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Unformatted text preview: Node Voltage Analysis Node Voltage Analysis In these lectures, you will learn: The difficulty in using only KVL, KCL, and Ohm’s Law in the analysis of large DC circuits.
How node voltage analysis simplifies the solution of large DC circuits.
How to apply NodeVoltage Analysis. Node Voltage Analysis is the most important technique we will cover in EE302. It is essential that you master this method!
2 Definitions What is a node? A node is a point in the circuit where two or more elements are connected What is an essential node? An essential node is a node where three or more elements are connected. (This concept is not used in your text, but it greatly simplifies the analysis.)
R3 R2 1A R1 R4 +
 R5 2V 2A How many essential nodes are there in this circuit?
4! 3 Example Circuit To get the power we need the voltage Calculate the power associated with each element in the circuit
below.
R3 R2 1A R1 R4 +
 R5 2V 2A and current for every element. What if we use KVL, KCL and Ohm’s Law to solve this? How many unknowns? 1 current for the voltage source. 2 voltages for the current sources. 5 voltages or 5 currents for the resistors. We need 13 equations! 4 How Many Equations? We have 13 unknowns so we need 13 Calculate the power associated with each element in the circuit
below.
R3 R2 1A R1 R4 +
 R5 2V 2A equations: We can write an Ohm’s Law equation for each resistor.
• 5 Equations. We can write 3 independent KCL equations. We can write 5 independent KVL equations. Now we are all done except for the algebra. There has got to be an easier way!
5 The Easier Way Systematic methods allow us to solve R3 R2 1A R1 R4 +
 R5 2V 2A circuits using the minimum number of equations. Systematic methods allow us solve every circuit in the same manner. We don’t have to be clever. We don’t have to see “the trick”. We will learn two systematic methods. The first is nodal analysis. With Nodal Analysis, we need only:
2 Equations with 2 unknowns
6 Concept of Nodal Analysis If I know all of the voltages in a circuit, then I know everything! Resistors • Current can be found via Ohm’s Law
• Power can be calculated by V2/R Current Sources • Power can be calculated once the voltage is known. Voltage Sources • Finding the current through the voltage sources is harder – often have to use KCL. Nodal analysis is a systematic method for finding all of the voltages in any circuit.
7 Finding Current from Voltages
How can we use Ohm’s Law to find the resistor current?
R Circuit
A + IR
+ VR + Va Vb  Circuit
B  What is the voltage across the resistor (VR)?
KVL ⇒ VR + Vb – Va = 0 VR = Va Vb What is the current through the resistor (IR)? IR = VR/R 8 Finding Current from Voltages
IS Circuit
A + + Va Vb  Circuit
A +
Va
 I  I = IS
+ Circuit
B VS I I = ??
Va = Vb + VS +
Vb Circuit
B  9 What is a Node Voltage? A voltage requires two points in space and a polarity. We can define the value of the voltage at a node if we have a point of reference. 30 V 20 V B 10 kΩ +
30 V C Let’s define the voltage at
node A to be 0 V… + +
Where are the nodes? 20 kΩ What is the value at node B?
What is the value at node C?
Called the reference node A 0 V The polarity is implied by the reference node convention.
10 Gravity Analogy Node voltages are analogous to altitude. By convention, the altitude reference is mean sea level.
Positive altitudes are above mean sea level:
• Denver is the “mile high city”
• Mt. Everest is +8,850 meters Negative altitudes are below mean sea level:
• Death Valley is 282 feet
• The Mariana Trench is 10,911 m The reference does not matter:
• Mean sea level
• Top of Mt. Everest • Bottom of the Mariana Trench It is still uphill from ENS to the UT Tower.
• The altitude number would be different, but not the relative values. 11 Methodology of Node Voltage Analysis
1. Find all of the essential nodes in the circuit.
2. Select one of these essential nodes as the reference node. We often call this the ground node.
Assign the reference node a value of 0 V.
In real circuits, the choice is often obvious – Chassis ground or the power supply ground. 1. Label each of the remaining essential nodes in the circuit Use A,B,C… or 1,2,3…. 1. Use the labels of each node to write the KCL equation for each node. Define all current as leaving the node. 12 Example Circuit
R3 R2 1A R1 R4 +
 R5 2A 2V Identify all of the essential nodes Next, pick a reference node. Which one to use? Rules: The one that has the most elements connected to it.
The one that has the most VOLTAGE sources connected to it. 13 Example Circuit R3 R2 A 1A R1 B +
 R4 C R5 2A 2V 0V Let’s make the bottom node our reference node….
Label all of the remaining essential nodes
Now we will write the KCL equations for each of these
three essential nodes…
14 KCL for Node A R3 R2 A 1A R1 B +
 R4 C R5 2A 2V 0V Write KCL for the currents leaving node A:
−1 A + VA
(V − V ) (V − V )
+A B+A C
R1
R2
R3 =0 Note that the number of terms is equal to the number of paths out of the node.
15 KCL for Node C
R3 R2 A 1A R1 B +
 R4 C R5 2A 2V 0V Write KCL for the currents leaving node C:
VC ( VC − VB ) ( VC − VA )
+
+
−2A +
R5
R4
R3 =0 Count the branches and count the terms!
16 KCL for Node B
R3 A 1A R2 R1 B +
 R4 C R5 2A 2V 0V Write KCL for the currents leaving node B: What is the current through the voltage source? We don’t know! We cannot write KCL for this node like we did for the others.
17 KCL for Node B
R3 A 1A R2 R1 B +
 C R4 R5 2A 2V 0V What do we do about Node B? Remember that we are looking for the voltage! The voltage source dictates that the voltage between B and the reference is 2 V. VB = 2 V  0V = 2V 18 KCL for the Reference Node
R3 R2 A 1A R1 B +
 R4 C R5 2A 2V 0V What about KCL for the Reference Node?
Since we have assigned it a value of zero, we do not need to write the KCL equation for that node – we already
know the voltage!
19 Example Circuit R3 A R2 R4 C R5 R1 1A B +
 2A 2V 0 V −1 + ( VA − VB ) ( VA − VC ) VA R2
VB = 2 V + R3 + R1 =0 −1 + ( VA − 2) + ( VA − VC ) + VA = 0 R2
R3
R1
( V − 2) ( VC − VA ) VC
( VC − VB ) ( VC − VA ) VC
2+ C
+
+
=0
2+
+
+
=0
R4
R3
R5
R4
R3
R5
20 Completing the Process
−1 + ( VA − 2) ( VA − VC ) VA
+ + =0 R2
R3
R1
( V − 2) ( VC − VA ) VC
2+ C
+
+
=0
R4
R3
R5 Best to write these equations in “standard form”. To do this, we combine like terms in each equation and move all constants to the righthand side. 1
1
2
1
1
+
+ VA − VC = 1 + R1 R 2 R3
R2 R3 1 V + 1 + 1 + 1 V = 2 + 2
− A C
R4 R3 R3 R 4 R5 21 Solving these equations
1
1
2
1
1
+
+ VA − VC = 1 + R1 R 2 R3
R2 R3 1 V + 1 + 1 + 1 V = 2 + 2
− A C
R3 R3 R 4 R5 R4 Let’s give the resistors values: R1 = R2=R3=2 Ω
R4=R5=4 Ω [1.5]VA − [ 0.5]VC = 2
− [ 0.5]VA + [1]VC = 2.5
22 Solving by elimination of variables
[1.5]VA − [ 0.5]VC = 2
− [ 0.5]V A + [1]VC = 2.5 Multiply the top equation by 2
[ 3]VA − [1]VC = 4 − [ 0.5]VA + [1]VC = 2.5 Now add the two equations together
2.5VA = 6.5
VA = 2.60 V
23 Solving these equations Con't
[1.5]VA − [ 0.5]VC = 2
− [ 0.5]VA + [1]VC = 2.5 Plug the value of VA into either of the equations above: VA = 2.60 V [1.5]( 2.6) − [ 0.5]VC = 2
VC = 3.80 V
24 The Final Result… R3 R2 A R1 1A B +
 C R4 R5 2A 2V 0V
R3 2.6 V 1A R2 R1 2V +
 R4 3.8 V R5 2A 2V 0V 25 Other Solution Techniques Substitution Express one unknown in terms of the others and substitute that expression into the remaining equations.
