll bbbbbbbbbbbrrrrrrrrrrrraaaaaaaaaannnnccccccccccccchhhhhhhhhhheeeeeeeeeeesssssssssss,,,,,,,

because they pppppppppprrrrrrrrrrrrroooooooooooooodddddddduuuuuuuccccccceeeeeeeeeeee ffffffffffeeeeeeeeeeeewwwwwwwwwwwwwweeeeeeeerrrr iiiinnnnnnnnnnndddddddddddeeeeeeeeeppppppppppppppeeeeeeeeeeeeennnnnnnnnnnndddddddddeeeeeeeeeeennnnnnnntttttttttttttt eeeeeeeeeeqqqqqqqqqqqqqquuuuuuuations to solve.


4.1. TerminologyThe Systematic Approach

o We write the equations on the basis of essential nodes and branches.

o The circuit has4 essential nodes and6 essential branches, denoted i1 - i6,

for which the current is unknown.

o We derive 3 of 6 simultaneous equationsby applying KCL to any 3 of 4 essential nodes.

o Using nodes b, c, and e:


The Systematic Approach

o We write the equations on the basis off essential nodes and branchheeeesssss...

o The circuit has4 essential nodes and6 essential brancheeesssssssssss,,,,, dddddddddddeeeeeeeeeeeennnnnnnnnnnnooooooooooottttttttttttteeeeeeeeeeeddddddddd iiii111111 --- iiiiiiii666666666,,,,,,,,,,,

for whicccchhhhhhhhhhhhhh ttthhhhhhhhheeeeeeeee ccccccuuuuuuuuuuuuurrrrrrrrrrrrrrrrrrreeeeeeeeeeeeeennnnnnnnnnttttttttttt iiiiisssss uuuuuuuuuunnnnnnnnnnnnnnnkkkkkkkkkknnnnnnnnnnnooooooooowwwwwwwwwwwwwnnnnnnnnnnnn....

o We derive 3 of 6 simuuuuuuuuuulllllllltttttttttttttaaaaaaaannnneeeeeeeoooooooooooouuuuuuuussssssssss eeeeeeeeeeeeeeqqqqqqqqqquuuuaaaaaaaaaatttttttttttiiiiiiiiiooooooooooonnnnnnnnnssssssssssssssby applying KCL to annnnnnnnnyyyyyyyyyyyyy 3333333333333 ooooooooooooffffffffff 44444444444 eeeeeeeeesssssssssssssssssssseeeeeennnnnnntttttttttiiiiiiiiiaaaaaaaaaaaaalllllll nnnnnnnnnnnooooooooooooddddddddddddeeeeeeeeeeeessssssss.

o Using nodes b, c, and e:



o We derive the remaining 3 equationsby applying KVL around 3 meshes.

o Because the circuit has 4 meshes,we need to dismiss one mesh.

o We choose R7 — I,because we don't know the voltage across I.

o Using the other 3 meshes:



o We derive the remaining 3 equationssby applying KVL around 3 mmmeeeesssssshhhhhhhheeeeeeessssss...

o Because the circuit has 4 mesheeeeeeessssss,,,,we need to dismiss one messhhhh....

o We choose R7 — I,Ibecause wwwweeeeeeeeeeee ddddddddddddooooooooooonnnnnnnnnnnn'''tttt kkkkkkkkkknnnnnnnnnnnoooooooooooooowwwwwwwwwwww tttthhhhhhhhhhheeeeeeeeeee vvvvvvvvvvvvvvoooooooooooooollllllllttttttttttaaaaaaaaagggggggggggggeeeeeeeeeeeeee aaaaaaaaaaaccccccccccrrrrrrrrrrrrooooooooooosssssssssssss IIIIIIIIII....

o Using the other 3 meeeeeeessssssshhhhhhhhhhhheeeeeesssss:::::



o By introducing new variables, we can describe a circuit with justn-1 equations orb-(n-1) equations.

o These new variables allow us to obtain a solutionby manipulating fewer equations.

o The new variables are known asnode voltages and mesh currents.

o Node-voltage method enables us to describe a circuit in terms ofne-1 equations.

o Mesh-current method enables us to describe a circuit in terms ofbe-(ne-1) equations.



o By introducing new variables, we cann ddescribe a circuit with justn-1 equations orb-(n-1) equations.

o These new variables allow us too oooobbbbbbbbtttttttttaaaaaaaaaaiiiiiinnnnnn aaaa ssolutionby manipulating feewwwwwwwwwwweeeeeeeeeerrrrrrrrrrr eeeeeeeeeeeeqqqqqqqqqqqqquuuuuuuuuuuaaaaaaaaaaaaaattttttttiiiiiiioooooooooonnnnssssss......

o The new variaaaabbbbbbbbbbbbbbllllleeeeeeeeeeeeessssssssss aaaaaaaaarrrreeeeeeeeeee kkkkkkkkkknnnnnnnnnnnnnnooooooooooooowwwwwwwwnnnnnnnnnnnn aaaaaaaaaaaaasssssssssnode voltages annnnnnndddddddddddddd mmmmmmmmmmeeeeeeessssssshhhhhhhhhhhhh ccccccccccuuuuuuuuuuuuurrrrrrrrrrrrrrrrrrrrrrreeeeeennnnnnnnnnntttttttsssssssssssss....

o Node-voltage method eeennnnnnnnnnaaaaaaaaaaabbbbbbbbbbbllllllllleeeeeeeesssss uuuuuuuusssssssss ttttttttttttooooo dddddeeeeesssssscccccccccrrrrrrrriiiiibbbbbbbbbeeeeeeeee aaaaaaa cccircuit in terms ofne-1 equations.

o Mesh-current method enables us to describe a circuit in terms ofbe-(ne-1) equations.


4.2. Introduction to Node-Voltage Methodo We introduce node-voltage method by

using essential nodes of circuit.

o The first step isto make a neat layout of circuit so that

no branches cross over andto mark clearly essential nodes on circuit diagram.

o This circuit hasne = 3 essential nodes.

o We need2 or (ne-1) node-voltage equations

to describe the circuit.


o We introduce node-voltage method byusing essential nodes of circuit.

o The first step isto make a neat layout of cirrrcccccccuuuuuuiiiiiitttt ssssoooo ttttthhhhhhhaaaaaaatttt

no branches cross over aannnnddddddto mark clearly esseeeeeennnnntttttttttiiiaaaalll nnnnnooooooooooddddeeeesssssss ooooooonnnnn cccciiiiiirrrrccccuuuiittt dddddddiiiaaaaggggrrram.

o This circuit haaassssssne = 3 essential nnnnoooooooooooooodddddddddddeeeeeeeessss.....

o We need2 or (ne-1) node-voltage equations

to describe the circuit.


4.2. Introduction to Node-Voltage Methodo The next step is

to select one of 3 essential nodes as a reference node.

o Although theoretically the choice is arbitrary,practically the choice for the reference node often is obvious.

o For example,the node with the most branches is usually a good choice.

o We flag the chosen reference node with the symbol .


o The next step isto select one of 3 essential nodes as a reference node.

o Although theoretically the choiicccceeeee iiiiiisssssss aaaaaarrrrrrrbbbbbbbiiiiittttrrrrary,practically the choice for theeeeeee rrrrrreeeeffffeeeerrrreeeennnnnnccccccceeeeee node often is obvious.

o For example,the node with the mmmmmmmmmmmoooooooooosssssssssssttttttttttt bbbbbbbbbbbrrrrrrrrrraaaaaaaaaaaaannnnnnnnnnncccccccchhhhhhhhhheeeeeeesssss iiiiiiisssssssss uuuuuuuuuuuuuuussssssssssuuuuuuuuuuaaaaaaaaalllllllllllyyyyyyyyyyyyy aaaaaaaa ggggggooood choice.

o We flag the chhhhooooooooooooossssssssssseeeeeeeeeeennnnnnnnnnnnnn rrrreeeeeeeffffffffeeeeeeeeeeeerrrrrrrrrrrrreeeeeeeeeeeeeennnnnnnnnccccceeeeeeeeeeee nnnnnnnnnnnnnooooooooooooodddddddddddddddeeeeeeeeeeee wwwwwwwwwwwwwiiiiiiiittttttttttttthhhhhhhhhhh ttttttthhhhhhhhhhhhheeeeeeeeee ssssyyyyyyyyyyyyymmmmmmmmmmmmmbbbbbbbbbbbooooooooooooolllllllll ...


4.2. Introduction to Node-Voltage Methodo After selecting reference node,

we define node voltages on circuit diagram,i.e., the voltage rise

from the reference node to a non-reference node.

o We are now ready to generate node-voltage equations.

o We do so by firstwriting the current leaving each branch

connected to a non-reference nodeas a function of the node voltages and then

summing these currents to zeroin accordance with KCL.


o After selecting reference node,we define node voltages on circuit diagram,

i.e., the voltage risefrom the reference noodddddddeeeeeee ttttttoooo aaaa nnnnnnnooooooonnnn-reference node.

o We are now ready to generate nnnnooooooodddddddeeeeeee----vvvvvvvoooooolllllltttttaaage equations.

o We do so by firstwriting theeee cccccccccccccuuuuuuuuurrrrrrrrrrrrrrrrreeeennnnnnnnnttttttttttt lllllleeeeeeeeeeeeaaaaaaaaaavvvvvvvvvvviiinnnnnnnnngggggggggggg eeeeeeeeeaaaaaaaaaaaaaaaccccccccccchhhhhhhhhhhh bbbbbbbbbbbrrrrrrrrrrrraaaaaaaaaaaannnnnnnnnnnccccccccccchhhhhhhhhhhh

connected to aaaaaaaaaaa nnnnnnnnnnnoooooooonnnn------rrrrrrrrreeeeeeeeeeeffffffffeeeeeeeeeeerrrrrrrrreeeeeeeeeeeeennnnnnnnnnnnncccceeeeeeeee nnnnnnnnnnnooooooooddddddddddddeeeeeeeeeeeas a function offff tttttttttttthhhhhhhhhhhheeeeeeeeeee nnnnnnnnnnnnooooooooooooodddddddddddddeeeeeeeeeeee vvvvvvvvoooooolllllllllllltttttttttttaaaaaaaaaaaaaggggggggggggggeeeeeeeeeeessssssssss aaaaaaaaannnnnnnnnnnnddddddddddddd ttthen

summing these currents to zeroin accordance with KCL.


