ECE 221
Electric Circuit Analysis I
Chapter 10
Circuit Analysis 3 Ways
Herbert G. Mayer, PSU
Status 10/21/2015
1
Syllabus
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Goal
Sample Problem 1
Solve by Substitution KCL
Using Cramer’s Rule
Solve by Node Voltage Method
Solve by Mesh Current Method
Conclusion Problem 1
Same for Problem 2
2
Goal
 We’ll analyze simple circuits, named Sample
Problem 1 and Sample Problem 2
 With various constant voltage sources and resistors
 Goal is to compute branch currents i1, i2, and i3
 First by using conventional algebraic substitution,
applying Kirchhoff’s Laws; we’ll need 3 equations
 Secondly, we use the Node Voltage Method
 Thirdly we compute fictitious currents ia and ib,
using the Mesh Current Method
 Any method may apply Cramer’s Rule to conduct the
arithmetic computations, once the equations exist
3
Problem 1
4
Circuit for Sample Problem 1
5
Solve Problem 1
Via KCL, KVL
Using Arithmetic Substitution
6
Sample Problem 1: 3 Equations
KCL at node n1:
(1)
i1 = i2 + i3
KVL in the left mesh, labeled ia:
(2)
R1*i1 + R3*i3 - v1
= 0
KVL in the right mesh, labeled ib:
(3)
R2*i2 + v2 - R3*i3
(3)’
i3 = (R2*i2)/R3 + v2/v3
7
= 0
Solve Problem 1 Arithmetic Substitution
(1) in (2)
R1*(i2+i3) + R3*i3
= v1
R1*i2 + R1*i3 + R3*i3
= v1
R1*i2 + i3*(R1+R3)
= v1
R1*i2 + (R2*i2 + v2)*(R1+R3)/R3 = v1
. . .
i2*(R1+R2*(R1+R3)/R3) = v1-v2*(R1+R3)/R3
. . .
i2*(100+2*400/3) = 10 - 20*(400/300)
i2 = -45.45 mA
8
Solve Problem 1 Arithmetic Substitution
i3
= i2 * R2/R3 + v2/R3 = -0.0303+0.066667
i3
= 0.03636 A
i3 = 36.36 mA
i1
= i2 + i3
i1 = -9.09 mA
9
Solve Problem 1
Via KCL, KVL
Using Cramer’s Rule
10
Solve Problem 1 Using Cramer’s Rule
i1
= i2 + i3
R1*i1 + R3*i3 - v1
= 0
R2*i2 + v2 - R3*i3
= 0
Normalized:
i1 – i2 - i3
=
0
R1*i1 + 0 + R3*i3
=
v1
0 + R2*i2 - R3*i3
= -v2
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Cramer’s Characteristic Determinant
Normalize i1, i2, i3 positions in matrix
Δ =
| 1 -1
| R1 0
| 0 R2
| 1
Δ = |100
| 0
S =
|
1
| -1
|
1
-1
R3
-R3
-1
0
200
-1
1
-1
|
|,
|
R =
-1 |
300 |
-300 |
1|
-1 |
1|
12
| 0 |
| v1 |
|-v2 |
Cramer’s Characteristic Determinant
Δ = 1
| 0
300 | -100 | -1
-1 | + 0
| 200 -300|
| 200 -300|
Δ = 1*( 0 – 60,000 ) - 100*( 300 + 200 )
Δ = -60k - 50k
Δ = -110,000
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Numerator Determinant N1, and i1
N(i1) = N1
N1 = -10
| 0 -1
= | 10
0
|-20 200
-1 |
300 |
-300|
| -1
-1 | -20|-1
| 200 -300|
| 0
-1|
300|
N1 =
-10 * (300+200) -20 * (-300 )
N1 =
-10*500 + 6,000
N1 =
1,000
i1=
1,000 / -110,000
i1 = -0.00909 A = -9.09 mA
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Numerator Determinant N2, and i2
N(i2) = N2
N2 = 1
| 1
0 -1
|
= | 100 10 300 |
| 0
-20 -300 |
| 10 300 | -100 | 0
-1 |
|-20 -300 |
| -20 -300|
N2 = -3,000 + 6,000 -100 * ( 0 - 20 )
N2 =
3,000 + 2,000 = 5,000
i2 =
5,000 / -110,000
i2 = -0.04545 A = -45.45 mA
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Numerator Determinant N3, and i3
N(i3) = N3
N3 = 1
|
1
= | 100
|
0
