Chapter 4: Methods of Analysis When SCT are not applicable, it s because the circuit is neither in series or parallel. There exist extremely powerful mathematical methods that use KVL & KCL as its basis that will make circuit soling easy! In fact, some students in the past called soling circuits "enjoyable" and they actually make you seem "smart." These methods are by far the most popular way to actually sole resistie circuits - they are called Node and Mesh analysis. NODE analysis uses KCL to find (node) oltages conert currents into oltages. MESH (or Loop) analysis uses KVL to find (mesh) currents conert currents into oltages. NODE ANALYSIS with CURRENT SOURCES ONLY Let me remind you of seeral preious concepts:. When we talk about oltages, we really mean electric potential differences, which reference a particular reference point (usually the ground). In circuit soling, you will choice the ground that is the most conenient for the particular circuit you are dealing with.. A node oltage is the potential defined at a single point. For a circuit with n nodes, there are n node oltages; howeer, since a ground has to be defined, there are actually n- node oltages to sole for. 3. Node analysis in its simplest form only has current sources present. General Idea Step : Determine the number of nodes n, label them and pick the reference. If the ground is around, choose it! Label all the other nodes n :, n-. Step : Pick the direction of the currents (usually from high to low potential). Be consistent! Step 3: Apply KCL and Ohm's law to each node and rewrite all of the currents in terms of node oltages. Step 4: Sole the equations for, n-. You should get n nodes and n equations to sole. Example:. There are 3 nodes including the ground, so there are 3 node oltages to sole for. Label them as A and B.. Pick current directions. 3. Apply KCL and Ohm's law to each node to get node oltages: A 0 A 0 A B KCL at Node : is i+ i3 + i and Ohm's law: i, i 3, i R R R Substitute back into KCL: 4 4 3 4 A A A B Mathematical i + i3 + i4 + + i S trick + + A B is R R3 R4 R R3 R4 R4 standard textbook method - not in matrix form Repeating the same process for Node : Carlos' method immediately gets it into matrix form 4.
KCL at Node : i i + i and Ohm's law: i, i B A B 4 S 4 R R4 Substitute back into KCL: (note the way I will write it out is i i + i i i i ) ' ' ' ' 4 S 4 S i i i + i ' ' 4 S S R R4 R R4 R4 Since we are gien current sources i S and i S as well as all resistor alues, we end up with two equations and two unknowns to sole and with. 4. Sole for and ; we can then back substitute to get all the primed currents using Ohm s law. + + i R R R R 3 4 4 A B S + i B A S R R R 4 4 Let s examine this problem again but this time note the following details. The current sources hae a (+) sign for the current source means it flows into the node (supplies the node) and the ( ) sign flows out of the node (draws from the node). NODE ANALYSIS with current sources only Step : Choose a reference, label all nodes (including the ground) and determine the number of unknown node equations. # of unknown nodes (# of node equations) ( ) Step : Apply the node equation to each unknown node oltage until the number of unknown node oltages equals the number of node equations: Sum of conductances Mutual (, 3,, n ) ± isource connected to node conductances Step 3: Sole for all node oltages using matrix techniques and answer the question/problem. This form of Node analysis is the same as that in the book, howeer, my method simplifies the book methods by one additional algebraic step so to get the node equations in matrix form quicker. SOLVING SYSTEMS OF EQUATIONS When using node analysis to sole circuit problem, one sets up a system of equations that must be soled, and therefore, you need to be really good at soling these. I am aware that some of you hae not had linear algebra; howeer, these techniques are straightforward enough that you can pick it up here. What does the calculator essentially do when you enter this into your calculator? Two equations, two unknowns Gien the system of equations, we can write this in matrix form: ax + by c a b x c ax c + by c a y b The solution is 4.