• Easy to do for 2 x 2 systems and works OK for some 3 x3 systems (e.g., if some coefficients are zero). Matrix Methods: Cramer’s Rule and matrix inversion
• Can be done by hand for 3 x 3 systems.
• See Appendix A for an explanation. OK to use your engineering calculator. • But be careful with both the algebra and the data entry. Verify your results! Are KCL and KVL respected? Is the answer reasonable? 26 Using Node Voltages
R3 2.6 V 1A R2 R1 2V +
 R4 3.8 V R5 2A 2V 0V What is the current flowing thru resistor R4 (= 4 Ω )? Assume that the current is flowing lefttoright.
I = (VBVC)/R4
I = (2V – 3.8 V)/4Ω = 1.8 V/ 4Ω = 0.450 A Thus the current is really flowing from right to left.
27 Calculations using Node Voltages
R3 R2 A 1A R1 B +
 R4 C R5 2A 2V 0V What is the current flowing thru resistor R5 (= 4 Ω )? Assume that the current is flowing down.
I = (VC0)/R5
I = 3.8 V/4Ω = 0.950 A 28 Calculations using Node Voltages
R3 R2 A 1A R1 B +
 R4 C R5 2A + 2V 0V What is the power being delivered by the 2A current source? Need the voltage with the positive terminal on the top.
This voltage is equal to the node voltage Vc. P2A = VI = (3.8)(2A) = 7.60 W or 7.60 W (delivered) 29 Calculations using Node Voltages
R3 R2 A 1A R1 B +
 R4 C R5 2A 2V 0V What is the power being absorbed or delivered by the 2V voltage source? Need the current flowing down thru the source.
Using KCL
Currents entering Node B are (VAVB)/R2 and (VCVB)/R4
Sum of these is 0.3A + 0.45A = 0.75A P2V = VI = 2(0.75) = 1.50 W (absorbed) 30 What about KCL and the Reference Node?
R3 2.6 V 1A R2 R1 2V R4 0.75A
+
 R1 = R2=R3=2Ω R4=R5=4Ω 3.8 V R5 2A 2V 0V Let’s verify KCL for the reference node: 1+ 0 − 2.6
0 − 3.8
− 0.75 +
+2 =0
2
4 1 − 1.3 − 0.75 − 0.95 + 2 = 0 ⇒ 0 =0 31 Other Cases
Consider the following….
A R1 R2 B C R3 R4 What would be the appropriate KCL equation for this case? ( VB − VA ) + ( VB − VC ) +
R1 R2 VB
=0
R3 + R 4
32 Other Cases
Consider the following….
A R1 R2 B C R3 +
 V1 What would be the appropriate KCL equation for this case? ( VB − VA ) + ( VB − VC ) + VB − V 1 = 0
R1 R2 R3 33 How Can All the Currents Leave all the Nodes?
A IAB R1 IBA B For node A, the current leaving toward node B is
I AB (VA − VB )
=
R1 For node B, the current leaving toward node A is
(VB − VA )
I BA =
= − I AB
R1
The method automatically takes care of the signs.
34 Use My Method and Not the Text’s
Alexander and Sadiku define reference directions for branch currents and write KCL equations with currents on both sides of the equal sign. While this method is valid, it is not helpful. This approach needlessly complicates the process of writing and solving the KCL equations.
I strongly urge you to use my approach to writing the KCL equations. With my approach you do not define branch currents. Just write the terms for all of the currents leaving each node and set the sum equal to zero. This method automatically takes care of the signs. 35 Important Points on Node Voltage Method
1. Identify all essential nodes (ne). The number of equations you will have is ne – 1. 1. Pick a good reference node. Should be an essential node connected to a voltage source, if possible, since this will simplify the equations. 1. Write KCL for the currents LEAVING the node. Current Sources already have their current defined; just use the value with the appropriate sign.
Use Ohm’s Law for resistors .
Voltage sources require special care.
The number of terms in each equation is equal to the number of paths out of the essential node.
• Count to be sure!
• Every term is a current! 36 Node Voltages Are Easy to Measure
Ground
Or Reference Node
Node
Voltage 37 Example Problem
Calculate the power delivered or absorbed by each source in this circuit when R = 5Ω
3 A A B C
R R R
R
5 V + R +
10 V 1 A 0 V 38 Example Problem: Equations
3 A A B C
R R R
R
5 V + R +
10 V 1 A 0 V VA = 5 (VB − 5) + ( VB − VC ) − 1 = 0 R
R
(VC − VB ) + VC + (VC − 10) − 3 = 0
R
R
R 39 Example Problem: Solution
VA = 5 ( VB − 5) + ( VB − VC ) − 1 = 0 R
R
( VC − VB ) VC ( VC − 10)
++
−3 = 0
R
R
R
with R = 5Ω 2VB − VC = 10
− VB + 3VC = 25 VB = 11.0 V and VC = 12.0 V
40 Example Problem: Calculations
3 A +
A B C
R R R
R
+ 5 V R +
10 V 1 A 0 V 3A current source Voltage is VC – VA = 12 – 5 = 7V P3A = VI = (7)(3) = 21.0 W or 21.0 W delivered 41 Example Problem: Calculations
3 A A B C
R R R
R
+ 5 V R +
10 V 1 A 0 V 10V voltage source Current is (10 – VC)/R = (10 – 12)/5 = 0.4 A P10V = VI = (10)(.4) = 4.00 W (absorbed) 42 Example Problem: Calculations
3 A A B R + 5 V + + C
R R
R
R +
10 V 1 A 0 V 1A current source Voltage is VR + VB = 5(1) + 11 = 16V P1A = VI = (16)(1) = 16.0 W or 16.0 W delivered 43 Example Problem: Calculations
3 A A B C I
5 V R R R
R
+
R +
10 V 1 A 0 V 5V voltage source solve for current I + (VAVB)/R + 3 = 0 I = (VAVB)/R + 3 = (5 11)/5 + 3 = 1.8A P5V = VI = (5)(1.8) = 9.00 W or 9.00 W delivered 44 Standard Example
Find the power dissipated by the 8ohm resistor.