4.2. Introduction to Node-Voltage Methodo Node-voltage equation at node 1 is:

o Node-voltage equation at node 2 is:


o Node-voltage equation at node 1 is:

o Node-voltage equation at nodeee 222222 iiiiiisssss:::::


4.2. Introduction to Node-Voltage Methodo Once the node voltages are known,

all branch currents can be calculated.

o Once these are known,the branch voltages and powers can be calculated.


o Once the node voltages are known,all branch currents can be calculated.

o Once these are known,the branch voltages and pooowwwwwwweeeeeerrrrssss ccccaaaannnnnn bbbbbbbeee calculated.


4.2. Introduction to Node-Voltage Method


4.3. NVM and Dependent Sourceso If the circuit contains dependent sources,

the node-voltage equations must be supplemented withthe constraint equations imposed by

the presence of the dependent sources.


o If the circuit contains dependent sources,the node-voltage equations must be supplemented with

the constraint equations immmmmmmppppppppoooooosssssseeeeeeddddddd bbbbythe presence of the deeeeeppppppppeeeeeennnnnnddddeeeeennnnnnnttttttt sssssoources.


4.3. NVM and Dependent Sources


4.4. NVM: Some Special Caseso When a voltage source is

the only element between 2 essential nodes,the node-voltage method is simplified.

o There are 3 essential nodes.

o 2 simultaneous equations are needed.

o A reference node has been chosen.

o Two other nodes have been labeled.

o The 100 V source constrainsthe voltage between node 1 and the reference node

to 100 V.

o There is only one unknown node voltage (v2).


o When a voltage source isthe only element between 2 essential nodes,

the node-voltage method iiiiiiissssss ssssssssiiiiimmmmmmmpppppppllllliiiiffffiiied.

o There are 3 essential nodes.

o 2 simultaneous equations are nneeeeeeedddddddddeeeeeeeeeeddddddddd....

o A reference node has bbbbeeeeeeeeeeeeeeeeeeeeeennnnnnnnnnnn cccccccccccchhhhhhhhhhhhhoooooooooooossssssssssseeeeeeeeeeeennnnnnnnnnn..

o Two other nooodddddeeeeeeeeeeessssssssssss hhhhhhhhhhhaaaaaaaaaavvvveeeeeeeee bbbbbbbbbbeeeeeeeeeeeeeeeennnnnnn llllllllllaaaaaaabbbbbbbbbbbbbeeeeeeeeeeellllllleeeeeeeeeeeeddddddddddd...

o The 100 V source connsssssssttttttrrraaaiiiiinnnnssssthe voltage betweenn nnnnoooddddee 1111 aannnndddd ttthhhheee rrreeefffeerrreeeennncce node

to 100 V.

o There is only one unknown node voltage (v2).


4.4. NVM: Some Special Cases

o Knowing v2,we can calculate the current in every branch.


o Knowing v2,we can calculate the current in every branch.


4.4. NVM: Some Special Caseso When you use NVM to solve circuits

that have voltage sources connected directly between essential nodes,the number of unknown node voltages is reduced.

Because the difference between the node voltages at these nodes equals the voltage of the source.

o The circuit contains 4 essential nodes.

o We anticipate writing3 node-voltage equations.

o 2 essential nodes are connected by an independent voltage source.

o 2 other essential nodes are connected by a dependent voltage source.

o There is only 1 unknown node voltage.


o When you use NVM to solve circuitsthat have voltage sources connected directly between essential nodes,

the number of unknown nooooooodddddddeeeeee vvvvvvooooooollllllttttaaaages is reduced.Because the differenceeeeeee bbbbbbbeeeeeettttwwwweeeeeeeeeeeeennnnnnn the node voltages at these nodesequals the voltage of ttthhhheeeeee ssssssoooooouuuuuuurrrrrrrccccccceeeee...

o The circuit connnntttttttttttaaaaaaaaaiiiinnnnnnnssssss 4444444 eeeeeeessssssseeeennnnnnnnnntttiiiiiiiiaaaaaaaaaallll nnnnnnnnnooooddddeeeessss.......

o We anticipate writingggggggggggggg3 node-voltage equaaaatttttttttiiiiiiiiiioooooooooooooonnnnnnnnnnnssssssssssss....

o 2 essential nodes are connected by an independent voltage source.

o 2 other essential nodes are connected by a dependent voltage source.

o There is only 1 unknown node voltage.



o We introduce current i becausewe cannot express it as a function of node voltages v2 and v3.

o When a voltage source is between 2 essential nodes,we can combine those nodes

to form a supernode.

o KCL must hold for the supernode:


o We introduce current iii bbbbbbbbbbbeeeeeeeeeeecccccaaaaaaauuuuuuuuuussssssseeeeeeeeeeeewe cannottt eeeeeeeeeeeeeeexxxpppppppppprrrrrrreeessssssssssssssssssss iiiittttttt aaaaaaaasssssssss aaaaaaa ffffffffffuuuuuuuunnnnnnnnnnnnncccccccccctttttiiiiiiiooooooooonnnnnnnnnn oooooooofffffffff nnnnnnnnoooooooooooodddddddeeeeeeeeeee vvvvvvvvoooooooooollllllllllttttttttaaaaaaaagggggggggggeeeeeeeeessssssss vvvvvvvv222222222 aaaand v3.

o When a voltage sourrrrrrrccccccccccccceeeeeeeeeeeee iiiiiissss bbbbbbbeeeeeeeeeeeeettttttttwwwwwwwwwwwwweeeeeeeeeeeeeeeeeeeeeeeeeennnn 22222222 eeeeeeeeeeeessssssssssssssssssseeeeeeeeeeeeeennnnnnnnnntttttttttttiiiiiiiaaaallllllll nnnnnnnnnnnnoooooooooooddddddddddddeeeeeeeeeeeeeessss,we can combine thossssssssssseeeeeeeeeeeeee nnnnnnnnnnnnooooooooooooddddddddddddeeeeeeeeeeeessssssssssssss

to form a supernode.

o KCL must hold for the supernode:



o When we used the branch-current method of analysis,we faced the task of writing and solving

6 simultaneous equations.

o Nodal analysis can simplify our task.


o When we used the brannnccccccccccchhhhhhhhhhhhhh----ccccccccuuuuuuuuuurrrrrrrrrrrrrrrrrreeeeeeeeennnnnnnnnnntttttttttttt mmmmmmeeeeeetttttthhhhhhhhhhhooooooooodddddddddddd oooooooooooffffffffffff aaaaaaaaaaannalysis,we faced the task of writing and solving

6 simultaneous equations.

o Nodal analysis can simplify our task.



o The circuit has 4 essential nodes.

o Nodes a and d are connected by an independent voltage sourceas are nodes b and c.

o The problem reduces to finding a single unknown node voltage.Electric Circuits 1 Chapter 4. Techniques of Circuit Analysis 33

o The circuit has 4 essential noddes.

o Nodes a and d are connected by an independent voltage sourceas are nodes b and c.

o The problem reduces to finding a single unknown node voltage.Electric Circuits 1 Chapter 4. Techniques of Circuit Analysis 33



4.1. Terminology4.2. Introduction to the Node-Voltage Method4.3. NVM and Dependent Sources4.4. NVM: Some Special Cases4.5. Introduction to the Mesh-Current Method4.6. MCM and Dependenttt SSSSoouuurrrrcceeeess4.7. MCM: Somee SSSSppppppeeecccciaaaalll CCCCCCCaaaasssseeeessss4.8. NVM Versusss MMMMMMCCCCCCCMMMMM4.9. Source Transformaatttiiioooooonnnsssss4.10. Thévenin and Nortonnn EEEEqqquuuuiiivvvaaallleeennnttsss4.11. More on Deriving a Thévenin Equivalent4.12. Maximum Power Transfer4.13. Superposition


4.5. Introduction to Mesh-Current Methodo Mesh-current method (MCM) describes a circuit

in terms of be-(ne-1) equations.

o A mesh is a loop with no other loops inside it.

o MCM is applicable only to planar circuits.

o The circuit shown contains7 essential branches with unknown currents and4 essential nodes.

o To solve it via the MCM, we must write4 or 7-(4-1) mesh-current equations.


o Mesh-current method (MCM) describes a circuitin terms of be-(ne-1) equations.

o A mesh is a loop with no other lllloooooooooooopppppppsssss iiiiiiinnnnnnnssssssiiiidddde it.

o MCM is applicable only to plannnnaaaaaaarrrr ccccciiiirrrrrrcccccccuuuuuiiiiittttttssssss...

o The circuit shown contains7 essential brancheeeesssssssssss wwwwwwwwwwwwwwiiiiiiiitttttttttttttthhhhhhhhhhhhh uuuuuuuuuuuunnnnnnnnnnnkkkkkkkkkkkknnnnnnnnnnnooooooowwwwwwwwwwwwnnnnnnnnnn ccccccccccccccccuuuuuuuuuuurrrrrrrrrrrrrrrrrrrrreeeeeeeeennnnnnnnnnnttttttttttttsssssssssss aaaaaaannnnd4 essentiaaallll nnnnnnnnnnnnoooooooooooodddddddddddddeeeeeeeeessss...

o To solve it via the MCCCCCCCCCCMMMMMMMMMMMMM,,, wwwwwwweeeeeeee mmmmmmmmuuuuuuuuuuussssssstttttttttt wwwwwwwrrriiiiiitttttttttttteeeeee4 or 7-(4-1) mesh-cuuuuurrrrrrrrrrrrrrrreeeeeeennnnnnnnntttttttt eeeeeeeeeeqqqqqqquuuuuuuaaaaaaaaaattttiiiiiiioooooonnnnnnnssssssss..