c b a c c b a c x, y D D where D is the determinant for a matrix: a b D ab ba a b Using these matrix techniques to sole node and mesh analysis setups, let s apply to the preious two examples we just looked at. Example 4. (a) Find the node oltages. (b) Find the powers of the ma and 4mA-current sources; are they actie or passie? Solution a. There should be a huge red flag being waed right in front of your eyes. Note that the units on the circuit are in Ω and ma. The easy way to compute circuits is either with (Ω & A) or (kω & ma) but not mixed together as shown in the circuit. So I will conert into kω & ma. Let s do the numbers: 3 nodes 0 current sources 3 node equations and 3 unknown node oltages. Define the nodes, and 3 and apply node analysis to get The system is ( + ) ( ) ( ) 3 0. 0. 0. 0. ( 0. 0. ) ( 0.) ( 0. ) 3 ( ) ( ) ( ) Node : 4 Node : + 4 Node 3: + + 0. 0. 0. 3 0. 0. 4 0 0 0 4 0 5 5 4 0 5 0 0. 0. 0. 3 3 matrix form 0.+ 0. 0. 3 4 solution 4 0. 0. + 0. 3 3 solutions 3 0. V 0.4 V 0. V b. The power of the ma-current sources is oltage across the source times it current alue. Since the current enters the negatie node, the masource is actie: PmA mai (( 0. 0 V) ma) 0. mw P ma For the left 4mA-source, the current enters from the ground, which is the higher potential. Therefore, the 4mA-source is passie: 4.3
P4mA 4mAi ((0 0. V) 4 ma) + 0. mw P 4mA Once again, it is required that you be able to sole system of equations with your calculator. If you do not know, I am sure that many students in this course already know how to do this or look online. Your ability to pass the first exam and quizzes will depend on this skill. Get it! Note that mathematics is a tool in this course; it is not our main direction. The bottom line is soling circuits and understanding them. NODE ANALYSIS WITH VOLTAGE SOURCES When using Node analysis to calculate node oltages, there are two oltage sources, defined by their locations relatie to the ground that one must understand: Voltage sources connected to the reference ground Voltage sources NOT connect to the reference ground. Voltage sources in a circuit requires one to modify how node analysis is applied. Consider the following circuit and let s analyze all of the node oltages without any calculations. As usual, we ask how many nodes are in the circuit, and we find that there are 3 nodes to sole for (not including the ground). Note that the two oltage sources in the circuit are different in how they are connected to the ground: the () 0V-source is connected to the ground whereas the () 8V-source is NOT connected to the ground.. 0V-source is connected to the ground The left top node must be 0 V since the 0V-source is directly connected to the ground and immediately determines the alue of the top left node. Therefore, the 0V-oltage source literally reduced the number of unknown node oltages by one. So the role of a oltage source in node analysis reduces the number of node oltages by the number of oltage sources. In this case, we hae (Number of node equations) (# of Nodes ) (# of oltage sources) 3 unknown nodes source unknown node oltages. 8V-source is NOT connected to the ground A oltage source not connected to the ground sets up a special situation that requires care in setting up the node equations. Wheneer a oltage source is NOT connected to the ground it forms a Supernode oltage; that is, a supernode is formed when two (or more) nodes are connected to each ia a oltage source NOT connected to the ground. Identify the Supernode by drawing a dashed circle around the two connected nodes. The supernode forces node oltages a and b to be dependent upon each other and do not act independently anymore. This can be seen by writing out the oltage across the 8V-oltage source in terms of nodes oltages: 8 + 8 a b If I know the alue of b, I will automatically know the alue of a. Haing the 8V-source in the circuit then further reduces the number of nodes we need to sole for: 3 nodes oltage sources unknown node oltage to sole for a b 4.4