|
0 10
| 200-20
-1 0 |
0 10 |
200 -20 |
| -100 | -1
0 |
|
| 200 -20 |
N3 =
-2,000 - 100 * (20 )= -4,000
i3=
-4,000 / -110,000
i3 = 0.0363636 A = 36.36 mA
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Solve Problem 1
Using NoVoMo
17
Solve Problem 1 by Node Voltage Method
Ignoring the current or voltage directions from
the substitution method, we use the Node
Voltage Method at node n1, currents flowing
toward reference node n2
We generate 1 equation with unknown V300,
voltage at the 300 Ω resistor, yielding i3
Once known, we can compute the voltages at
R1 and R2, and thus compute the currents i1
and i2, using Ohm’s law
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Solve Problem 1 by Node Voltage Method
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Solve Problem 1 by Node Voltage Method
3 currents flowing from n1 toward reference node n2:
V300/300 + (V300-10)/100 + (V300-20)/200 = 0
V300 + 3*V300 + V300*2/3
V300*( 1 + 3 + 2/3 )
= 30 + 3*20/2
= 60
Students Compute V300
20
Solve Problem 1 by Node Voltage Method
3 currents flowing from n1 toward reference node n2:
V300/300 + (V300-10)/100 + (V300-20)/200 = 0
V300 + 3*V300 + V300 * 3/2 = 30 + 3*20/2
V300*( 1 + 3 + 3/2 )
= 60
V300
= 60 * 2 / 11
V300
= 10.9090 V
Students Compute i3
21
Solve Problem 1 by Node Voltage Method
3 currents flowing from n1 toward reference node n2:
V300/300 + (V300-10)/100 + (V300-20)/200 = 0
V300 + 3*V300 + V300 * 3/2 = 30 + 3*20/2
V300*( 1 + 3 + 3/2 )
= 60
V300
= 60 * 2 / 11
V300
= 10.9090 V
i3 = V300 / 300
i3 = 36.363 mA
22
Solve Problem 1 by Node Voltage Method
V(R1)
= v1 - V300
V(R1)
= 10 - 10.9090 = -0.9090 V
i1
= V(R1) / R1
i1
= -0.9090 / 100
i1
= -9.09 mA
Students Compute i2
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Solve Problem 1 by Node Voltage Method
V(R1)
= v1 - V300
V(R1)
= 10 - 10.9090 = -0.9090 V
i1
= V(R1) / R1
i1
= -0.9090 / 100
i1
= -9.09 mA
From this follows i2 using KCL:
i2
= i1 - i3
i2
= -9.0909 – 36.3636
i2
= -45.45 mA
24
Solve Problem 1
Using MeCuMo
25
Solve Problem 1 by Mesh Current Method
 The mesh current is fictitious, one such current
associated with its own individual mesh
 Fictitious in the sense as if it were uniquely tied to a
mesh; yet depending on the branch of the mesh,
mesh currents from other parts flow though that
very mesh as well
 Kirchhoff’s current law is trivially satisfied, but mesh
currents are not everywhere measurable with an
Ampere meter: not measurable, when currents from
other meshes super-impose
 In Sample Problem 1 we have 2 meshes, with mesh
currents indicated as ia and ib
 But we must track that, R3 for example, has both
flowing though it in opposing directions
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Solve Problem 1 by Mesh Current Method
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Solve Problem 1 by Mesh Current Method
KVL for mesh with ia yields:
(1)
R1*ia + R3*(ia-ib)
=
v1
KVL for mesh with ib yields:
(2)
R3*(ib-ia) + R2*ib
= -v2
Students Compute (1) for ib