4.5. Introduction to Mesh-Current Methodo A mesh current is current that

exists only in perimeter of a mesh.

o On a circuit diagram, it appears as eithera closed solid line or an almost-closed solid line

that follows perimeter of appropriate mesh.

o An arrowhead on solid line indicatesreference direction for mesh current.

o Mesh currents automatically satisfy KCL.At any node in circuit,

a given mesh currentboth enters and leaves node.


o A mesh current is current thatexists only in perimeter of a mesh.

o On a circuit diagram, it appearsss aaaaasssssss eeeeeeeiiiiittttttthhhhhhheeeeeeerrrrra closed solid line or an almost-closed solid line

that follows perimmmmmmmmmeeeeeeeeettteeeerrrrrr oooooooffffff aaaapppppppppppppprrrroooooppppprrrrrriiiiaaaattteee mmmmmmmeeeesssshhh..

o An arrowheadddd oooooooooooonnnnnnnnnnn ssssssssssooolllliiiiiiiiddddddddddd llllllliiiiiiiiinnnnnnnnnnnnneeeeeeeeee iinnnnnnnnndddddddddddiiiiiiiccccccccccccaaaaaaaaaaaattttttttttteeeeeeeeeeeesssssssreference directioooooooooooooonnnnnnnnnnn ffffffffoooorrrrrrr mmmmmmmmmmmmeeeeeeeeeeessssssssshhhhhhhhhhhhh cccccccccuuuuuuuuurrrrrrrrrrrrrrrrreeeeeeeeeeeennnnnnnnnttttttttt....

o Mesh currents automatiiiicccccccccccccaaaaaaaaaaaaallllllllllllllllllyyyyyyyyyyyy sssssssssssaaaaaaaaaatttttttttttiiiiiiiiiiisssssssssffffffffffffffyyyyyyyyyyyy KKKKKKKKCCCCCCCCCCLLLLLLLLLLLLL....At any node in circuit,

a given mesh currentboth enters and leaves node.


4.5. Introduction to Mesh-Current Methodo Identifying a mesh current in terms of a branch current

is not always possible.

o i2 is not equal to any branch current,whereas i1, i3, and i4 can be identified with branch currents.

o Measuring a mesh current is not always possible.

o There is no place where an ammeter can be inserted to measure i2.

o The fact that a mesh current can be a fictitious quantitydoesn't mean that it is a useless concept.


o Identifying a mesh current in terms of a branch currentis not always possible.

o i2 is not equal to any branch cuuurrrrrrrrreeeeeeennnnnnnttttttt,,,,whereas i1, i3, and i4 can beee iiiiiiiidddddddeeeennnnttttiiiiffffiiiiieeeeeeddddddd wwwith branch currents.

o Measuring a mesh current is noott aaaaaaalllllllwwwwwwwwwwwaaaaaaaayyyyyyyssss ppossible.

o There is no place wherrrreeeeeeeeee aaaaaaaaaaannnnnnnnnnnn aaaaaaaaaaaammmmmmmmmmmmmmmmmmmmmmmmeeeeeeeeettttttttttteeeeeeerrrrrr ccccccccccaaaaaaaaaaannnnnnnnnnnnnnnn bbbbbbbbbbbbbbeeeeeeeeeeee iiiinnnnnnnnnnnssssssssseeeeeeeeeeeeeerrrrrrrtttted to measure i2.

o The fact that aaaa mmmmmmmmmmmmeeeeeeeeesssssssssssshhhhhhh ccccccccuuuuuuuurrrrrrrrrrrrrrrrrreeeeeeeeennnnnnntttt cccccccaaaaaaaaaaaannnnnnnnn bbbbbbbbbbbbbbeeeeeeeeee aaaaaaaaa fffffffffffiiiiiiccccccccttttiiiiiiiiitttttttttttiiiiiiiiiooooooouuuuusssssss qqqqqqqqqquuuuuuuuuaaaaaaaaaaannnnnnnnnnnttttttttttiiiiiiiittttttttttttyyyyyyyyyyyydoesn't mean thaaaaaaatttttttttt iiiiiiittttttt iiissssss aaaaaaaa uuuuusssssssssseeeeeeeeellllllllleeeeeessssssssss cccccccoooooooonnnnnnncccccccccceeeeeeeeepppppppppttttttttt....


4.5. Introduction to Mesh-Current Method


o Above equations are identical in form,with mesh currents ia and ib replacing branch currents i1 and i2.

o MCM of circuit analysisevolves quite naturally from branch-current equations.

o MCM is equivalent to a systematic substitution ofne-1 current equations into be-(ne-1) voltage equations.



o Above equatioonnnsss aaarreeeeee iiiiiiddddeeennnnntttttttiiiiiccccaaaalllllll iiinnnn fffffffffoooooorrrrrmmmmm,,,with mesh currennttsss iiiaaaa aaaaaannnndddddd iiiibbbbbbbbbb rrrreeeeepppppppppppllllaaaaaaacccccccccccciiiinnnnggggggggggg bbbbbbbrrrraaaannnncccchhhh cccuuuurrents i1 and i2.

o MCM of circuit analysisevolves quite naturally from branch-current equations.

o MCM is equivalent to a systematic substitution ofne-1 current equations into be-(ne-1) voltage equations.



o The circuit has7 branches (5 essential branches)

where current is unknown and5 nodes (3 essential nodes).

o Thus, we need3 orb-(n-1)=7-(5-1) orbe-(ne-1)=5-(3-1) mesh-current equations to describe circuit.


o The circuit has7 branches (((5 essennnnttttttttttiiiiaaaaaaaaaaaaalllllllll bbbbbbbbbbbbrrrrrrrrraaaaaaaaaaannnnnnnnnnccccccccccchhhhhhhhheeeeeeeesssssssss)))))

where ccccuuuuuuuuuurrrrrrrrrrrrrrrrrrrreeeeeeeennnnnnnnnnnnttttt iiiisssssssss uuuuuuuuuuunnnnnnnnkkkkkkkkkknnnnnooooooooooowwwwwwwwwwwwnnnnnnnnn aaaaaaaaaaaannnnnnnnnndddddddddddd5 nodes (3 essenttttttttiiiiiiiiaaaaaaaaaaallll nnnnooooooddddddddddddeeeeeeesssssss)))))))...

o Thus, we need3 orb-(n-1)=7-(5-1) orbe-(ne-1)=5-(3-1) mesh-current equations to describe circuit.




4.1. Terminology4.2. Introduction to the Node-Voltage Method4.3. NVM and Dependent Sources4.4. NVM: Some Special Cases4.5. Introduction to the Mesh-Currrrrrrreeeeeeennnnnntttttt MMMMMMMMeeeeeeettttthhhod4.6. MCM and Dependent Sources4.7. MCM: Somee SSSSppppppeeecccciaaaalll CCCCCCCaaaasssseeeessss4.8. NVM Versusss MMMMMMCCCCCCCMMMMM4.9. Source Transformaatttiiioooooonnnsssss4.10. Thévenin and Nortonnn EEEEqqquuuuiiivvvaaallleeennnttsss4.11. More on Deriving a Thévenin Equivalent4.12. Maximum Power Transfer4.13. Superposition


4.6. MCM and Dependent Sourceso If the circuit contains dependent sources,

mesh-current equations must be supplemented byappropriate constraint equations.


o If the circuit contains dependent sources,mesh-current equations must be supplemented by

appropriate constraint eqqqquuuuuuuaaaaaaatttttttiiiiiiooooooonnnnnnsssssss...


o Circuit has6 branches with unknown currents and4 nodes.

o We need 3 mesh currents:

4.6. MCM and Dependent Sources


o Circuit has6 branches with unknown currrrrreeeeeeeeeennnnnnnnnnnnnntttttttttttsssssssss aaaand4 nodes.

o We need 3 meeeessssssssssshhhhhhhhhhh cccccccuuuuuuuuuuuurrrrrrrreeeeeeeeennnnnnnnnntttttttttttssssssss:::::


4.6. MCM and Dependent Sourceso What if you had not been told to use MCM?

o Would you have chosen NVM?

o It reduces problem to finding 1 unknown node voltagebecause of the presence of 2 voltage sources between essential nodes.

o We present more aboutmaking such choices later.


o What if you had not been told to use MCM?

o Would you have chosen NVM?

o It reduces problem to finding 1 uuuuunnnnnnnkkkkkkknnnnnoooooowwwwwwwnnnnnnn nnnode voltagebecause of the presence of 2222222 vvvvvvoooollllltttttttaaaaaaagggggggeeeeee sssssssooources between essential nodes.

o We present more aboutmaking such choiceeeesssssssssss lllllaaaaaaaaaaaaaattttttttttteeeeeeeeeeeeeerrrrrrrrr....rrrr




4.1. Terminology4.2. Introduction to the Node-Voltage Method4.3. NVM and Dependent Sources4.4. NVM: Some Special Cases4.5. Introduction to the Mesh-Currrrrrrreeeeeeennnnnntttttt MMMMMMMMeeeeeeettttthhhod4.6. MCM and Dependenttt SSSSoouuurrrrcceeeess4.7. MCM: Some Special Cases4.8. NVM Versusss MMMMMMCCCCCCCMMMMM4.9. Source Transformaatttiiioooooonnnsssss4.10. Thévenin and Nortonnn EEEEqqquuuuiiivvvaaallleeennnttsss4.11. More on Deriving a Thévenin Equivalent4.12. Maximum Power Transfer4.13. Superposition


4.7. MCM: Some Special Caseso When a branch includes a current source,

MCM requires some additional manipulations.

o We definedmesh currents ia, ib, and ic, andvoltage across 5 A current source.

o Circuit contains5 essential branches

with unknown currents and4 essential nodes.

o We need to write 2 or 5-(4-1) mesh-current equations to solve circuit.


o When a branch includes a current source,MCM requires some additional manipulations.

o We definedmesh currents ia, ib, and ic, aaaaaaannnnnndddddvoltage across 5 A current ssssoooouuuuuuurrrrrrrccccccceeeeeee....

o Circuit contains5 essentiaaallll bbbbbbbbbbbbrrrrrrrrraaaaaaaaannnnnncccchhhhhhhhhhhheeeeeeeeeeessssssssss

with unknownn ccccccccccccuuuuuuuuuuurrrrrrrrrrrreeeeeeeeennnnnnnnnnttttttttttsssssssss aaaaaaaaaannnnnnnnnnnnnddddddddddddd4 essential nodes.

o We need to write 2 or 5-(4-1) mesh-current equations to solve circuit.