20 Ω 10 Ω
+ 1A 20 Ω 8Ω 10 V
 45 4.5 W Review of Node Voltage Method
1. Identify all essential nodes (ne). The number of equations you will have is ne – 1. 1. Pick a good reference node. Should be an essential node connected to a voltage source, if possible, since this will simplify the equations. 1. Write KCL for the currents LEAVING the node. Current Sources already have their current defined; just use the value with the appropriate sign.
Use Ohm’s Law for resistors .
Voltage sources require special care.
The number of terms in each equation is equal to the number of paths out of the essential node.
• Count to be sure!
• Every term is a current! 46 Questions?
Node Voltage Analysis Practice Problem 1
Determine the power being delivered or absorbed by
each source in the circuit
.075 A 8 Ω
y 20 Ω 4 Ω 10 Ω
8 V 16 W 3 A x
2 V P3A = 61.4W, P2V=3.10W, P.075=.980W, P8V=1.50W 48 Practice Problem 2
Use Node Voltage Analysis to determine the value of R such
that V = 5 V.
5 A 1 A R
R + R R + 10 V R + 20 V R v
49 R=7.20Ω Node Voltage Analysis Supernodes A More Challenging Circuit
R2
6V C B +
 A R4 R5
R1 R3 +
 8V R6 This circuit is more challenging than we have done so far. Why? Two voltage sources which DO NOT share a node. How do we deal with the 6V source between nodes A and B? 51 Node A
R2
6V C B +
 A R4 R5
R1 R3 +
 8V R6 What is the equation for node A?
VA VA − VC
+
+ ?? = 0
R1 R 2 + R 4
We do not know the current thru the voltage source…
52 The Unknown Current Method Define an additional unknown: R2 The current IX flowing from A to B. we know how to write the KCL equations for nodes A, B and C. A
R1 C B +
 Now we have four unknowns, but at least 6V R4 Ix R5
R3 +
 First, note that VC = 8 volts.
Then let’s look at nodes A and B: VA
VA − 8
+
+ Ix = 0
R1 R 2 + R 4
VB − 8 VB
+
− IX = 0
R5
R3
53 8V R6 The Unknown Current Method
R2 IX appears in both equations:
A
R1 C B +
 VA
VA − 8
+
+ Ix = 0
R1 R 2 + R 4
VB − 8
VB
+
− IX = 0
R5
R3 6V R4 Ix R5
R3 +
 Add these equations to eliminate IX: VA
VA − 8 VB − 8 VB
+
+
+
=0
R1 R 2 + R 4
R5
R3
54 8V R6 The Unknown Current Method
R2 VA
VA − 8
VB − 8 VB
+
+
+
=0
R1 R 2 + R 4
R5
R3 6V
A C B +
 Now we have one equation in two unknowns: R4 R5
R1 +
 R3 Where can we get another equation?
Because of the 6volt source, there is a constraint between node A and node B: VB = VA + 6 volts. (VA + 6) − 8 + VA + 6 = 0
VA
VA − 8
+
+
R1 R 2 + R 4
R5
R3
1
1
1
8
2
6
1
VA +
+
+ = R 2 + R 4 + R5 − R 3 R1 R 2 + R 4 R5 R 3 55 8V R6 A Better Approach
R2 Can we avoid that? 6V
A
R1 C B +
 The unknown current IX seems like an unnecessary complication. R4 Ix R5
+
 R3 Yes, we can consider KCL for the combination of nodes A and B.
This is called the supernode method. 56 8V R6 What is a Supernode?
Consider any two adjacent nodes:
A R1 B KCL applies to node A and to node B. The sum of the currents leaving each node is zero.
KCL also applies to the combination of node A and node B. The sum of the currents leaving the combined node is zero.
This is called a supernode. 57 A Past Example – Super Node Approach
12A
6A 3A IX IX=? 2.5A 2A  I X + 3 + 6  2 + 2.5  12 = 0
I X = 2.50 A
58 The Supernode Method
As before, we know that Vc = 8 V R2 A 6V 8V VA R4 6V
+
B C +
R5 We also know that R1 +
 R3 VB = VA + 6 Volts We can now write KCL for the currents leaving the supernode AB. (V A + 6 ) − 8 + V A + 6 = 0
VA
VA − 8
+
+
R1 R 2 + R 4
R5
R3
We get to the final equation in one step! 59 8V R6 The Supernode MethodSummary Use the supernode method only if there are two or more voltage sources that do not share a common node. Otherwise use one of the voltage source nodes as the reference node. Each voltage source yields a constraint equation. The constraint equations are used to eliminate variables. You do not know the current through a voltage source and cannot write KCL in the normal node voltage fashion. But, you cannot ignore the voltage source current. Instead, create a supernode from the nodes at either end of the voltage source. Write KCL for this combined supernode. If you prefer, you can define an unknown to represent the current through the voltage source. It will cancel out in the algebra. 60 Node Voltage Analysis Dependent Sources What About Dependent Sources?
A 10Ω 1Ω B
IC +
10V 5Ω
5IC 2A Dependent sources are treated just like independent sources. The control variable must be expressed in terms of the node voltages.
IC = (VA – VB)/10 62 What About Dependent Sources?
10Ω A 1Ω B
IC +
10V 5Ω
5IC 2A Node A:
VA = 10 V 10 − VB VB IC =
=1− 10 10 Node B: VB − 10 VB V
+
+ 51 − B = 0
10
5 10 63 What About Dependent Sources?
A 10Ω 1Ω B
IC +
10V 5Ω
5IC 2A VB − 10 VB VB +
+ 51 − = 0
10
5 10 VB VB VB
+
−
= +1 − 5 = −4
10 5
2 1 1 1
VB + − = −0.2VB = −4 ⇒ VB = 20.0 V
10 5 2 64 Summary Write a KCL equation for each essential node other than the reference node. One equation for each essential node.
One term for each branch connected to the node. Voltage sources imply constraints: VA = 10 volts Express the dependent source control variable in terms of the node voltages. IC = (VA – VB)/10 Solve for the node voltages. Then compute the source control variable if you need it. 65 In Class Exercise  + a) Write the node voltage equations for this circuit.
b) Determine the power being delivered by the 20V voltage source. 35ix
2Ω 1Ω
+
+
 4Ω
 vx
20 V 3.125vx
40 Ω 20 Ω 80 Ω ix 66 (P=602.5W) Summary of the Node Voltage Method Node equations have a term for every branch connected to the node. Count the branches; count the terms! Each term in a node equation is a branch current. Think carefully about the current in each branch.
Learn how to express the current for every kind of branch – resistors, current sources, voltage sources, combinations of these. Define node voltages only for essential nodes. Voltages for nonessential nodes can be computed later. A voltage source between essential nodes yields a constraint equation. Use supernodes when needed. You may not need all the node voltages to answer the question. Only solve for the ones you need. Think ahead! 67 Node Voltage Analysis Focus on writing the KCL equations correctly.
If the equations are wrong, the answer is sure to be wrong!
Any branch with only a voltage source requires special attention! 68 Questions?