That is, although we started with 3 nodes, we now hae 3 unknown node equations to sole. Howeer, since nodes a and b are dependent on each other, we write a Supernode Condition SC. The SC tells us how these two nodes are connected to each other ia the oltage source: SC : 8 Oerall summary of Supernodes. The effect of a oltage source in a circuit is to reduce the number of unknown node oltages required to calculate. (Number of node equations) (# of nodes ) (# of oltage sources). If a oltage source is NOT connected to the ground, then that oltage source forms a SUPERNODE and one has to modify the node equations. A Supernode is a node equation where seeral node equations are combined into one single supernode equation. A Supernode is typically circled with dashed lines to identify it. SN equation: (node equation for node a) + ( node equation for node b) ± i source a b supernode equation 3. Along with eery Supernode equation, there is a SUPERNODE CONDITION ( SC) that explicitly shows how these two nodes are dependent on each other ia the oltage source not connected to the ground: SC: a b S 4. Tricky scenario: if after the counting there are no node equations to be had (that is, the number of nodes is equal to the number of oltage sources in the circuit), then it is the collection of SC equations that still exist and are used to sole for the node oltages. Node Analysis with Current and Voltage Sources Step : Choose a reference and label all nodes including the ground, and determine the number of unknown nodes equations: (Number of node equations) # of nodes # of oltage sources ( ) ( ) Step : Apply the node equation to each unknown node oltage until the number of unknown node oltages equals the number of node equations: Sum of conductances Mutual (, 3,, n ) ± isource connected to node conductances Step 3: If a oltage source is NOT connected to the ground, then that oltage source forms a SUPERNODE (and identify it by drawing a dashed loop). The two nodes that form a supernode must be written as a Supernode (SN) equation plus a Supernode Condition (SC): SN equation: (node equation for node a) + ( node equation for node b) ± i source SC: a b S supernode equation Step 4: Sole for all node oltages and appropriate quantities. If we apply this process to this circuit, let s go through these steps. Write down the Supernode equation for this circuit: ( + ) a 00 00 ( 00 ) 8+ ( 00 ) b 0 Apply the Supernode condition: Writing this into matrix form, we get node a + 8 a b node b 4.5
+ 6V a b 8 b 8 a V 8 8 00 a 00 b 00 matrix form 00 00 a 00 solution a Example 4. Determine the node oltage and i for the circuit. Solution After choosing the bottom node to be ground, there are 3 unknown nodes with one oltage source not connected to the ground, so that the nodes associated with this oltage source forms a supernode. Due to the supernode, there is automatically an associated SC. Let s do the numbers: regular node 3 nodes source node eqs SN ( B, C) SC (, ) 3 eqs & 3 unknows ( A, B, C) Applying our rules for node analysis, B ( ) A ( ) The system of equations is C + B Node A: C + B SN (, ): ( ) ( ) ( ) ( ) ( ) 3 A B 0 3 C 0 A 7 matrix 7 solution A B 4 C form 4 B A 0 A- B C 0 + + C + + B C 7 B C + B A 5 5 C + + 5 4 C 5 B + + Node B Node C SC (, ): A.9V For the current i, we don t need to sole any equation, and immediately apply simple circuit techniques to get it. The V-source is parallel to the 5Ω resistor, and therefore, has the same oltage drop. Applying Ohm s law: Ri i 0.4A 5 MESH or LOOP ANALYSIS Node Analysis equations are generated by KCL. In Mesh Analysis the equations are generated instead by KVL to create loop currents. The easiest situation to start with in Mesh Analysis is haing all oltage sources and later including current source into our equations. MESH ANALYSIS with oltage sources only The general idea is 4.6
Step : Identify all independent loops and assign a loop current to all loops in the same direction (either CW or CCW). Here I choose CW directions. Step : From the current directions, these define the resistor polarities and label them. Step 3: Apply KVL in the CW direction to each loop and rewrite all oltages into currents using Ohm s law. Step 4: Sole the resulting equations for each loop currents. There should be equation for each independent loop. Example: KVL for loop-a: + + 4+ 0 350 00 Ohm's law 350 350ia 00 00(ia i b ) Substitute back into KVL: Mathematical + 350ia + 4 + 00 00( ia i b ) 0 trick ( 350 + 00) ia ( 00) ib 4 standard textbook method - not in matrix form Repeating this process again for loop-b with Carlos trick, 400 + 00 00 i 4 8 ( ) i ( ) b a Carlos' method - immediately matrix form Summarizing these system of equations, we get ( 350 + 00) ia ( 00) ib 4 matrix 450 00 ia solution ia 6.5 ma form ( 400 + 00) ib ( 00) ia 4 8 00 500 i b 4 ib 9.3 ma Let s step back and analyze this result: Sum of the resistors for the current flowing in loop- If another loop is mutual, subtract those resistor alues multiplied by the appropriate current. Place oltage source alues on the other side with the understanding that MESH or LOOP ANALYSIS with oltage sources only Step : For each loop assign a current loop and define the polarities. Determine the number of unknown loop currents: (# of loop equations) # of unknown loop currents ( ) Step : Apply the loop equation to each unknown loop current until the number of unknown loop currents equals the number of loop equations: Sum of resistors Mutual i (i,i 3,,i n) ± connected with loop- resistors Step 3: Sole for all loop currents and appropriate quantities. sources Example 4.3 Determine the mesh currents (i, i, and i 3) and the oltage across 3Ω-resistor for the circuit shown. 4.7