4.7. MCM: Some Special Caseso Presence of current source reduces

3 unknown mesh currents to 2 such currents.

o It constrainsdifference between ia and ic to equal 5 A.

o If we know ia,we know ic, and vice versa.


o Presence of current source reduces3 unknown mesh currents to 2 such currents.

o It constrainsdifference between ia and iiicccccc ttttttoooooo eeeeqqqqqqquuuuaaaaaalllllll 555555 A.

o If we know ia,we know ic, and viccceeeeeeeee vvvvvvvvvvveeeeeeeeeeeerrrrrrrrrrrrsssssssssaaaaaaaaaa....


4.7. MCM: Some Special Caseso For mesh a:

o For mesh c:

o Adding the above equations:

o For mesh b:

o For current source branch:

o Solving equations:


o For mesh a:

o For mesh c:

o Adding the above equations:

o For mesh b:

o For current source brancccchhhhhhhhhhhh:::::::::

o Solving equations:


4.7. MCM: Some Special Cases

o We can solve the problemwithout introducing unknown voltage v

by using the concept of a supermesh.

o To create a supermesh,we mentally remove current source from circuit

by simply avoiding this branchwhen writing mesh-current equations.


o We can solve ttthhhheeee ppppprrrooobbbblllleeeemmmwithout innnttttrrrooooddddddduuuucccciiiinnnnggg uuuunnnnkkkknnnnoooowwwwnnnn vvvvoooolllllllttttaaaaaaaaagggggeeeee vvvv

by using the coonnnnnnncccceeepppppppppptttt ooooffff aaaa ssssuuuppppppppeeeerrrrmmmmeeeessssshhhhhhhhhh..

o To create a supermesh,we mentally remove current source from circuit

by simply avoiding this branchwhen writing mesh-current equations.



o We express voltages around supermesh in terms of original mesh currents:


o We express vooollllttttaaaagggggeeeesss aaaarrrroooouuunnnddddddddddd ssssuuuuppppppppeeeerrrrmmmmeeeesssssssshhhhhhhhhh iiiiiinnnn ttttttteeeerrrrmmmmssssss ooooffff oooorrrrigggiinnnaal mesh currents:


4.7. MCM: Some Special Caseso Circuit has

4 essential nodes and5 essential branches

with unknown currents.

o Circuit can be analyzed in terms of5-(4-1) or 2 mesh-current equations.

o Although we defined 3 mesh currents,dependent current source forces

a constraint between mesh currents ia and ic .

o We have only 2 unknown mesh currents.


o Circuit has4 essential nodes and5 essential branches

with unknown currents.

o Circuit can be analyzed in termssss ooooooofffffff5-(4-1) or 2 mesh--ccccccccccuuuuuuuurrrrrrrreeeennnnnnnnnntttt eeeeeeqqqqqqqquuuuuuuaaaaaattttiiiiioooonnnnssssss...

o Although we ddddeeeeeeeeeeeeeeffffffiiinnnnnnnnneeeeeeeeedddd 333333333333 mmmmmmmmmmmmmmeeeeeeeeeessssssssssshhhhhhhhh ccccccccccccuuuuuuuuuurrrrrrrrrrrrrrrrrrrrrreeeeeeeeeeennnnnnnnnnntttttttssssssssss,,,,,,,,dependent curreeeennnnnnnnnnnnnnttttttttttt ssssssssooooooouuuuuuurrrrrrrrrrrcccccccceeeeeeeeeee fffffffffooooooooooooorrrrrrrrrrccccceeeeeeeeesssssssssss

a constraint betweeeeeeeeeeeeeeeeeeeeennnnnnnnnnnn mmmmmmmmmmmmmmeeeeeeeeeeesssssssssshhhhhhhhhhh cccccccccccuuuurrrrrrrrrrrrrrrreeeeeeeeeeeennnnnnnnnnnttttttttttttssssssssssss iiiiiiiiaaaaaaaaaa aaaaaaaaaaaannnnnnnnnnnnnddddd ic .

o We have only 2 unknown mesh currents.




4.1. Terminology4.2. Introduction to the Node-Voltage Method4.3. NVM and Dependent Sources4.4. NVM: Some Special Cases4.5. Introduction to the Mesh-Currrrrrrreeeeeeennnnnntttttt MMMMMMMMeeeeeeettttthhhod4.6. MCM and Dependenttt SSSSoouuurrrrcceeeess4.7. MCM: Somee SSSSppppppeeecccciaaaalll CCCCCCCaaaasssseeeessss4.8. NVM Versus MCM4.9. Source Transformaatttiiioooooonnnsssss4.10. Thévenin and Nortonnn EEEEqqquuuuiiivvvaaallleeennnttsss4.11. More on Deriving a Thévenin Equivalent4.12. Maximum Power Transfer4.13. Superposition


4.8. NVM Versus MCMo The greatest advantage of both NVM and MCM is that

they reduce the number of simultaneous equationsthat must be solved.

o They also require the analyst to be quite systematic in terms oforganizing and writing these equations.

o When is NVM preferred to MCM and vice versa?

o There is no clear-cut answer.

o Asking a number of questions,may help you identify the more efficient method.


o The greatest advantage of both NVM and MCM is thatthey reduce the number of simultaneous equations

that must be solved.

o They also require the analyst tooooo bbbbbbbeeeee qqqqquuuuiiiitttttteeeeeee sssssyyystematic in terms oforganizing and writing theseeee eeeeeeqqqqqqqquuuuuuuaaaaaaattttttiiiiioooooonnnnsss.

o When is NVM preferreeedddddddddd ttttttttttooooooooooo MMMMMMMMMMMMCCCCCCCCCCCMMMMMMMMMMMM aaaaaaaaaannnnddddddd vvvvvvvvviiiiiiccccccccccceeeeeeeeeee vvvvvvvveeeeeeeeerrrrrrrsssssssssaaaaaaaaaaa?????????

o There is no cleeeeaaaaaaaaaaaaarrrrrrrrrrr-------ccccccccccuuuuuuuuuuuuuttttt aaaaaaaaaaaannnnnnnnnnnssssssssssssswwwwwwwwwwwwwwweeeeeeeerrrrrrrrr....

o Asking a number of qqqqqqquuuuuuuuuuueeeeeeesssstttttttiiiiioooooooonnnnnnnnsssssss,,,,may help you identiffffyyyyyyyyyy tttttttttttthhhhhhhhhheeeeeeeee mmmmmmmmmoooooooorrrrrrrreeeeeeeeeee eeeffffffffffffiiiicccccciiiiiiiieeeeeeeeennnnnnnnnttttttt mmmmmmmmmmeeeeeeeeetthhod.


4.8. NVM Versus MCMo Does one of the methods result in fewer simultaneous equations to solve?

o Does the circuit contain supernodes?If so, using NVM will permit you

to reduce the number of equations to be solved.

o Does the circuit contain supermeshes?If so, using MCM will permit you

to reduce the number of equations to be solved.

o Will solving some portion of the circuit give the requested solution?If so, which method is most efficient

for solving just the pertinent portion of the circuit?

o Some time spent thinking about the problem in relation to the various analytical approaches available is time well spent.


o Does one of the methods result in fewer simultaneous equations to solve?

o Does the circuit contain supernodes??If so, using NVM will permittt yyyyyooooooouuuuuuu

to reduce the number of eeeeeeeqqqqqquuuuaaaattttiiiioooonnnnnnnssssss to be solved.

o Does the circuit contain supermmeeessssshhhhhhhhhheeeeeeeeeessssssss???????If so, using MCM wwwiillllllllllllllll ppppppppppppeeeeeeeeeeeeerrrrrrrrmmmmmmmmmmmiiiiiitttttttttt yyyyyyyyyyoooooooooouuuu

to reduucccceeeeeeeeeeeeee tttttttthhhhhhhhheeeeee nnnnnnnnnuuuuuuuuuummmmmmmmmmmmbbbbbbbbbbeeeeeeeeeeerr oooooooooooofffffffff eeeeeeeeeeeeqqqqqqqqqqqqquuuuuuuuuuuaaaaaaaaatttttttttttiiiiiiiiiooooooooooooonnnnnnnnnnnssssssss tttttttttttoooooooooo bbbbbbbbbbbeeeeeeeeeeee sssssssssoooooooooooolllllllvvvvvvvvvvvvveeeeeeeeeedddddddddddd.......

o Will solving some poooooorrrrrrrrrrtttttttttttttiiiiiiiiiiiioooooooonnnnnn ooooooooooofffffffffffff tttttttttthhhhhhhhhhhhhheeeeeeeeeeeeee ccccccciiiirrrrrrrrrrcccccccuuuuuuuuuuuuuiiiiiiiitttttttt ggggggggggggggiiiiiivvvvvvvvvveeeeeeeeeeee ttthhhhhhhhhhhheeeeeee rrrrrrrrreeeeeeeeeeqqqqqqqqqqqqqquuuuuuested solution?If so, which method iiiisssssssssssss mmmmmmmmmmmmmmoooooooooooossssssssssstttttttttt eeeeeeeeeeeeffffffffffffffffffffffiiiiiiiiiiiiccccccciiiieeeeeeeennnnnnnnttttttttttttt

for solving just the pertinent portion of the circuit?

o Some time spent thinking about the problem in relation to the various analytical approaches available is time well spent.


4.8. NVM Versus MCMo Find the power dissipated in R = 300 .

o We need to find eithercurrent or voltage of resistor.

o MCM yields current in the resistor.

o MCM requires solving5 simultaneous mesh equations.

o In writing the 5 equations,we must include

the constraint i = -ib.


o Find the power dissipated in R = 300 .

o We need to find eithercurrent or voltage of resistor.

o MCM yields current inn ttttttttthhhhhheeeeeeeeeeeeee rrrrrrrrrrrrreeeeeeeeeeeeessssssssssssiiiiiisssssssssssstttttttttttoooooooooooorrrrrrrrrr.......

o MCM requireeessss sssssssssooooooooooooolllllvvvvvvvvvvviiiiiinnnnngggggg5 simultaneous mmmmmmmeeeeeeeeeeessssshhhhhhh eeeeeeqqqqqqqquuuuuuuaaaaaaaatttttttttiiiiiiiiiiioooooooooonnnnsssssss..

o In writing the 5 equationnnssss,we must include

the constraint i = -ib.