Node Voltage Analysis Node Voltage Analysis
A Difficult Dependent Source Example Example Circuit with Dependent Sources
R2 R4
ix C B
A
R1 R3 R5
+
 6V R6 2ix R7 D Identify all of the essential nodes for this circuit Next, we pick a reference node. Which one to use?
This is a hard problem: Three unknown node
voltages and a dependent source.
71 Equations for Nodes A & B
R2 R4
ix C B
A
R1 R3 R5
+
 6V R6 VA VA − VB VA − VC
+
+
=0
R1
R3
R2 + R4 2ix R7 D VB = 6 Substitute for VB and gather terms: 1
1
1
6
1 R1 + R 2 + R 4 + R3 VA + − R 2 + R 4 VC = R 3 72 Equations for Node C
R2 R4
ix C B
A
R1 R3 R5
+
 6V 2ix R6 R7 D VC − VD VC − VB VC − VA
− 2ix +
+
+
=0
R7
R5
R2 + R4
Express ix in terms of the node voltages and substitute. ix = VC − VB
V −6
=C
R2 + R4 R2 + R4 12 − 2VC VC − VD VC − 6 VC − VA
+
+
+
=0
R2 + R4
R7
R5
R2 + R4 73 Equations for Node C (cont.)
R2 R4
ix C B
A
R1 R3 R5
+
 6V R6 2ix R7 D 12 − 2VC VC − VD VC − 6 VC − VA
+
+
+
=0
R2 + R4
R7
R5
R2 + R4
1
1
1
12
6 1 1
−
VA + +
−
VC + −
VD = −
+ R2 + R4 R 2 + R 4 R5 R 7 R5 R 2 + R 4 R7 74 Equations for Node D
R2 R4
ix C B
A
R1 R3 R5
+
 6V R6 VD VD − VC
2ix +
+
=0
R6
R7 R7 2ix D ix = VC − VB
V −6
=C
R2 + R4 R2 + R4 2VC − 12 VD VD − VC
+
+
=0
R2 + R4 R6
R7
2
1
1
12 1 R 2 + R 4 − R 7 VC + R 6 + R7 VD = R 2 + R 4 75 Final Node Equations R2 R4 ix C B
A Four node voltages and four equations. R3 R1 R5
+
 6V 2ix R6 R7 D A) 1
1
1
6
1 +
+
V A + −
VC = R1 R 2 + R 4 R3 R3 R2 + R4 B) VB = 6 C) 1
1
1
12
6 1 1
−
VA + +
−
VC + −
VD = −
+ R2 + R4 R 2 + R 4 R5 R 7 R5 R 2 + R 4 R7 D) 2
1
1
12 1 R 2 + R 4 − R 7 VC + R 6 + R7 VD = R 2 + R 4 76 Solving… We know VB so that is done: VB = 6 volts The equations for nodes A and D have only two unknowns The equation for Node C has three. Because of this structure, we can solve by substitution. Solve the Node A equation for VA in terms of VC. Solve the Node D equation for VD in terms of VC. Substitute these expressions for VA and VD into the Node C equation and solve for VC. 77 Solve for VA and VD in terms of VC
Solve the Node A equation for VA in terms of VC:
1
1
1
6
1 +
+
V A + −
VC = R1 R 2 + R 4 R3 R3 R2 + R4 6 1
1
1 1
VA = +
VC +
+ R 3 R 2 + R 4 R1 R 2 + R 4 R3 Solve the Node D equation for VD in terms of VC:
2
1
1
12 1
−
VC + +
VD = R2 + R4 R7 R2 + R4 R6 R7 12
2
1
1 1
VD = +
−
VC R6 + R 7 R2 + R4 R7 R2 + R4 78 Now we are Done:
We have one equation in one unknown, VC.
6 1
1
1 1
VA = +
VC +
+ R 3 R 2 + R 4 R1 R 2 + R 4 R3 1
1
1
12
6 1 1
−
VA + +
−
VC + −
VD = −
+ R2 + R4 R 2 + R 4 R5 R 7 R5 R 2 + R 4 R7 12
2
1
1 1
VD = +
−
VC + R 2 + R 4 R7 R 2 + R 4 R6 R7 79 Key Points This problem was hard: Four node voltages and a dependent source.
A lot of algebra.
• Please let me know if you find I made an error! The method is simple: Write a KCL equation for each essential node:
• One equation for each essential node.
• One term for each branch connected to the node.
Voltage sources imply constraints:
• VB = 6 volts
Express the dependent source control variable in terms of the node voltages.
• ix = (VC6)/(R2+R4)
Solve. 80 Summary of the Node Voltage Method Node equations have a term for every branch connected to the node. Count the branches; count the terms! Each term in a node equation is a branch current. Think carefully about the current in each branch.
Learn how to express the current for every kind of branch – resistors, current sources, voltage sources, combinations of these. Define node voltages only for essential nodes. Voltages for nonessential nodes can be computed later. A voltage source between essential nodes yields a constraint equation. Use supernodes when needed. You may not need all the node voltages to answer the question. Only solve for the ones you need. Think ahead! 81 Node Voltage Analysis Focus on writing the KCL equations correctly.
If the equations are wrong, the answer is sure to be wrong!
Any branch with only a voltage source requires special attention! 82 Questions?
Node Voltage Analysis Mesh Current Analysis Mesh Current Circuit Analysis In these lectures, you will learn: How to apply Mesh Current Analysis: • Learn to identify and assign mesh currents. • Learn to write the set of simultaneous algebraic equations. • Learn to express the quantities to calculate in terms of the mesh currents in the circuit. To identify and analyze circuits requiring a “Supermesh” approach. 85 Definitions Branch: a path that connects two nodes. Essential branch: a branch that connects two essential nodes. Loop: any closed connection of branches. Mesh: a loop that does not contain other loops. 86 Example
R3 Mesh 3 1A Outer Loop is not a Mesh R2 R4 +
 Mesh 1 Why? R1 2V Mesh 2 87 Mesh Current Analysis How many essential nodes (ne)? 4 How many essential branches (be)? 6 Number of mesh currents? 3 In general, number mesh currents = be (ne 1)
In our case, number mesh currents = 6 (4 1) = 3
88 Mesh Current Method Mesh current method provides another systematic tool to solve circuit problems. Only applicable to planar circuits.
• Not an issue in EE302. Use mesh currents as the independent variables and write KVL equations around each mesh. 89 More on Mesh Current Method Because of the passive sign convention, the direction of the mesh current flow automatically defines the polarity of the branch voltages. Mesh currents are not necessarily the same as branch currents. Mesh currents automatically satisfy KCL since each mesh current both enters and leaves each node it encounters. 90 Mesh Current Analysis Methodology 1. Identify all of the meshes in the circuit.
2. Define the current in each mesh as flowing clockwise.
3. Assign each mesh current a name: (I1, I2, I3) or (IA, IB, IC) 1. Using KVL, sum the voltage drops around the mesh and set them = 0. Passive sign convention is used for all voltage and are determined in relation to the mesh current value
In some branches, the current you will use is the difference of two mesh currents. 1. Current sources imply constraints on the mesh currents. More on this later. 91 Example Assume the mesh currents flow R1 clockwise. Apply KVL around the mesh Mesh equation becomes: V + IR1 + IR2 = 0 V R2 I Same sign convention as KVL approach studied earlier. 92 A More Complex Example
R1 V1 I1 R3 R2 I2 V2 Define two mesh currents I1 and I2. Let’s look at mesh 1. Current in R1 is I1.