Solution There are 3 unknown loop currents (which I draw in the CW directions) and we want to sole for 3Ω. Now that we are more sophisticated, we should start adding a soler equation for 3Ω. Since we do not know the directions of the loop currents, we guest that the oltage 3Ω is positie when 3Ω 3(i i ) or 3i 3i 3Ω 0 soler equation I now add this to my calculations for the mesh circuit. Let s do the numbers: loops (i, i, i 3) 3 unknow loops 3 loop currents soler ( 3 ) 4 eqs & 4 unknows (i, i, i 3, 3) Writing these out, gies ( + 3 + 9) i ( 3) i ( 9) i3 0 4 3 9 0 i 0 i 3 A ( 3 6) i ( 3) i ( 6) i3 5 i + matrix 3 9 6 0 5 solutions i A form ( 6 + 9) i3 ( 9) i ( 6) i 9 6 5 0 i 3 i3 4 A 3i 3 3 0 3i 3 0 3 0 3 V 3 MESH ANALYSIS with oltage and current sources When using mesh analysis to calculate loop currents, there are two flaors of current sources in circuits that one must accommodate for: current sources that are isolated from other loops current sources CONNECTING two loops together. Current sources in a circuit requires one to modify how mesh analysis is applied. Consider the following circuit and let s determine all of the loop currents. As usual, we ask how many loops are in the circuit, and we find that there are 3 loops to sole for. Note that the two current sources in the circuit are different in how they are connected in the circuit: the () 0mA-source is isolated from the other loops whereas the () 0mA-source is connected between two loops and NOT isolated.. 0mA-source is isolated from the rest of the loops The left most loop current must be 0mA since the 0mA-source is isolated from the rest of the loops and immediately determines the alue of the left most loop. Therefore, the 0mA-current source literally reduced the number of unknown loop currents by one. So the role of a current source in mesh analysis reduces the number of loop currents by the number of current sources. 4.8
In this case, we hae (Number of loop equations) (# of loops ) (# of current sources) 3 unknown loops source unknown loop currents. 0mA-source IS connected to other loops A current source connected to other loops sets up a special situation that requires care in setting up the mesh equations. Wheneer a current source is CONNECTS two loops together it forms a Superloop (or Supermesh) current; that is, a superloop is formed when two loops are connected to each ia a current source. Identify the Superloop by drawing a dashed circle around the two connected loops. The superloop forces loop currents i and i 3 to be dependent upon each other now and do not act independently anymore. This can be seen by writing out the current through the 0mA-current source: i i 0 i i + 0 3 3 If I know the alue of i, I will automatically know the alue of i 3. Haing the 0mA-source in the circuit then further reduces the number of loops we need to sole for: ( ) ( ) (Number of mesh equations) # of loops # of current sources 3 loops current sources unknown loop currents to sole for That is, although we started with 3 loops, we now hae 3 unknown loop equation to sole. Howeer, since loops i and i 3 are dependent on each other, we write a Superloop Condition SC. The SC tells us how these two loops are connected to each other ia the current source: SC : i3 i 0 If I know the alue of b, I will automatically know the alue of a. Haing the 8V-source in the circuit then further reduces the number of nodes we need to sole for: 3 nodes oltage sources unknown node oltage to sole for That is, although we started with 3 nodes, we now hae 3 unknown node equations to sole. Howeer, since nodes a and b are dependent on each other, we write a Supernode Condition SC. The SC tells us how these two nodes are connected to each other ia the oltage source: SC : 8 Analogy: Superloops are analogous to Supernodes in Node Analysis. When a oltage source was included in the circuit and Node Analysis is done, the number of node oltages that one needed to sole for was reduced by the number of oltage sources in the circuit. A similar situation exists for Mesh Analysis. Oerall summary of Supernodes 5. The effect of a oltage source in a circuit is to reduce the number of unknown node oltages required to calculate. (Number of node equations) (# of nodes ) (# of oltage sources) 6. If a oltage source is NOT connected to the ground, then that oltage source forms a SUPERNODE and one has to modify the node equations. A Supernode is a node equation where seeral node equations are combined into one single supernode equation. A Supernode is typically circled with dashed lines to identify it. a b 4.9