4.8. NVM Versus MCMo Find the power dissipated in R = 300 .

o The circuit has 4 essential nodes.

o Only 3 NV equations are required.

o Because of dependent voltage sourcebetween 2 essential nodes,

we have to sum currents at only 2 nodes.

o Problem is reduced to writing2 NV equations anda constraint equation.

o NVM requires only 3 simultaneous equations,it is the more attractive approach.


o Find the power dissipated in R = 300 .

o The circuit has 4 essential nodeeeesssss..

o Only 3 NV equations are requirrreeeeedddddd..

o Because of dependent voltage soouuuuuuuuurrrrrrrrrrrrrcccccccccccccceeeeeeeeeeebetween 2 essentiaaaallll nnnnnnnnnnooooooooooodddddddddddddeeeeeeeeesssssssssss,,

we haveeee tttttttttttooooooooooooo ssssssssuuuuuuuuuuummmmm cccccccccuuuuuuuuuuurrrrrrrrrrrrrrreeeeeeennnnnnnnnnttttttttttssssssssss aaaaaaattttttttttt oooooooooonnnnnnnlllllllllyyyyyyyyyyy 22222222222222222 nnnnnnnooooooooooodddddddeeessssssssss..

o Problem is reduced tttooo wwwwrrriiiiittttiiiinnnngggggg2 NV equations anda constraint equation.

o NVM requires only 3 simultaneous equations,it is the more attractive approach.


4.8. NVM Versus MCM


4.8. NVM Versus MCMo Find the voltage vo in the circuit.

o Circuit has4 essential nodes and2 voltage-controlled dependent sources.

o NVM requires solving3 NV equations and2 constraint equations.


o Find the voltage vo in the circuit.

o Circuit has4 essential nodes and2 voltage-controlleeeddddddddd dddddddddddeeeeeeeeeeeeeppppppppppppppeeeeeeeeeeeennnnnnnndddddddddddeeeeeeeeeeeennnnnnnnnnntttt ssssssoooooooooooooouuuuuuuuuuurrrrrrrrrrrrrrrcccccccccceeeeeeeeeeeeessssssssssss..

o NVM requiresss sssssssssssoooooooooooolllllllvvvvvvvvviiiiiiiinnnnnnnngggg3 NV equations aaaaaaannnnnnnnnnndddddddddd2 constraint equatioooonnnnnnnnnnsssssssssss..


4.8. NVM Versus MCMo Find the voltage vo in the circuit.

o Circuit contains 3 meshes.

o We can use leftmost one to calculate vo.

o If ia denotes leftmost mesh current,vo = 193-10ia.

o Presence of 2 current sources reduces problem to solvinga single supermesh equation and2 constraint equations.

o MCM is the more attractive technique here.


o Find the voltage vo in the circuit.

o Circuit contains 3 meshes.

o We can use leftmost one to calculllaaaaaaaaatttttttttttteeeeeeeeeeeee vvvvvvvvvoooo.

o If ia denotes leeeffffttttmmmmoooossstttt mmmmeeeessshhhhh ccccuuuurrrrrrreeeennntttt,,,vo = 193-111100000000iiiiiiia.

o Presence of 2 currentt soooouuuuuuuuuurrrrrrrccccccceeeeeeeesssssssss rrrrrrrreeeeeeedddddddduuuuuuuuucccceeeeeeeeessssssssss pppppppppppprrrrrrrrrrooooooooobbbbbbblllleeeeeeemmmmmmmmm tttto solvinga single supermesh equatiion anddd2 constraint equations.

o MCM is the more attractive technique here.


4.8. NVM Versus MCM




4.1. Terminology4.2. Introduction to the Node-Voltage Method4.3. NVM and Dependent Sources4.4. NVM: Some Special Cases4.5. Introduction to the Mesh-Currrrrrrreeeeeeennnnnntttttt MMMMMMMMeeeeeeettttthhhod4.6. MCM and Dependenttt SSSSoouuurrrrcceeeess4.7. MCM: Somee SSSSppppppeeecccciaaaalll CCCCCCCaaaasssseeeessss4.8. NVM Versusss MMMMMMCCCCCCCMMMMM4.9. Source Transformations4.10. Thévenin and Nortonnn EEEEqqquuuuiiivvvaaallleeennnttsss4.11. More on Deriving a Thévenin Equivalent4.12. Maximum Power Transfer4.13. Superposition


4.9. Source Transformationso Even though NVM and MCM are powerful techniques,

we are still interested in methodsthat can be used to simplify circuits.

o Series-parallel reductions and -to-Y transformations arealready on our list of simplifying techniques.

o We begin expanding this list withsource transformations.

o A source transformation, allowsa voltage source in series with a resistor to be replaced by

a current source in parallel with same resistor or vice versa.

o A source transformation is bilateral.


o Even though NVM and MCM are powerful techniques,,we are still interested in methods

that can be used to simplifffffffyyyyyyyy cccccccciiiiirrrrrccccccuuuuuuuiiiiittttssss.

o Series-parallel reductions and ----ttttttoooo----YYYY ttttrrrrraaaaaaannnnnnsformations arealready on our list of simpliffffyyyyiiiiinnnnnnnngggggggg tttttttteeeeeeecccccchhhhhhnnniques.

o We begin expanding thhhiiiissssssss llllllliiiiiiisssssssssttttttttttt wwwwwwwwwwwiiiiiittttttttttthhhhhhhhsource traaannnnnnsssssssssssssffffoooooooooooorrrrrrrrrrmmmmaaaaaaaaaaaattttttttttiiiiiioooooooooooonnnnnnnnnnssssssssss..

o A source transformaaaaatttttttttttttiiiiiiiiiioooooooooooooonnnnnnnn,,, aaaaaaallllllllllllllloooooooowwwwwwwwwwwwwsssssssssssssa voltage source in sssseeeeeeeeeeeeerrrrrrrrrrrriiiiiiiiiiiieeeeeeeeeeessssssssssss wwwwwwwwwwwwwiiiiiitttttttttttttthhhhhhhhhhh aaaaaaaaaaaa rrrrrrreeeeeeessssssssiiiiiissssssssssssttttttttoooooooooooorrrrrrrrrr ttttttttttttooooooooooooo bbbbbbbbeee replaced by

a current source in parallel with same resistor or vice versa.

o A source transformation is bilateral.


4.9. Source Transformationso We need to find relationship between vs and is

that guarantees 2 configurations are equivalentwith respect to nodes a, b.

o Equivalence is achieved ifany RL experiences same current, and same voltage,

if connected between nodes a, b:

o If polarity of vs is reversed,orientation of is must be reversed to maintain equivalence.


o We need to find relationship between vsv and isithat guarantees 2 configurations are equivalent

with respect to nodes a, b...

o Equivalence is achieved ifany RL experiences same cuurrrrrrrrreeeeeeennnnnnntttttt,,,, aaaaaannnnnndddd ssame voltage,

if connected betwwwwwwwwweeeeeeeeeeeennnn nnnnnooooooooodddddeeeesssssss aaaaaaa,,,, bbbb::

o If polarity of vsv is reversed,orientation of isi must be reversed to maintain equivalence.


4.9. Source Transformations

o Circuit has4 essential nodes and6 essential branches with unknown currents.

o We can find current in branch containing 6 V source by solving either3 = [6-(4-1)] mesh-current equations, or 3 = [4-1] node-voltage equations.

o We can also simplify circuit by using source transformations.

o We must reduce circuit in a way thatpreserves identity of branch containing 6 V source.


o Circuit has4 essential nodes and6 essential brancheeeesssssssssss wwwwwwwwwwwwwwwiiiiiiiiiiiittttttttttthhhhhhhhhhhh uuuuuuuuuuuunnnnnnnnnnnkkkkkkkkkkknnnnnnnnnnnoooooooowwwwwwwwwwwwnnnnnnnnnn ccccccccccccccuuuuuuuuuuuuurrrrrrrrrrrrrrrrrreeeeeeeennnnnnnnnnntttttttttttttssssssssss...

o We can find cccuuuuurrrrrrrrrrrrrrrrrrrrrreeeeeeeeeeennnnnnnnntttttttttt iiinnnnnnnn bbbbbbbbbbbrrrrrrrraaaaaaaannnnnnnccchhhhhhhhhhhh cccccccccccoooooooooonnnnnnnnnnnttttttttaaaaaaaaaiiiiiiinnnnnnnnnnniiiinnnnnnnnnngggggg 66666666666666 VVVVVVV ssssssssssooooooooouuuuuuuuuurrrrrrrrcccccccccccceeeeeeeeee bbbbbbbbbbbbbyyyyyyyyyy sssssssssssooooooooooolllllvvving either3 = [6-(4-1)] mmmmmmeeeeeeessssssssssshhhhhhh-cccccccuuuuuuuurrrrrrrrrrrreeeeeeeeennnnnnnnnttttttttttt eeeeqqqqqqquuuuuuuaaaaaaattttiiiiiiioooooooooonnnnnnnnnsssssssss,,,, oooorrrrr 3 = [4-1] node-volttttttaaaaaaagggggggggggeeeeeee eeeeeqqqqquuuuaaaaaattttttttttttiiiiiiiiioooonnnnsssss..

o We can also simplify circuit by using source transformations.

o We must reduce circuit in a way thatpreserves identity of branch containing 6 V source.



is = (19.2-6)/16 = 0.825 A


isi = (19.2-6)/16 = 0.825 A


4.9. Source Transformationso What happens if

there is an Rp in parallel withvoltage source?

o What happens ifthere is an Rs in series with

current source?

o In both cases,resistance has no effect on equivalent circuit that

predicts behavior with respect to terminals a, b.


o What happens ifthere is an Rp in parallel with

voltage source?

o What happens ifthere is annn RRRRRRRRRRRRRRssssssss iiiiiiinnnnnnnnnnnnn ssssseeeeeeerrrrrrrriiiiiiiiiieeeeeeeeeeeeesssssssssssss wwwwwwwwwwiitttttttttttthhhhhhhhhhhh

current sourceeeeeeeee?????????????

o In both cases,resistance has no effect on equivalent circuit that

predicts behavior with respect to terminals a, b.




p8A(developed) = -(-60)(8) = 480 WElectric Circuits 1 Chapter 4. Techniques of Circuit Analysis 72

p8A(developed) = -(-60)(8) = 480 W



o 125 and 10 resistorsdo not affect the value of vo butdo affect the power calculations!


o 125 and 10 resistorsdo not affect the value of vo butdo affect the power calculations!