Sign convention follows I1.
What is the current in R3?
• Net current in R3 = (I1 I2). KVL around mesh 1: V1 + I1R1 + (I1I2)R3 = 0 93 A More Complex Example
R1 V1 I1 R3 R2 I2 V2 Now, let’s look at mesh 2. Current in R2 is I2.
Sign convention follows I2.
What is the current in R3? • Net current in R3 = (I2 – I1). KVL around mesh 2: (I2I1)R3 +I2R2 + V2 = 0 94 A More Complex Example
R1 V1 I1 R3 R2 I2 V2 KVL around mesh 1: V1 + I1R1 + (I1I2)R3 = 0 KVL around mesh 2: (I2I1)R3 +I2R2 + V2 = 0 For each mesh, the sign convention follows that mesh current. 95 A More Complex Example
R1
V1 R2
R3 I1 I2 V2 V1 + I1R1 + (I1I2)R3 = 0
(I2I1)R3 +I2R2 + V2 = 0 Group terms to get equations in standard form:
I1(R1 + R3) I2 R3 = V1 I1(R3) + I2(R2 + R3) = V2
96 Branch Currents vs. Mesh Currents
R1 R2 A V1 I1 R3 I2 V2 If we use an ammeter to measure the current in R1, we would find IR1 = I1 I1 is the same as the branch current in R1 and is a physical current If we measure the current in R3, it would be equal to (I1 I2) which equals the branch current. The ammeter could not directly read I1 or I2. 97 Branch Currents vs. Mesh Currents
R10 R1 1 R12
R4 S1 R7 R1
R1 5 R14 R13
R5 S2
R2 I5
R16 R17 R18 R8 The mesh current I5 does not correspond to any physical branch current and cannot be measured with an ammeter. S3
R3 R6 R9 98 Mesh Currents and KCL
R10 R1 1 R12
R4 S1 I1 R1
R1 5 I2 R7 I3 R14 R13
R5 S2 I4 R2 I5 KCL is inherently
satisfied. I6
R16 R17 R18 R8 Note that each mesh current both enters and exits every node it encounters. S3 I7 R3 R6 I8 I9 R9 99 What about current sources?
R1 Suppose we have a current source included in the mesh. Can we use KVL and sum 80 V
the voltage drops for that mesh? R2 i2
R3 +
 R4 i1 3A
i3
R5 100 Current Source Continued What is the value of mesh current i3? R 3A! Why? The 3A source tells us the direction of current in that mesh.
It is opposite to the defined direction of i3.
80 V Current sources create a constraint equation in the mesh current method. i3 = 3 amps R i2
R +
 R i1 3A
i3
R 101 Current Source Example
Mesh 1: (R2 + R4)I1 R2I2 – R4I3 80 = 0 R1 Mesh 2: R2I1 + (R1+R2+R3)I2 – R3I3 = 0
Mesh 3: Cannot write KVL, but I3 = 3
Substitute for I3 and simplify:
(R2 + R4)I1 R2I2 = 80 – 3(R4)
R2I1 + (R1+R2+R3)I2 = 3(R3) R2 I2
R3 80 V +
 R4 I1 3A I3
R5 102 Mesh Current Summary Mesh equations have a term for every element in the mesh. Count the elements; count the terms! Each term in a mesh equation is a voltage. Think carefully about the voltage across each element.
The voltage polarity corresponds to the direction of the mesh current.
Learn how to express the voltage for every kind of element – resistors, current sources, voltage sources, combinations. Each current source yields a constraint equation. 103 Standard Example
Find the power dissipated by the 8ohm resistor. 20 Ω 10 Ω
+ 1A 20 Ω 8Ω 10 V
 104 4.5 W Practice Problem 1
Find IX 1A
8Ω
100V +
Ix 4Ω 2Ω 10Ω
3Ω 6A 105 Mesh Current Analysis
Supermeshes Example ProblemSupermesh Suppose a circuit has a current source between two meshes as shown. If we try to use KVL to sum the voltages around a loop, we encounter an additional unknown: the voltage across the current source. 10 Ω 3Ω ib
2Ω 100 V +
 ia
6 Ω 5A +
?
 +
 ic 4 Ω 107 50V The Unknown Voltage Approach Define unknown voltage Vx. Write mesh equation for mesh A: 10 Ω 100 + 3(iaib) + Vx + 6ia = 0 Write mesh equation for mesh C: Vx + 2(icib) +50 + 4ic = 0 3Ω If we add these equations, Vx 2Ω drops out. 100 + 3(iaib) + Vx + 6ia = 0
2(icib) + 50 Vx + 4ic = 0 ib 100 V +
 ia
6 Ω 5A +
Vx
ic
4 Ω 100 + 3(iaib) + 2(icib) +50 + 4ic + 6ia = 0
108 +
 50V The Supermesh Approach Mentally remove the current source . Write mesh current equations using the original mesh currents around the resulting mesh100 V ib 3Ω 2Ω
+
 ia 5A 6 Ω +
Vx
ic
4 Ω 100 + 3(iaib) + 2(icib) +50 + 4ic + 6ia = 0 Same as before!
9ia 5ib + 6ic = 50 grouping terms in standard form 109 +
 50V The SupermeshApproach Current in the current source is constrained: (icia) = 5A 10 Ω ⇒ ic = 5 + ia Mesh B:
3ia + (3 + 10 + 2) ib – 2ic = 0
Substituting for ic: 3Ω 100 V +
 ia
6 Ω 5A ib
+
?
2Ω
+
 ic
4 Ω 3ia + 15ib – 10 2ia = 0
5ia + 15ib = 10 110 50V The Supermesh Approach 10 Ω For the supermesh:
9ia 5ib + 6ic = 50 3Ω Substituting for ic:
9ia 5ib + 6(5+ia) = 50
15ia – 5ib = 20
Final Equations:
15ia 5ib = 20
5ia + 15ib = 10 100 V +
 ia
6 Ω 5A ib
+
?
2Ω
+
 ic
4 Ω ic = 5 + ia Solving, ia = 1.75A, ib = 1.25A, ic = 6.75A
111 50V The Supermesh Approach 10 Ω We use the original mesh currents to write the KVL equation! But we apply KVL around the supermesh. Remember, we don’t know the voltage across the current source. 3Ω 100 V +
 ia
6 Ω 5A ib
+
?
2Ω
+
 ic
4 Ω ic = 5 + ia 112 50V Practice Problem Find the power supplied by the 2.2V source.
2Ω
+
 4.5A 9Ω
3V 3Ω +
 2.2V 5V 4Ω
2A 3 Ω +
 6Ω 1 Ω 113 Mesh Current Analysis
Dependent Sources Dealing with Dependent Sources We need to express the dependent source value in terms of the mesh current values The mesh approach is exactly the same as the previous examples except for this change.