+ ( ± i SN equation: (node equation for node a) node equation for node b) supernode equation 7. Along with eery Supernode equation, there is a SUPERNODE CONDITION ( SC) that explicitly shows how these two nodes are dependent on each other ia the oltage source not connected to the ground: SC: x y S 8. Tricky scenario: if after the counting there are no node equations to be had (that is, the number of nodes is equal to the number of oltage sources in the circuit), then it is the collection of SC equations that still exist and are used to sole for the node oltages. Oerall summary of Superloops. The effect of a current source in a circuit is to reduce the number of unknown loop currents required to calculate. (Number of mesh equations) (# of loops ) (# of current sources). If a current source is NOT isolated from other loops, then that current source forms a SUPERLOOP and one has to modify the mesh equations. A superloop is a mesh equation where seeral mesh equations are combined into one single superloop equation. A Superloop is typically circled with dashed lines to identify it. SL equation: (mesh equation for loop-) + ( mesh equation for loop-) ± source superloop equation 3. Along with eery Superloop equation, there is a SUPERLOOP CONDITION ( SC) that explicitly shows how these two loops are dependent on each other ia the current source connecting the two loops together: SC: i i i x y S 4. Tricky scenario: if after the counting there are no loop equations to be had (that is, the number of loops is equal to the number of current sources in the circuit), then it is the series of SC equations that still exist and are used to sole for the currents. Mesh (or Loop) Analysis with Current and Voltage Sources Step : For each loop assign a current loop and define the polarities. Determine the number of unknown loop currents: (Number of mesh equations) # of loops # of current sources ( ) ( ) Step : Apply the loop equation to each unknown loop current until the number of unknown loop currents equals the number of loop equations: Sum of resistors Mutual i (i,i 3,,i n) ± sources connected with loop- resistors Step 3: If a current source is NOT isolated from other loops, then that current source forms a SUPERLOOP (and identify it by drawing a dashed loop) and the two loops from Step must be written as a Superloop (SL) equation plus a Superloop Condition (SC): SL equation: (mesh equation for loop-) + ( mesh equation for loop-) ± source SC: i i i x y S superloop equation Step 4: Sole for all loop currents and appropriate quantities. Example 4.4 Find the current i, x and z using Mesh Analysis. source 4.0
Solution There are 3 loops in this circuit and there is one current source connecting two loops that forms a superloop. Due to the superloop, this automatically implies that there is also a SC associated with the superloop. Let s do the numbers: regular loop i 3 loops (i, i, i 3) current source mesh equations SL(i & i 3) SC(i & i ) 3 So there are loop equations (regular plus a superloop) and SC. Writing these out, we get S(i, i 3): (+ + )i 3 ()i 0 0 3 0 i SC (i, i ): i i 0 0 i Loop-: 3 3 matrix form (+ + 0)i ()i ()i3 0 7 4 0 i 3 0 x 3 i x Soler: i solution i3 i i x 4.44 ma 0.656 ma 3.44 ma.78 V NODE & MESH ANALYSIS with DEPENDENT SOURCES When there are circuits with dependent sources, we treat the dependent source exactly as any other source when performing node or mesh analysis. Howeer, an additional equation must be generated ia the dependent source condition ( DC) in order to sole the circuit. Here is the process:. Apply regular Node or Mesh analysis. If a circuit contains a dependent source, apply the DC: Go to the controlling element and see how it is defined The controlling current or oltage of the dependent source MUST be written as a function of the Node oltages in node analysis Loop currents in mesh analysis As before, usually it DC comes in the form of Ohm s law but also expect KCL and KVL. Node Analysis with Dependent Sources Determine 0 using Node analysis. There are nodes but there is one oltage. Howeer, we hae a dependent current source that automatically implies there is also a DC. Let s do the numbers: regular node 0 nodes source node equation DC (i x ) eqs and unknowns ( & ) 0 x 4.