4.1. Terminology4.2. Introduction to the Node-Voltage Method4.3. NVM and Dependent Sources4.4. NVM: Some Special Cases4.5. Introduction to the Mesh-Currrrrrrreeeeeeennnnnntttttt MMMMMMMMeeeeeeettttthhhod4.6. MCM and Dependenttt SSSSoouuurrrrcceeeess4.7. MCM: Somee SSSSppppppeeecccciaaaalll CCCCCCCaaaasssseeeessss4.8. NVM Versusss MMMMMMCCCCCCCMMMMM4.9. Source Transformaatttiiioooooonnnsssss4.10. Thévenin and Norton Equivalents4.11. More on Deriving a Thévenin Equivalent4.12. Maximum Power Transfer4.13. Superposition


4.10. Thévenin and Norton Equivalentso Sometimes, we want to concentrate on

what happens at a specific pair of terminals.

o For example, when we plug a toaster into an outlet,we are interested primarily in

voltage and current at terminals of toaster.

o We have little or no interest ineffect that connecting toaster has on

voltages or currents elsewhere in circuit supplying outlet.

o We then are interested inhow voltage and current delivered at outlet change

as we change appliances.

o We want to focus on the behavior of circuit supplying outlet,but only at outlet terminals.


o Sometimes, we want to concentrate onwhat happens at a specific pair of terminals.

o For example, when we plug a tooooaaaaassssssstttttttteeeeeeerrrrrr iiiiiinnnnnnnttttttoooo an outlet,we are interested primarily iiiiinnnnnnn

voltage and current at teerrrrmmmmmmmiiiiiinnnnnnnaaaaaaallllllssssss oooooofff toaster.

o We have little or no innttteeeeeeeeeeerrrrrrrrreeeeeeeeeeeessssssssssssstttttttt iiiiinnnnnnnnneffect thattt ccccccccccccccoooooonnnnnnnnnnnnnnnnnneeeecccccccccctttttttiiiiiiiiiiinnnnnnnnnnggggggggggggg ttttttttttoooooooaaaaaaaaassssssssssstttttttttteeeeeeeeeeeerrrrrrrrrrrr hhhhhhhhhhhhhaaaaaaaaaasssssssssss ooooooooooonnnnnnnnnnn

voltages or currrrrrrrrrrrrrrrrreeeeeeeeeeennnnnnnnttttsssssssss eeeeeeeeeeeelllllssssssssseeeeeeeeeeeeewwwwwwwwwwwwwwhhhhhhhhhhhhheeeeeerrrrreeeeeeeeeee iiiiiiiinnnnnnnnn cccccccccccccciiiiiiiiiirrrrrrrccccccccccccccuuuuiiiitttttttttt ssssssssssuuuuuuupppppppppppppppppppppppppllllying outlet.

o We then are interested iiiinnnnnnnnnnnnnhow voltage and current delivered at outlet change

as we change appliances.

o We want to focus on the behavior of circuit supplying outlet,but only at outlet terminals.


4.10. Thévenin and Norton Equivalentso Thévenin and Norton equivalents are

circuit simplification techniquesthat focus on terminal behavior.

o They are extremely valuable aids in circuit analysis.

o Here we discuss resistive circuits,but Thévenin and Norton equivalent circuits may be used to represent

any circuit made up of linear elements.


o Thévenin and Norton equivalents arecircuit simplification techniques

that focus on terminal behhhhhaaaaaaavvvvvvviiiiiiiooooooorrrrrr...

o They are extremely valuable aiddddddddsssss iiiiinnnn cccciiiirrrrcccccuuuuuuuiiiiitttt analysis.

o Here we discuss resistive circuittsss,but Thévenin and NNNooooooooooorrrrrrrrrrtttttttttttooooooooooooonnnnnnnnnnn eeeeeeeeeeeqqqqqqqqqqqqqquuuuuuuuuiiiiiiivvvvvvvvvvaaaallllleeeeeeeeeeennnnnnnnnttttttttt cccccccccciiiiiirrrrrrrrcccccccccuuuuuuiiiiiiiittttttttttttssssssss mmmmmmmmmaaay be used to represent

any circccuuuuuuiiiiiiiiiiitttttttt mmmmmmmmmmmaaaadddddddddeeeeeeeeeeee uuuuuuuuuuppppppppppppp ooooooooooofffff lllliiiiiinnnnnnnnnneeeeeeeeeeeeeeeaaaaaaaaaaarrrrrrrrrrr eeeeeeeellllllleeeeeeeeeeeemmmmmmmmmmmeeeeeeeeeennnnnnnnnnntttttttttsssss...


4.10. Thévenin and Norton Equivalents

o Letters a and b denotepair of terminals of interest.

o A Thévenin equivalent circuit isan independent voltage source VTh in series witha resistor RTh ,

which replaces an interconnection of sources and resistors.

o If we connect same load across terminals a and b of each circuit,we get same voltage and current at terminals of the load.

o This equivalence holdsfor all possible values of load resistance.


o Letters a and b denotepair of terminals of interestttt...

o A Thévenin equivalent circuit isan independent vooooltttttttttttaaaaaaaaaggggggggggggggeeeeeeeeeeeeee ssssssssssssoooooooooooouuuuuuuuuuuuuurrrrrrrrcccccccccccceeeeeeeeeee VVVVVVVVTTTTTTTTThhhhhhhhhhVVVV iiiiiiiiinnnnnnnnnnnnn ssssssssssseeeeeeeeeeeeeerrrrrrriiiiiieeeeeeeeeeeesssssssss wwwwwwwwwwwwwiiiitha resistor RRRRRRRRTTTTTTTTTTThhhhhhhhhhh ,,,,,,,

which replacessssssssss aaaaaaaaaaaannnnnnnn iiinnnnnnnnttttttttttteeeeeeeeeeeeerrrrrrrrccccccccccccoooooooooooonnnnnnnnnnnnnnnnnneeeeeeeeeeeccccccccccctttttttttttttiiiiiiiiioooooooooonnnnnnnnnnnnnn oooooooooooffffffffffff ssssssoooooooouuuuuuuurrrrrrrrrrrrrrccccccccccceeeeeeeeeessssssssssssss aaand resistors.

o If we connect same loadddd aaaaacccccccrrrrrooooosssssssss tttttttttteeeeeerrrrrrrmmmmmiiiiinnnnnaaaallllllllsssssssss aaaaaaaaa aaaaaannnnnndddddddddddd bbb of each circuit,we get same voltage and current at terminals of the load.

o This equivalence holdsfor all possible values of load resistance.



o We must be able todetermine VTh and RTh.

o If load resistance is infinitely large,we have an open-circuit condition.

The open-circuit voltage at the terminals a and b is VTh in circuit (b).It must be same as open-circuit voltage at terminals a and b inoriginal circuit.

o To calculate VTh,we simply calculate open-circuit voltage in original circuit.


o We must be able todetermine VThVV and RTh.

o If load resistance is inffffiiiinnnnnnnnnniiiiittttttttttteeeeeeeeellllllllllllyyyyyyyyyyy llllllllllaaaaaaaaarrrrrrrrrrggggggggggeeeeeeeeee,,,we have aaannnnnn oooooooooooppppppppppeeeeeeeeeeeennnnnn---cccccccccciiiiiiiiiirrrrrrrrcccccccccccuuuuuuuuiiiiittttttt cccccccccccoooooooooonnnnnnnnnnndddddddddddddiiiiiiiiiiittttttttiiiiiiiiioooooooonnnnnnnnnnn....

The open-circuuuuuuuiiiiiiiitttttttttt vvvvvvooooolllllttttaaaaaagggggggeeeeeeeeee aaaaaattttttttt ttthhhhhhhhhhheeeeeee tttttttteeeeeeerrrrrrrrrrmmmmmmmmmiiiiiiiiinnnnnnnnnaaaallllllssssss aaaaaaaaaaaa aaaaaannnnnnnndd b is VThVV in circuit (b).It must be same aasssss oooooooppppppeeeeennnn--cccciiiiiiirrrrrrcccccccccuuuuuiiiitttttt vvvvoooolllllttttttttaaaaaaagggggeeeeee aaaaatttttt tterminals a and b inoriginal circuit.

o To calculate VThVV ,we simply calculate open-circuit voltage in original circuit.



Reducing load resistance to zerogives us a short-circuit condition.

o In circuit (b), short-circuit current directed from a to b is:

o This short-circuit current must be identical toshort-circuit current that exists in

a short circuit placed across terminals a to b of original network.