Then substitute ib in (2)
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Solve Problem 1 by Mesh Current Method
KVL for mesh with ia yields:
(1) R1*ia + R3*(ia–ib)
=
v1
KVL for mesh with ib yields:
(2) R3*(ib-ia) + R2*ib
= -v2
From (1) follows:
(1) ib
= ( R1*ia + R3*ia - v1 ) / R3
Substitute ib in (2):
(2) -v2
-v2
= ib*(R2+R3) - R3*ia
= ia*(R1+R3)*(R2+R3)/R3 v1*(R2+R3)/R3 - R3*ia
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Solve Problem 1 by Mesh Current Method
v1*(R2+R3)/R3 - v2 =
ia*( (R1+R3)*(R2+R3)/R3 – R3)
-20 + 10*5/3 = ia*(400*500/300 – 300)
ia = -10 / 1100
ia = -0.00909 A = -9.09 mA
Since ia = i1:
i1 = -9.09 mA
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Solve Problem 1 by Mesh Current Method
Recall (1):
(1) R1*ia + R3*(ia–ib)
= v1
R3*ib= ia*(R1+R3) - v1
ib
= ia*(R1+R3)/R3 - v1/R3
ib
= -10*400/(1,100*300) - 10/300
ib
= -0.04545 A = -45.45 mA
since i2 = ib:
i2 = -45.45 mA
31
Conclusion Problem 1 via Mesh Current
Since i3 = i1 - i2,  i3 = -9.09 mA - -45.45 mA
it follows:
i3 = 36.36 mA
We see consistency across 3 different
approaches to circuit analysis 
32
Problem 2
33
Sample Problem 2
 We’ll analyze another, similar circuit, named Sample
Problem 2
 With 2 constant voltage sources of 3 V and 4 V
 Plus 3 resistors at 100, 200, and 300 Ohm
 Again we compute 3 branch currents i1, i2, and i3
 Using 3 methods:
 First we use substitution, applying Kirchhoff’s Laws
 Then we use the Node Voltage Method
 Thirdly the Mesh Current Method
 Any of these methods may use Cramer’s Rule
34
Circuit for Sample Problem 2
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Sample Problem 2: Three Equations
KCL states:
(1)
i1
= i2 + i3
KVL in the upper mesh labeled ia yields:
(2)
i1*100 + i2*200-3
= 0
KVL in the lower mesh, labeled ib yields:
(3)
-i2*200 + i3*300
36
+ 4 + 3= 0
Solve Problem 2 by Substitution
-200*i2 + (i1-i2)*300
= -7
// (1)in(3)
-500*i2 + 300*i1
= -7
// (3’)
100*i1
+ 200*i2
=
3
// (2)*3
300*i1
+ 600*i2
=
9
// (2’)
- 600*i2
= -7 -9= -16
(3’)-(2’)
-500*i2
i2*1,100
= 16
i2 = 16 / 1,100
i2 = 14.54 mA
37
Solve Problem 2 by Substitution
i1*100 + i2*200 = 3
i1*100
= 3-200*(16/1,100)
i1*100
= 100/1,100
i1
= 1 / 1,100
i1 = 0.91 mA
i3
= i1 - i2
i3
= -15 / 1,100
i3 = -13.63 mA
38
Solve Problem 2
Via KCL, KVL
Using Cramer’s Rule
39
Solve Problem 2 Using Cramer’s Rule
i1
= i2 + i3
i1*100 + i2*200-3
-i2*200 + i3*300
= 0
+4 +3= 0
Normalized:
i1
- i2
- i3
100*i1 + 200*i2
0
- 200*i2
= 0
+ 0
= 3
+ 300*i3
= -7
40
Cramer’s Characteristic Determinant
Normalize i1, i2, i3 positions
D =
| -1
1
| 100 200
|
0
-200
1 |
0 |,
R =
300 |
S =
|
1
| -1
| 1
1
-1
1
-1
1
-1
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|
|
|
| 0 |
| 3 |
| -7|
Cramer’s Characteristic Determinant
Δ = -1 | 200 0 | -100 |
1
1 | + 0
| 200 -300|
|-200 300 |
Δ = -60,000 – 50,000
= -110,000
Δ = -110 k
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Numerator Determinant N1, and i1
| 0
N(i1) = N1 = | 3
| -7
1
200
-200
1 |
0 |
300 |
N1 = 0 - 3|
1
1 | -7 | 1