3Vx
+
2Ω 14Ω ib 3Ω
+ Vx 5Ω 5A
ia +
10V Vx = 3(ic – ib)
1Ω ic 115 Dependent Source Example Find the power delivered by the dependent source
3Vx
+
2Ω 14Ω ib 3Ω
+ Vx 5Ω 5A
ia +
10V 1Ω ic 116 Practice Problem Find the power being dissipated in the 2Ω resistor. ia 5Ω
2Ω
60V 3Ω +
6ia +
4A
4Ω 5A 117 Shortcut and Check Method Add all resistors contained in a mesh together This is the number for that mesh current variable 30 Ω For each resistor that is common to more than one mesh 5 Ω Put negative that value in front of the other mesh current variable Place the negative sum of the voltage sources on the RHS of the equation 80 V i2
90 Ω +
 26 Ω i1 8 Ω i3 118 Shortcut and Check Method Let’s look at Mesh 1 Add all resistors contained in a mesh together 30 Ω • Sum is 31Ω 31i1 For each resistor that is common to more than one mesh, Put negative that value in from of the other mesh current variable
• 5Ω 5i2
• 26Ω 26i3 80 V Place the negative sum of the voltage sources on the RHS of the equation 5 Ω i2
90 Ω +
 26 Ω i1 8 Ω i3 • Sum is 80 V 31i1 5i226i3 =80
119 Hints for Success Mesh equations have a term for every element in the mesh. Count the elements; count the terms! Each term in a mesh equation is a voltage. Think carefully about the voltage across each element.
The voltage polarity corresponds to the direction of the mesh current.
Learn how to express the voltage for every kind of element – resistors, current sources, voltage sources, combinations. Each current source yields a constraint equation. Use supermeshes when needed. You may not need all the mesh voltages to answer the question. Only solve for the ones you need. Think ahead! 120 Mesh Current Analysis Focus on writing the KVL equations correctly.
If the equations are wrong, the answer is sure to be wrong!
Remember that current sources require extra care. 121 Questions?
Mesh Current Analysis Mesh Current Analysis
A Three Mesh Example Mesh Current Example
Find the power associated with each source.
6 Ω
2 Ω +
40V
Ia 8 Ω Ib 4 Ω
6 Ω +
20V
Ic Mesh a: 40 + 2 Ia + 8 (Ia – Ib)= 0 Mesh b: 8 (Ib – Ia) + 6 Ib + 6 (Ib – Ic) = 0 Mesh c: 6 (Ic – Ib) + 4 Ic + 20 = 0 124 Example Continued
Original Equations:
(Eq. 1)
40 + 2 Ia + 8 (Ia – Ib)= 0
(Eq. 2) 8 (Ib – Ia) + 6 Ib + 6 (Ib – Ic) = 0 (Eq. 3) 6 (Ic – Ib) + 4 Ic + 20 = 0 Rewriting in standard form:
(Eq. 1)
10 Ia – 8 Ib + 0 Ic = 40
(Eq. 2) 8 Ia + 20 Ib – 6 Ic = 0 (Eq. 3) 0 Ia – 6 Ib + 10 Ic = 20 125 Example Continued
Multiply Eq. 1 by 8/10 and add to Eq. 2: 8 Ia – 6.4 Ib + 0 Ic = 32
8 Ia + 20 Ib – 6 Ic = 0 13.6Ib 6Ic = 32 Multiply by 10/6 and add to Eq. 3
0 Ia + 22.67 Ib – 10 Ic
= 53.33 0 Ia 6.00 Ib + 10 Ic = 20 16.67 Ib = 33.33 Ib = 2.00 A 126 Final Solution
Substituting Ib = 2.0A into Eq. 3:
0 Ia – 6 (2) + 10 Ic = 20
Ic = 0.800 A
Substituting into Eq. 1:
10 Ia – 8 (2) + 0 Ic = 40
Ia = 5.60 A
127 Now Find Power from Sources
+
40V
6 Ω 2 Ω 4 Ω
+
20V
+ Ia Vo
8 Ω Ib 6 Ω Ic P40V = VIa = (40)(5.6) = 224. W (224. W delivered) P20V = VIc = (20)(.8A) = 16.0 W (16.0 W delivered) Find Vo. Vo = 8(IaIb) = 8(3.6) = 28.8V
128 Questions?
Mesh Current Analysis Node and Mesh Circuit Analysis
When should we use each method? Solving Problems Since the first exam, we have introduced two systematic methods: Node Voltage Analysis allows you to assign voltage values to each essential node once a reference node is defined.
Mesh Current Analysis allows you to assign current values to all mesh currents in the circuit. Now that we have these two tools which one should we use? 131 Issue #1: What circuit am I analyzing? In terms of the number of equations you need to write out and solve, either node voltage analysis or mesh current analysis could be simpler to apply. How should I evaluate this? Do a Quick Inspection before you start.
Spend 6090 seconds on each circuit and ask the following: • If I use node voltage analysis, how many equations do I have to solve?
• If I use mesh current analysis, how many equations do I have to solve? • Does either method seem particularly difficult or particularly easy for this circuit?
• What is the answer I need? 132 Node vs. Mesh Example
R3 R2 1A R1 R4 +
 R5 2A 2V How many unknown node voltages? 2 How many unknown mesh currents? 3 133 Example Circuit #2
4Ω
3Ω 20 Ω
5A 6Ω iy 12 A
13 Ω 7Ω ix If I use node voltage analysis, how many equations do I have to solve? I am left with 3 nodes for which I need to write a KCL equations. This means I have 3 equations and 3 unknowns 134 Example Circuit #2 Con’t
4Ω
3Ω 20 Ω
5A 6Ω iy 12 A
13 Ω 7Ω ix If I use mesh current analysis, how many equations do I have to solve? I am left with 4 meshes for which I need to write a KVL equations This means I have 4 equations and 4 unknowns
Although one pair is a super mesh So this reduces to three meaningful unknowns. 135 Example Circuit #3 Which method would you use for this circuit? 136 Points to Remember Quick Inspection is designed to make your life easier. Suggestions for Node Voltage • Identify the essential nodes
• Pick an appropriate ground!! Suggestions for Mesh Current • Identify the meshes
• Determine which ones are known!! In both cases: You do not need to name the nodevoltages or mesh currents until you pick your method.
Learn to do this by “looking” at the circuit and not writing anything down until you decide on a method… 137 Example Circuit #4 In terms of the number of equations I would need to write, which method is best: mesh current or node voltage?
y 8 kΩ
0.2 kΩ
20 mA 4 kΩ 0.5 kΩ 1 kΩ 25 V x 2 kΩ 138 Equal Number of Equations This will happen for many circuits you will analyze. Now which method should you choose? Things to consider: Do I have to use a “supernode” or “supermesh”?
• In general, these are harder to set up… Which method is most comfortable for me?