Reducing load resistance to zerogives us a short-cirrccccuuuuuuuuuuiiiiiiiiittttttttttttt ccccccccccccooooooooooonnnnnnnnnnndddddddddddiiiiiiittttttttttiiiiiiiooooooooooonnnnnnnn...

o In circuit (b), ssshhhhhhhhoooorrrrttt--ccciiiirrrrccccuuuiiiiiiitttt ccccuuuurrrrrrrreeeennnntttt dddddddddiiiirrrreeeecccctttttttttteeeedddddddd ffffffffrooooommmmm aaaa ttttoooo bbbbbbb iiiiiiiissssssss::::::

o This short-circuit current must be identical toshort-circuit current that exists in

a short circuit placed across terminals a to b of original network.



o Thévenin resistance isratio of open-circuit voltage to short-circuit current:

o If short-circuit current is directed from b to a is,a minus sign must be inserted in the equation.


o Thévenin resissttttaaaannnnccceeee iiissssratio of opppeeennnn---ccciiiiiiiirrrccccuuuuiiittttt vvvvoooolllllttttaaaagggggggggeeee ttttoooo sssshhhhhhhhoooorrrrrrrrrtttttt---ccccciiiiirrrcuuuuuiiiiittttt ccccuuuurrrrrrrreeeenntttttt::::

o If short-circuit current is directed from b to a is,a minus sign must be inserted in the equation.



o When terminals a and b are open,there is no current in 4 resistor, andvab is identical to v1.

o Thévenin voltage for circuit is 32 V.


o When terminals a and b are openn,,,there is no currentt iiiinnnnnnnn 4444444444444 rrrrrrrreeeeeeeessssssssiiiiiiiiiiisssssssstttttttttttoooooooorrrrrr,, aaaaaaaaaannnnnnnnnnnnnddddddddddddvab is identtttiiiiicccccccccccaaaaaaaaaaaalllllll ttttttttttttooooooooo vvvvvvvv1111.

o Thévenin voltage for circuit is 32 V.



o The short-circuit current (isc) is found easilyonce v2 is known.

o Problem reduces tofinding v2 with the short-circuit in place:


o The short-circcuuuuiiittttt ccccuuuurrrrrrreeeennnntttt (((((((iiisssssssscccccc)))))) iiissss ffffffffoooouuunnnndddddddd eeeeeeeeaaaaaasssiiiiilllllyyyyyyyonce v2 is kkkknnnoooowwwwnnn.....

o Problem reduces tofinding v2 with the short-circuit in place:



o If a 24 resistor is connectedacross terminals a and b in original circuit,

voltage across resistor is 24 V andcurrent in the resistor is 1 A,

as would be the case with Thévenin circuit.

o This same equivalence between circuits holdsfor any resistor value connected between nodes a and b.


o If a 24 resisstttoooorrrr iiissss cccoooonnnnnnnneeeccctttteeeedddddddddddacross termmmiiinnnnaaaallllllllsss aaaa aaannnddddddd bbbbbb iiiiinnnn oooorrrriiiiiiiggggggggiiiiinnnnaaaallllll ccccciiirrrrrccccuuuitttt,,,,,,,

voltage across rrreeeeeeeesssiiisssssttttooorrrr iiiissss 222224444 VVVV aaaaannnnddddddcurrent in the resiisssstttooorrrr iiiss 111 AAAA,

as would be the case with Thévenin circuit.

o This same equivalence between circuits holdsfor any resistor value connected between nodes a and b.


4.10. Thévenin and Norton Equivalentso A Norton equivalent circuit consists of:

an independent current source in parallel withNorton equivalent resistance.

o We can derive it froma Thévenin equivalent circuit simply by

making a source transformation.

o Norton current equalsshort-circuit current at terminals of interest.

o Norton resistance is identical toThévenin resistance.


o A Norton equivalent circuit consists of:an independent current source in parallel withNorton equivalent resistance....

o We can derive it froma Thévenin equivalent circuuuiiiittttt sssssssiiiimmmmmmmppppppplllllyyyyyy bbbyyy

making a source ttttttttttrrrraaaaaaaaannnnssssffffffoooorrrrrrrrrrmmmmaaaaaaattttiiiiooooonnnn...

o Norton currennnttttttttt eeeeeeqqqqqqqqqqqquuuuuuuuuuaaaallllssssssssssssshort-circuit currrrrreeeeeeeeeeeeeennnnnnnnnnntttttttt aaaaaattttttt ttttttttttteeeeeeeeerrrrrrrrrmmmmmmmmmmmmmmiiiiiinnnnnnnnnnnnaaaaaalllsssssssssss ooooooooooofffffffff iiiiinnnnnnnnnnnttttttteeeeeeeeeeeeerrrreeeeessssssssssssttttttttt......

o Norton resistance is idennnnnnnnnnnnnttttttttttttiiiiiiiiiiccccccccccccaaaaaaaaaaalllllllllll ttttttttttooooooooooThévenin resistance.


4.10. Thévenin and Norton Equivalentso Sometimes, we can make effective use of source transformations

to derive a Thévenin or Norton equivalent circuit.

o This technique is most useful whenthe network contains only independent sources.

o The presence of dependent sources requiresretaining identity of controlling voltages and/or currents.

o This constraint usually prohibitscontinued reduction of circuit by source transformations.


o Sometimes, we can make effective use of source transformationsto derive a Thévenin or Norton equivalent circuit.

o This technique is most useful wwwhhhheeeeeeennnnnnnthe network contains only innnnnnndddddddeeeepppppeeeennnndddddddeeeeeeennnnt sources.

o The presence of dependent souurrrcccceeeeeeeesssssssssss rrrrrrreeeeeeqqqquuiiiresretaining identity ooffff ccccccoooooooooooonnnnnnnnnnnnttttttttttrrrrrrrrrrooooooooooolllllllllllllllllliiiiiiinnnnnnnnngggggggggg vvvvvvvooooooooooolllllllttttttttaaaaaaaaaaaaagggggggggggggeeeeeeeeeeessssssss aaaaaaannnnnnnnnnnnddddddddd//////////ooooorr currents.

o This constrainnnttttt uuuuuuuuuuussssssssssssuuuuuuuuuuuuaaaaaaaaaaallllllllyyyyyyyyyyyy ppppppppppprrrrrrrrrrroooooooooooohhhhhhhhhhhiibbbbbbbbbbbbiiiiiiiiiiitttttttttttssssssssssscontinued reducttttttttttttiiiiiiiiiiooooooooooooonnnnnnnn oooooooofffffff ccccccccccccciiiiirrrrrrrrrrccccccccccccuuuuuuuuuuuuuiiiiiiiiittttttt bbbbbbbbbbbyyyyyyyyyyy ssssssssssoooooooooouuuuuuuuuuuuuurrrrrrrrccccccccccceeeeeeeeeeee tttrrrrrrrrrrraaaaaaaannnnnnnnnnnnnnnsssssssssfffffffffoooooooooooooorrrrrrmations.



o Thévenin Equivalent:

o Norton Equivalent:


o Thévenin Equuiiivvvvaaalllleeennnnttt::::

o Norton Equivalent:


4.10. Thévenin and Norton Equivalentso Find the Thévenin equivalent for the

circuit containing dependent sources.

o Current ix must be 0.Note the absence of a return path for ix to enter left-hand portion of circuit.


o Find the Thévenin equivalent for the circuit containing dependent sources.

o Current ixi must be 0.Note the absence oooffffffffffff aaaaaaaaaaa rrrrrrrrrreeeeeeeeeeettttttttttttuuuuuuuuuuurrrrrrrrrrrrrnnnnnnnnnnn ppppppppppaaaattttttthhhhhhhhhhhh fffffffffooooooooooooorrrrrrrrrr iiiiiiixxxxxxxxxxxxxi tttttttooooooooo eeeeeeeeennnnnnnnnttttteeeer left-hand portion of circuit.


4.10. Thévenin and Norton Equivalentso Find the Thévenin equivalent for the

circuit containing dependent sources.

o With short circuit shunting 25 ,all current from dependent current source appears in short circuit:


o Find the Thévenin equivalent for the circuit containing dependent sources.

o With short circuit shunting 25 ,,,,all current from dependent ccccuuuuuurrrrrrrrrrrrreeeeeeennnnnnntttttt source appears in shhhoooorrtt ccciiirrcccuuiittt::::


4.11. More on Deriving a Thévenin Equivalento Technique for determining RTh we discussed is

not always the easiest method available.

o 2 other methods generally are simpler to use.

o The first is useful if network containsonly independent sources.

o To calculate RTh for such a network, we:deactivate all independent sources,calculate resistance seen

looking into network at designated terminal pair.

o A voltage source is deactivated by replacing it with a short circuit.

o A current source is deactivated by replacing it with an open circuit.


o Technique for determining RTh we discussed isnot always the easiest method available.

o 2 other methods generally are ssssiiimmmmmmmppppppplllllleeeeeeerrrrr ttttoooo use.

o The first is useful if network connnnnnntttttttaaaaaaiiiinnnnsssssssonly independent sources.

o To calculate RTh for succcchhhhhhhhhhh aaaaaaaaaaa nnnnnnnnnnnnnneeeeeeeeeeeeettttttttttttwwwwwwwwwwwwwwooooooooooorrrrrrrrrkkkkkkkkkkk,,, wwwwwwwwwwwweeeeeeeeeee::::::::::deactivateeee aaaaaaaaaaaallllllllllll iiiiiiiinnnnnnnnnnnnnnndddddeeeeeeeeeeepppppppppppeeeeeeeeeeeeennnnnnnnnnnnnddddddddddddeeeeennnnnnnnnnntttttttttt ssssssssssooooooooooooooouuuuuuuuuuuurrrrrrrrrccccccccccccceeeeeeeeeeeeessssssssssss,,,,,,,calculate resistannnnnnnnccccccccccceeeeeeeeeeeeee ssssseeeeeeeeeeeeeeeeeennnnnnnnnnnnn

looking into netwooooooooooorrrrrrrrrkkkkkkkkkkkkkk aaaaaaaaaaaattttttttttttt ddddddddddddeeeeeeeeeessssssssssiiiiiiiiiiigggggggggggggnnnnaaaaaaaattttttteeeeeeeeeeeeeddddddddddddd tttttttttttteeeeeeeeeeerrrrrrrrrrrmmmmmmmmmmmmmiiiiiiiiinnnnnnnnaaal pair.

o A voltage source is deactivated by replacing it with a short circuit.

o A current source is deactivated by replacing it with an open circuit.


4.11. More on Deriving a Thévenin EquivalentFirst alternative procedure to find RTh


First alternative procedure to find RTh


4.11. More on Deriving a Thévenin Equivalento If circuit or network contains dependent sources,

an alternative procedure for finding RTh is as follows:We first deactivate all independent sources.We apply either

a test voltage source ora test current source

to Thévenin terminals a and b.Thévenin resistance equals

ratio of voltage across test source to current delivered by test source.


o If circuit or network contains dependent sources,an alternative procedure for finding RTh is as follows:

We first deactivate all indeeeeeeeppppppppeeeeeeeennnnnndddddddeeeeeeennnnntttt sources.We apply either

a test voltage source oooorrrra test current sssoooouurrrccceee

to TTTThhhhéééévvvveeeennnniiiinnnn ttttteeerrrmmmmiiiiinnnnaaaallllllssss aaaa aaaannnnddddddddddd bbbbbbbbbbbbbb...Théveniiiinnn rrrreeeessssiiiiiiissssttttaaaannnccceeee eeeeqqqquuuuaaaalllllssss

ratio of voltttaaggggggggeee aaaaaccccrrrroooossssssss tttteeeessstttt sssssooooouuuurrrrcccceeeeeeee tttttoooo ccccuuuurrrrrreeennnnt delivered by test source.