Apply the regular node analysis steps (& treating the dependent source as a regular source) and write out Node : ( + ) ( ) ( ) 0 5i + 5i 0 4 4 x 4 0 x unknowns and eq The way it stands this is an unsoled equation, finding another equation remedies the problem. The dependent source condition gies us that second equation to make it possible to sole for 0. To apply the DC, we look at how the controlling element i x is defined; it s the current through the 4kΩ resistor. This implies that we rewrite i x in terms of the node oltages for node analysis 0. Using Ohm s law, we rewrite i x in terms of 0: 4i Writing down our two equations gies ( ) 4 0 x matrix 4 0 solution form 4 i 0 + 4ix x 0 + 5i + 5 0.5V Example 4.5 Determine the oltage x for this node circuit. x Solution Where should I place the ground? There are 3 loops in this circuit and there is one current source connecting two loops that forms a superloop. Due to the superloop, this automatically implies that there is also a SC associated with the superloop. Let s do the numbers: SN (, ) SC (, ) 4 nodes 3 sources node equation DC (i) Soler ( x ) 4 eqs and 4 unknowns (,, i, x) Note that this is a supernode since it is not connected to the ground. Speaking of the ground, which ground do I choose? The most conenient node is usually the one where most of the oltage sources are connected. I would choose the node where the 8V and 4V node is located. Step : Apply regular node analysis treating the dependent source as a regular source SN (, ): 8 + 4 ( ) ( ) ( ) SC (, ) : 6i + Step : Dependent Source Condition The controlling element i is defined as the current through the Ω resistor. This implies that we rewrite this controlling current in terms of node oltages for node analysis. Using Ohm s law, we hae DC (, i): 0 i 4.
Putting this altogether, gies + + 0 8 matrix form + 6i 0 0+ i 0 0 8 6 0 3V 8 4V solution 0 i 0 MESH ANALYSIS WITH DEPENDENT SOURCES Analyze the simple circuit using Mesh analysis determine the loop current i a for the circuit. Let s do the numbers: loops source loop equation + DC equations + unknowns Apply regular mesh analysis treating the dependent source as a regular source. There is only one independent loop. ( ) ( ) ( ) ( ) Loop-i : 00 + 00 i 00 4i 8 300 i 400 i 8 0 0 x 0 x Dependent Source Condition: The controlling element i b is defined as the current through the 00Ω resistor. Since we are not interested in oltages but loop currents, this implies that we rewrite this controlling current in terms of loop currents for mesh analysis. Using KCL, we hae Putting this altogether gies DC: 4i i + i i + 3i 0 x x 0 0 x 300i0 400ix 8 matrix 300 400 i0 8 form i0 + 3ix 0 3 ix 0 Example 4.6 Determine the loop current i x and the oltage c across the CCCS for the mesh circuit. Solution Label the loops starting from the left as loop, upper one, lower one 3, and the right one i X going in the CCW direction. Note that oltage C is the oltage drop across the VCCS and i X is one of the loop currents. Let s do the numbers: solution i0 48mA 4.3
regular loop (i ) SL (i, i 3, i x) 4 loops sources loop equations SC: (i, i 3) & (i 3, i x) DC ( ix, y ) Soler ( C ) 6 eqs and 6 unknowns (i, i, i, i, 3 x y Apply mesh analysis note that because the i X current is going CCW, I am going to choose the currents in those directions so that I don t need to setup another equation. ( ), ) Loop : + + i i i 3i Superloop: i i + i 3 i + i x 3ix loop loop 3 3 C loop ix loop & 3: i3 i+ 9 Superloop conditions loop x & 3: i3 y + ix Dependent Source Condition there are two of them. The CCVS is written in terms of the loop current i X. This one is already met when we chose the loop currents to moe in the CCW direction. The second one is the VCCS which depends on y the oltage across the left Ω resistor. In terms of loop currents, y is Dependent Source Condition: y (i i ) Soler equation: applying KVL to get C, we get Putting this altogether gies c ix 3ix i i i3 3ix 0 3 0 0 i 0 i 3i + i3 + 5ix 3 5 0 0 i 0 i i3 9 matrix 0 0 0 0 i 3 9 form solutions ix A 9 i3 ix y 0 0 0 0 ix 0 C 5V 0 i 0 0 0 + i y 0 y 0 0 0 0 0 5 0 c + 5ix 0 0 C x 4.4