What am I solving for? • In general, if I need a current, mesh current analysis will be easier; if I need a voltage, node voltage analysis will be easier. 139 Example Circuit # 3 Will I require either a “supermesh” or “supernode” in this circuit? y
8 kΩ
0.2 kΩ
20 mA 4 kΩ 0.5 kΩ 1 kΩ No!!! 25 V 2 kΩ x 140 Example Circuit #3 Con’t
y
8 kΩ
0.2 kΩ
20 mA 4 kΩ 0.5 kΩ 1 kΩ 25 V 2 kΩ x What if I am asked to solve for the power being delivered by the 25V source? Need a current to calculate
Use mesh current analysis 141 Example Circuit #3 Con’t
y
8 kΩ
0.2 kΩ
20 mA 4 kΩ 0.5 kΩ 1 kΩ 25 V 2 kΩ x What if I am asked to solve for the power being delivered by the 20 mA current source? Need a voltage to calculate
Use node voltage analysis 142 Example Circuit #3 Con’t
y
8 kΩ
0.2 kΩ
20 mA 4 kΩ 0.5 kΩ 1 kΩ 25 V 2 kΩ x What if I am asked to solve for the power being absorbed by the 4kΩ resistor? Could use either a voltage or a current
Use method I am most comfortable with… 143 Additional Comments: Node vs. Mesh Circuits with series elements tend to be easier using mesh analysis. Circuits with parallel elements are almost always easier using nodal analysis. Some students prefer mesh and use it even when nodal analysis is MUCH easier. They almost always get the wrong answer! Exam problems are often designed to be easier with one method than with the other. Always do the Quick Inspection before you start.
All of my EE302 exam problems that request a numeric answer can be solved using two simultaneous equations (plus constraints). 144 Practice Circuits 145 Practice Circuit 1
20 V 100 Ω
10 V 1A 200 Ω 0.2 A 146 Practice Circuit 1 20 V
20V 100 Ω
10 V 1A 200 Ω
A 0.2 A 147 Practice Circuit 1 20 V 100 Ω
10 V 1A 200 Ω 0.2 A 148 Practice Circuit 2
How many node and mesh equations are needed to solve this circuit?
10i1 +
R2 vx  + R3 R1 R4 8.5 V R7 +
R5 R6 Vx/5
i1 149 Practice Circuit 2 10i1 +
R2 vx  + R3 R1 R4 8.5 V R7 +
R5 R6 Vx/5
i1 150 Practice Circuit 2 10i1 +
R2 vx  + R3 R1 R4 8.5 V R7 +
R5 R6 Vx/5
i1 151 Quick Inspection Quick Inspection will make your life easier. Suggestions for Node Voltage • Identify the essential nodes
• Pick an appropriate ground!! Suggestions for Mesh Current • Identify the meshes
• Determine which ones are known!! In both cases: Do not name the node voltages or mesh currents until you pick your method.
Learn to do this by “looking” at the circuit and not writing anything down until you decide on a method… 152 Equal Number of Equations This will happen for many of the circuits you analyze. Now which method should you choose? Things to consider: Do I have to use a “supernode” or “supermesh”?
• In general, these are harder to set up… Which method is most comfortable for me?
What am I solving for? • In general if I need a current, mesh current analysis will be easier; if I need a voltage, node voltage analysis will be easier. 153 Questions?
Node Voltage Analysis vs. Mesh Current Analysis Method Selection Practice For each of the 21 circuits you will be given: Identify the number of equations you would need to write down and solve the problem using node voltage analysis. This means that the node voltage value is unknown and you need to solve for it.
Identify the number of equations you would need to write down and solve the problem using mesh current analysis. This means that the mesh current value is unknown and you need to solve for it. Consider other difficulty factors. 155 Figure 1 156 Figure 2 157 Figure 3 158 Figure 4
5 mA 1 mA
R3 R6 R1
10
10 V R2 R4 20 V R5 159 Figure 5
1.8 A 7Ω 2.8 Ω 8V
3V 8Ω 4Ω
1Ω 5Ω
2Ω 4Ω
3A 1 A
9Ω 7Ω 160 Figure 6 20 Ω
2A 10 Ω 80 V 4A 120 V
40 Ω 161 Figure 7 V1 R2 +
 +
 V3
R4 R1 +
 V2
R3 162 Figure 8
8Ω 4Ω 2Ω 10 Ω 6Ω 5A 12
12 V 163 Figure 9
075 A 8Ω
y 20 Ω 4Ω 10 Ω
8V 16 Ω 3A x
2 V 164 Figure 10 2.5 A 6Ω
R4
I RL 15Ω
20Ω
10 V 165 Figure 11
y
8 kΩ
0.2 kΩ 20
20 mA
4 kΩ 1 kΩ 0.5 kΩ 25 V 2 kΩ x 166 Figure 12 2V
20 Ω 5Ω 5V 20 Ω 9Ω 15 Ω 1A 167 Figure 13 R4
R2
2A R1 R3 1A
7V 168 Figure 14 I2 R3 R5 R2
I1
I1 R1 R4
V1 RL R6
I3 169 Figure 15 R1 I1
I1 R2 R4
R3
V2 RL V1 170 Figure 16
+ +
3.5 mA vx V1 12 mA 1 kΩ 10 kΩ 0.03vx  3 mA
1 mA  171 Figure 17
ix
6A 3Ω 9Ω 0.9Ix
6Ω 15 Ω
4A
6Ω 172 Figure 18
io R1
R1 V1 R2 +
 R3 R4
3io 173 Figure 19 V2 R3 io +
+
+
+
 R1 + R2
V1 +
 4vx +
 2io R4 vx  174 Figure 20 + vo  R2
2vo 4io R5 R1 +
+
+
+
+  I1 R3 R4 +
 V1 io 175 Figure 21 ix R1 R3
Vx/4 I1
V1 +
 + +
vx  4ix R2  176 Questions?
Node Voltage Analysis
Mesh Current Analysis
Selecting the Better Method Practice Figures with Solutions
(Not posted with Handouts) Example Circuit #3 Which method would you use for this circuit?
Node: 1 Eqn
Mesh: 2 or 3 Eqn 179 Example Circuit #3: Node
VA
VA − V 1
− I1 +
=0
R1 + R 2 + R3
R5 + R 6 V1 A 180 Example Circuit #3: Mesh
I must write equations for each mesh!!
And deal with a supermesh!! 181 Example Circuit #4 In terms of the number of equations I would need to write, which method is best: mesh current or node voltage?