4.11. More on Deriving a Thévenin EquivalentSecond alternative procedure to find RTh


Second alternative procedure to find RTh


4.11. More on Deriving a Thévenin Equivalento We can use a Thévenin equivalent to

reduce one portion of a circuit togreatly simplify analysis

of larger network.


o We can use a Thévenin equivalent toreduce one portion of a circuit to

greatly simplify analysisof larger network.


4.11. More on Deriving a Thévenin Equivalento This modification has no effect on branch currents i1, i2, iB, and iE.

o We replace circuit made up of VCC, R1, and R2

with a Thévenin equivalent,with respect to terminals b and d.


o This modification has no effect on branch currents i1, i2, iB, and iE.

o We replace circuit made up of VCCVV , CC RR11, and R2

with a Thévenin equivalent,,,,with respect to terminalssssss bbbbbbb aaaannnndddd ddddddd....


4.11. More on Deriving a Thévenin Equivalent


4.12. Maximum Power Transfero We assume

a resistive networkcontaining independent and dependent sources and

a designated pair of terminals a and b,to which a load RL, is to be connected.

o Problem is to determine value of RL thatpermits maximum power delivery to RL.


o We assumea resistive network

containing independent annnnnnnddddddd dddddddeeeeeeepppppppeeeeeennnndddent sources anda designated pair of terminaaaaaaalllllllsssssss aaaaaa aaaannnnnnddddddd bbbbbbb,,,

to which a load RL, is to bbbbeeeeee ccccccooooooonnnnnnnnnnnneeeeeeecccccttteeed.

o Problem is to determinneeeeeeeeee vvvvvvvvvaaaalllluuuuuuuuueeee ooooooffff RRRRRRLLLLLLL ttttthhhhaaaattttttpermits mmmaaaaaxxxxxxxxxxxxiimmmmmmmmmmmuuuummmmmmmmmmm pppppppppoooowwwwwwwwweeeeeerrrrrrrrr ddddddeeeeeeeeelllliiiivvvveeeerrryyyyyyyyyyy tttttoooo RRRRLLLLLLLLLL...


4.12. Maximum Power Transfero Replacing original network by its Thévenin equivalent

greatly simplifies task of finding RL.

o Derivation of RL requiresexpressing power dissipated in RL

as a function of 3 circuit parameters VTh, RTh, and RL:


o Replacing original network by its Thévenin equivalentgreatly simplifies task of finding RL.

o Derivation of RL requiresexpressing power dissipatedddddddd iiiinnnn RRRRLLLLL

as a function of 3 circuit ppppaaaaaaarrrrrrraaaaaaammmmmmmeeeeeetttttteeeeerrrsss VThVV , RTh, and RL:


4.12. Maximum Power Transfer

Derivative is 0 and p is maximized when:


Derivative is 0 and p is maximizedddddd wwwwwwwwwwwwwwwhhhhhhhhhhheeeeeeenn:




RL = 25


RL = 25

4.13. Superpositiono A linear system obeys principle of superposition, i.e.,

whenever a linear system is excited, or driven,by more than one independent source of energy,

total response is sum of individual responses.

o An individual response is result ofan independent source acting alone.

o Because we are dealing withcircuits made up of interconnected linear-circuit elements,

we can apply superposition directly to analysis of such circuits.

o At present, we restrict the discussion to simple resistive networks.

o Principle is applicable to any linear system.


o A linear system obeys principle of superposition, i.e.,whenever a linear system is excited, or driven,

by more than one indepennnnnnndddddddeeeeeeeennnnnntttttt ssssssoooooouuuurce of energy,total response is sum ooooooofffffff iiiiiinnnnnnddddiiiivvvvvviiiiiiiddddddduuuuual responses.

o An individual response is resulttt oooooofffffffan independent souuuuuuuurrrrccccccccceeee aaaaaaccccccttttiiiiiiiiinnnngggg aaaallllllooooonnnneeee....

o Because we arrreeeeeeeeeeeeee dddddddddeeeeeeeeeaaaaallliiiinnnnnnnnnnnngggggggggggg wwwwwwwwwwwwwwwiiiitttttttttthhhcircuits made up oooooooooooooofffffffffff iiiiiinnnnntttttttteeeeeeeeerrrrrrrrrrrccccccccooooooooooooonnnnnnnnnnnnnnnnnnnnnneeeeeeeeeccccccccctttttteeeeeeeeeeeeddddddddddd lllllllliiiiiinnnnnnnnnnnneeeeeeeeeeeaaaaaaaaaaaarrr---ccccccccccccciiiiiiiirrrrrrrrrrrccccuuuuuuuuuuuiiiiiiiiiiiittttt elements,

we can apply supeeeerrrrrrrrrrrpppppppppppppppoooooooooooossssssssssssiiiiiiiitttttttttiiiiiiiioooooooooonnnnnnnnnnn dddddddddddiiiirrrrrreeeeeeeeeeecccccccccttttttttttttllllllllyyyyyyyyyyyy tttttttttoooooooooo aaaaaaaaaaaaannnnnnnnnnnnnnaaaaaaalllysis of such circuits.

o At present, we restrict the discussion to simple resistive networks.

o Principle is applicable to any linear system.


4.13. Superpositiono Superposition is applied in both analysis and design of circuits.

o In analyzing a complex circuitwith multiple independent voltage and current sources,

there are often fewer, simpler equations to solveby applying superposition.

o Applying superposition can simplify circuit analysis.

o Sometimes, applying superposition actually complicates analysis,producing more equations to solve.

o Superposition is required only ifindependent sources in a circuit are fundamentally different.

o When all independent sources are dc sources,superposition is not required.


o Superposition is applied in both analysis and design of circuits.

o In analyzing a complex circuitwith multiple independent vvvvooooolllllllttttttttaaaaaaaggggggeeeeeee aaaaaannnnddd current sources,

there are often fewer, simmmmmmmpppppplllllleeeerrrr eeeeqqqqqquuuuuuuaaaaaations to solveby applying superpositttiiiioooooonnnnnnnn..

o Applying superpositionnn cccccccaaaaaaaaaaannnnnnnnn sssssssssssiiiiiimmmmmmmmmmmmmpppppppppppllllllliiiiiiiiffffffffffyyyy ccccccciiiiiiirrrrrrrrrcccccccccuuuuuuuuuuuiiiiiittttttttt aaaaaaaaannnnnnnaaaaaaaaaaaalllllyyyyyyyyyyyyyssssssssiiissss.

o Sometimes, aaapppppppppppppppppppppppllllllyyyyyyyyyyyyiiiiiiiiiinnnnnnnnnnngggg sssssssssssuuuuuuuuuuuupppppppppppppppeeeeeeeeeeeeerrrrrrrrppppoooooooooooossssssssssiiiiiiiiiittttttttiiiiiiiiiiiioooooooooooooonnnnnnnnnnn aaaaaaaaaaaaacccccccccccctttttttttttttuuuuuuuuuuuaaaaalllllllllllllllllllllyyyyyyyyyyy ccccooooooooooooommmmmmmmmmmmmpppppppppppllllllllllliiiiiiiiiccccccccccccaaaaaaaaaaatttttttttttteeeeeeeeeeeeesssssssssssss aaaaaaaaaaaannnnalysis,producing more eeeeeeeeeeqqqqqqqqqqqqquuuuuuuuuuuuaaaaatttttttiiiiioooooooooonnnnnnnnnssssssssss tttttttttttoooooooooooooo ssssooooooooooollllllvvvvvvvvvvvveeeeeeeeeee...

o Superposition is requireeddddddddddddd oooooooooonnnnnnnlllllllyyyyyyyy iiiiiiiiifffffffffindependent sources in a circuit are fundamentally different.

o When all independent sources are dc sources,superposition is not required.


4.13. Superpositiono Superposition is applied

in design to synthesize a desired circuit response thatcould not be achieved in a circuit with a single source.

o If desired circuit response can be written asa sum of two or more terms,

response can be realized by includingone independent source for each term of response.

o This approach to design of circuits with complex responsesallows a designer to

consider several simple designsinstead of one complex design.


o Superposition is appliedin design to synthesize a desired circuit response that

could not be achieved in a ccccccciiiiiirrrrrrrrccccccuuuuuuuiiiiiitttttttt wwwwwiiith a single source.

o If desired circuit response can bbbbbbbeeeeeee wwwwrrrriiiitttttttteeeeennnnnnn aaasa sum of two or more termssss,

response can be rrrrrrrrreeeeeeeaaaaaaallliiiizzzzeeeeeeeeeeddddddddd bbbbyyyy iiiinnnnnnnccccclllluuuddddddiiinnnnggggone iiinnnnnnnddddddddddeeeeeeeeeeppppppppeeennnnnnnnnddddddddeeeennnntttt ssssssooooooooouuuuuuuuuurrrrccccccccceeee ffffoooorrr eeeeeeeeaaaaccchhhh tttttttteeeeeeeeerrrrrrrrmmmmmmm oooooooooffff rrrreeeesssssspppppppoooonnnsssseeee.

o This approach to dessssssiiiiiiiiiigggggggggggggnnnnnnnnnnn oooooofffffff cccccccccccciiiiiiirrrrrrrrrccccccccccccuuuuuuuuuiiiiiiiitttttttttttsssssssss wwwwwwwwwwwwwiiiiiiiiittttttttttthhhhhhhh cccccccccccccoooooooooommmmmmmmmmmmmppppllllllleeeeeeeeeeeeexxxxxxxxxxx rrrrrreeeeeeeeeeesssssssssppppponsesallows a designer to

consider several simple designsinstead of one complex design.


4.13. Superposition


4.13. Superpositiono When applying superposition to linear circuits

containing both independent and dependent sources,you must recognize that dependent sources are never deactivated.


o When applying superposition to linear circuitscontaining both independent and dependent sources,

you must recognize that deeeeeeeppppppppeeeeeeeennnnnndddddddeeeeeeennnnntttt sources are never deactivated.


4.13. Superposition


Chapter 4. Techniques of Circuit Analysis - KNTU …wp.kntu.ac.ir/faradji/EC1/EC1_Ch4.pdf ·...

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