|-200 300 |
|200
1|
0|
Students Compute N1,
Given Δ = -110 k
43
Numerator Determinant N1, and i1
| 0
N(i1) = N1 = | 3
| -7
1
1 |
200
0 |
-200 300 |
N1 = 0 - 3|
1
1 |
|-200 300 |
-7 | 1 1 |
|200 0 |
N1 = -3*(300+200) -7*(-200) =
N1 = -1,500 + 1,400
N1 = -10
Now Students Compute i1
44
Numerator Determinant N1, and i1
| 0
N(i1) = N1 = | 3
| -7
1
1 |
200
0 |
-200 300 |
N1 = 0 - 3|
1
1 |
|-200 300 |
-7 | 1 1 |
|200 0 |
N1=
-3*(300+200) -7*(-200) =
N1=
-1,500 + 1,400
N1=
-100
i1 = -100 / -110,000
i1 = 0.000909 A
i1 = 0.91 mA
45
Numerator Determinant N2, and i2
| -1
N(i2) = N2 = |100
| 0
N2 = -1 | 3
| -7
0
1 |
3
0 |
-7 300 |
0 | -100 | 0
300 |
| -7
1 | + 0
300|
N2 = -(900) - 100* (7)= -1,600
i2 = -1,600 / -110,000
i2 = 14.54 mA
With i3 = i1 - i2 it follows:
i3 = -13.63 mA
46
Solve Problem 2
Using NoVoMo
47
Solve Problem 2 by Node Voltage Method
48
Solve Problem 2 by Node Voltage Method
There are 2 essential nodes, n1 and n2
One will be selected as reference node: pick n2
Compute 3 currents from n1 to n2, express as
function of v200
Students compose single KCL equation
For node n1, using single unknown v200
49
Solve Problem 2 by Node Voltage Method
Use KCL to compute 3 current from n1 toward reference
node n2:
V200/200 + (V200-3)/100 + (V200-3-4)/300= 0
Students compute v200, and i2
50
Solve Problem 2 by Node Voltage Method
Use KCL to compute 3 current from n1 toward reference
node n2:
V200/200 + (V200-3)/100 + (V200-3-4)/300= 0
V200*(3/2 + 3 + 1 ) = 9 + 7
V200*11/2
= 16
V200
= 2.9090 V
i2
= V200 / 200
i2 = 14.54 mA
51
Solve Problem 2 by Node Voltage Method
KVL in the lower mesh, with V300 being the voltage drop
across the 300 resistor, yields:
V300
= -7 + V200 = -7 + 2.9090 = -4.091 V
i.e. i3 = V300/300= -0.013637 mA
i3 = -13.63 mA
i1 = i2 + i3 = 14.54 - 13.63
i1 = 0.91 mA
52
Solve Problem 2
Using MeCuMo
53
Solve Problem 2 by Mesh Current Method
 Again we analyze 2 meshes, with fictitious
currents ia and ib
 Circuit is repeated below for convenience
54
Mesh Current In Sample Problem 2
55
Solve Problem 2 by Mesh Current Method
KVL for mesh with ia yields:
(1) 100*ia + 200*( ia-ib ) = 3
(1) 300*ia - 200*ib
= 3
(1) ib
= (300*ia-3)/200
KVL for mesh with ib yields:
(2)
300*ib+200*( ib – ia ) = -7
(2)
500*ib-200*ia
= -7
56
Solve Problem 2 by Mesh Current Method
Substitute ib from (1) in (2):
500*(300*ia - 3)/200 - 200*ia = -7
ia = 1/1,100 = 0.91 mA
i1 = ia, hence:
i1 = 0.91 mA
ib = 3*ia/2-3/200 = 3/(1,100 * 2) - 3/200
ib = -13.63 mA
i3 = ib, hence
i3 = -13.63 mA
57
Solve Problem 2 by Mesh Current Method
With i2 = i1 - i3, it follows:
i2 = 14.54 mA
58
Which Method is easiest?
• It seems the Node Voltage Method is simplest
for these problems
• With the smallest number of equations
• Mesh Current method has smaller number of
equations than pure KCL and KVL
• Small number of equations yields less
chances for sign confusion
• But for a large number of unknowns
Cramer’s Rule is THE methodical way to
compute
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