y 8 kΩ
0.2 kΩ Mesh = 2 Eqns. & Node = 2 Eqns. 20 mA 4 kΩ 0.5 kΩ 1 kΩ 25 V x 2 kΩ 182 Practice Circuits
Figure 1 Figure 8 Figure 15 Figure 2 Figure 9 Figure 16 Figure 3 Figure 10 Figure 17 Figure 4 Figure 11 Figure 18 Figure 5 Figure 12 Figure 19 Figure 6 Figure 13 Figure 20 Figure 7 Figure 14 Figure 21 183 Figure 1 Node: 2
Mesh: 3 184 Figure 1: Node 185 Figure 1: Mesh 186 Figure 2 Node: 2
Mesh: 1 187 Figure 2: Node 188 Figure 2: Mesh 189 Figure 3 Node: 3
Mesh: 3
(supermesh) 190 Figure 3: Node 191 Figure 3: Mesh 192 Figure 4
5 mA 1 mA Node: 2 R3
R6 R1
10
10 V R2 R4 Mesh: 3
20 V R5 193 Figure 4: Node
5 mA 1 mA
R3 R6 R1
10
10 V R2 R4 20 V R5 194 Figure 4: Mesh
5 mA 1 mA
R3 R6 R1
10
10 V R2 R4 20 V R5 195 Figure 5
1.8 A 7Ω 2.8 Ω Node: 3
8V
3V 8Ω 4Ω
1Ω 5Ω
2Ω (supermesh) 4Ω
3A 1 A
9Ω Mesh: 2 7Ω 196 Figure 5: Node
1.8 A 7Ω 2.8 Ω 8V
3V 8Ω 4Ω
1Ω 5Ω
2Ω 4Ω
3A 1 A
9Ω 7Ω 197 Figure 5: Mesh
1.8 A 7Ω 2.8 Ω 8V
3V 8Ω 4Ω
1Ω 5Ω
2Ω 4Ω
3A 1 A
9Ω 7Ω 198 Figure 6 20 Ω
2A 10 Ω Node: 2
80 V Mesh: 2
(supermesh) 4A 120 V
40 Ω 199 Figure 6: Node 20 Ω
2A 10 Ω 80 V 4A 120 V
40 Ω 200 Figure 6: Mesh 20 Ω
2A 10 Ω 80 V 4A 120 V
40 Ω 201 Figure 7 V1 R2 +
 Node: 1
+
 V3
R4 R1 +
 Mesh: 3 V2
R3 202 Figure 7: Node V1 R2 +
 +
 V3
R4 R1 +
 V2
R3 203 Figure 7: Mesh V1 R2 +
 +
 V3
R4 R1 +
 V2
R3 204 Figure 8
8Ω 4Ω Node: 2
2Ω 10 Ω 6Ω 5A Mesh: 2 12
12 V 205 Figure 8: Node
8Ω 4Ω 2Ω 10 Ω 6Ω 5A 12
12 V 206 Figure 8: Mesh
8Ω 4Ω 2Ω 10 Ω 6Ω 5A 12
12 V 207 Figure 9
075 A 8Ω Node: 2 y
20 Ω 4Ω 10 Ω
8V 16 Ω Mesh: 2
3A x
2 V 208 Figure 9: Node
075 A 8Ω
y 20 Ω 4Ω 10 Ω
8V 16 Ω 3A x
2 V 209 Figure 9: Mesh
075 A 8Ω
y 20 Ω 4Ω 10 Ω
8V 16 Ω 3A x
2 V 210 Figure 10 2.5 A 6Ω
R4
I RL 15Ω Node: 1
Mesh: 2 20Ω
10 V 211 Figure 10: Node 2.5 A 6Ω
R4
I RL 15Ω
20Ω
10 V 212 Figure 10: Mesh 2.5 A 6Ω
R4
I RL 15Ω
20Ω
10 V 213 Figure 11
y
8 kΩ
0.2 kΩ 20
20 mA
4 kΩ 1 kΩ Mesh: 2 0.5 kΩ 25 V Node: 2 2 kΩ x 214 Figure 11: Node
y
8 kΩ
0.2 kΩ 20
20 mA
4 kΩ 1 kΩ 0.5 kΩ 25 V 2 kΩ x 215 Figure 11: Mesh
y
8 kΩ
0.2 kΩ 20
20 mA
4 kΩ 1 kΩ 0.5 kΩ 25 V 2 kΩ x 216 Figure 12 2V
20 Ω 5Ω Node: 2
(supernode)
Mesh: 3 5V 20 Ω 9Ω 15 Ω 1A 217 Figure 12: Node 2V
20 Ω 5Ω 5V 20 Ω 9Ω 15 Ω 1A 218 Figure 12: Mesh 2V
20 Ω 5Ω 5V 20 Ω 9Ω 15 Ω 1A 219 Figure 13 Node: 1 R4
R2
2A R1 R3 Mesh: 1
1A 7V 220 Figure 13: Node R4
R2
2A R1 R3 1A
7V 221 Figure 13: Mesh R4
R2
2A R1 R3 1A
7V 222 Figure 14 Node: 3 I2 Mesh: 5 R3 R5 R2
I1
I1 R1 R4
V1 RL R6 (supermesh) I3 223 Figure 14: Node I2 R3 R5 R2
I1
I1 R1 R4
V1 RL R6
I3 224 Figure 14: mesh I2 R3 R5 R2
I1
I1 R1 R4
V1 RL R6
I3 225 Figure 15 R1 I1
I1 R2 Node: 1 R4 Mesh: 2 R3
V2 RL V1 226 Figure 15: Node R1 I1
I1 R2 R4
R3
V2 RL V1 227 Figure 15: Mesh R1 I1
I1 R2 R4
R3
V2 RL V1 228 Figure 16
+ +
3.5 mA vx V1 12 mA 1 kΩ 10 kΩ 0.03vx 3 mA
1 mA   Node: 2
Mesh: 0!!!
229 Figure 16: Node
+ +
3.5 mA vx V1 12 mA 1 kΩ 10 kΩ 0.03vx  3 mA
1 mA  230 Figure 16: Mesh
+ +
3.5 mA vx V1 12 mA 1 kΩ 10 kΩ 0.03vx  3 mA
1 mA  231 Figure 17
ix
6A 3Ω 9Ω 0.9Ix
6Ω 15 Ω
4A
6Ω Node: 2
Mesh: 4
(supermesh) 232 Figure 17: Node
ix
6A 3Ω 9Ω 0.9Ix
6Ω 15 Ω
4A
6Ω 233 Figure 17: Mesh
ix
6A 3Ω 9Ω 0.9Ix
6Ω 15 Ω
4A
6Ω 234 Figure 18
io R1
R1 V1 R2 +
 R3 Node: 2
R4
3io Mesh: 3
(supermesh) 235 Figure 18: Node
io R1
R1 V1 R2 +
 R3 R4
3io 236 Figure 18: Mesh
io R1
R1 V1 R2 +
 R3 R4
3io 237 Figure 19 V2 R3 io +
+
+
+
 Node: 2 R1 + R2
V1 +
 4vx +
 2io R4 Mesh: 3
(supermesh) vx  238 Figure 19: Node V2 R3 io +
+
+
+
 R1 + R2
V1 +
 4vx +
 2io R4 vx  239 Figure 19: Mesh V2 R3 io +
+
+
+
 R1 + R2
V1 +
 4vx +
 2io R4 vx  240 Figure 20
Node: 2
+ vo  Mesh: 4
(supermesh) R2
2vo 4io R5 R1 +
+
+
+
+  I1 R3 R4 +
 V1 io 241 Figure 20: Node + vo  R2
2vo 4io R5 R1 +
+
+
+
+  I1 R3 R4 +
 V1 io 242 Figure 20: Mesh + vo  R2
2vo 4io R5 R1 +
+
+
+
+  I1 R3 R4 +
 V1 io 243 Figure 21
Node: 2
ix R1
Vx/4 I1
V1 Mesh: 3
(supermesh) R3 +
 + +
vx  4ix R2  244 Figure 21: Node ix R1 R3
Vx/4 I1
V1 +
 + +
vx  4ix R2  245 Figure 21: Mesh ix R1 R3
Vx/4 I1
V1 +
 + +
vx  4ix R2  246 ...
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 Spring '08
 Preston
 Volt

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