ECE 221
Electric Circuit Analysis I
Chapter 10
Circuit Analysis 4 Ways
Herbert G. Mayer, PSU
Status 11/23/2014
For use at Changchun University of Technology CCUT
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Syllabus
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Goal
Sample Problem1
Solve by Substitution KCL
Solve by Cramer’s Rule
Solve by Node-Voltage Method
Solve by Mesh-Current Method
Conclusion Problem1
Same for Problem2
2
Goal
 We’ll analyze simple circuits, named Sample
Problem1 and Sample Problem2
 Both with 2 constant voltage sources, 3 resistors
 Computing 3 branch currents i1, i2, and i3
 First by using conventional algebraic substitution,
applying Kirchhoff’s Laws; we’ll need 3 equations
 Secondly, we use Cramer’s Rule, with these 3
equations, normalizing them to determinant form
 Then, we use the Node-Voltage Method
 Finally we compute fictitious currents ia and ib, using
the Mesh-Current Method
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Problem 1
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Circuit for Sample Problem1
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Sample Problem1: 3 Equations
KCL at node n1 states:
(1)
i1 = i2 + i3
KVL in the left mesh labeled ia yields:
(2)
R1*i1 + R3*i3 - V1
= 0
KVL in the right mesh, labeled ib:
(3)
i.e.
R2*i2 + V2 - R3*i3
= 0
i3 = (R2*i2)/R3 + V2/R3
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Solve Problem1 by 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
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Solve Problem1 by Substitution
i3
= i2 * R2/R3 + V2/R3 = -0.0303+0.066667
i3
= 0.03636 A
i1
i3
= 36.36 mA
i1
= -9.09 mA
= i2 + i3
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Solve Problem1 by Cramer’s Rule
Using 3 Equations from Substitution
i1
= i2 + i3
R1*i1 + R3*i3 - V1
= 0
R2*i2 + V2 - R3*i3
= 0
Normalized:
i1
- i2
- i3
=
R1*i1
+ 0
+ R3*i3
=
0
+ R2*i2
- R3*i3
= -V2
9
0
V1
Cramer’s Characteristic Determinant
Normalize i1, i2, i3 positions in matrix
Δ:
| 1
| R1
| 0
-1
0
R2
Δ# =
| 1
|100
| 0
-1
0
200
S =
|
1
| -1
|
1
-1
1
-1
-1
R3
-R3
|
|,
|
-1 |
300 |
-300 |
1
-1
1
10
|
|
|
R =
| 0 |
| V1 |
|-V2 |
Cramer’s Characteristic Determinant
Δ =
1 | 0
300 | -100 | -1
| 200 -300|
| 200
-1
| + 0 =
-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):
| 0
| 10
|-20
-1
0
200
-1 |
300 |
-300|
N1 = 0 -10 | -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
| 1
0
-1 |
N(i2): | 100 10 300 |
| 0
-20 -300 |
N2 = 1
| 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
|
1
-1
0 |
N(i3): | 100 0
10 |
|
0
200 -20 |
N3 = 1
|
0 1
| 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 Problem1 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, generating 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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Problem1 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
V300
= 60 * 2 / 11
V300
= 10.9090 V
i3 = V300 / 300
i3 = 36.36a mA
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Problem1 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
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Solve Problem1 by Mesh-Current Method
 The mesh current is fictitious, one each associated
with each individual mesh
 Fictitious in the sense that we act 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 directly measurable with an Ampere
meter, when currents from other meshes superimpose
 In the Sample Problem 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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Sample Problem1: Two Mesh Currents
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Solve Problem1 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 = ib*(R2+R3) - R3*ia
-V2 = ia*(R1+R3)*(R2+R3)/R3 V1*(R2+R3)/R3 - R3*ia
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Solve Problem1 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 Problem1 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
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Conclusion Problem1 via Mesh-Current
Since i3 = i1 - i2,  i3 = -9.09 mA - -45.45 mA
it follows:
i3 = 36.36 mA
We see consistency across 4 different methods
of circuit analysis 
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Problem 2
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Sample Problem2
 We’ll analyze a similar circuit, named Sample
Problem2
 With 2 constant voltage sources of 3 V and 4 V
 Plus 3 resistors at 100, 200, and 300 Ohm
 But now the elements are connected differently
 Again we compute 3 branch currents i1, i2, and i3
 Using 4 methods: substitution method, using
Kirchhoff’s Laws
 Then we use Cramer’s Rule
 Thirdly we use the Node-Voltage Method
 Finally the Mesh-Current Method
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Circuit for Sample Problem2
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Sample Problem2: 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
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+ 4 + 3= 0
Solve Problem2 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
i2
= 16
= 16 / 1,100
i2 = 14.54 mA
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Solve Problem2 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
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Solve Problem2 by Cramer’s Rule
i1
= i2 + i3
i1*100 + i2*200-3
-i2*200 + i3*300
= 0
+4 +3
= 0
Normalized:
i1
- i2
- i3
= 0
100*i1 + 200*i2
+ 0
= 3
0
+ 300*i3
= -7
- 200*i2
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Cramer’s Characteristic Determinant
Normalize i1, i2, i3 positions
| -1
1
D# = | 100 200
|
0
-200
S =
|
1
| -1
|
1
-1
1
-1
1 |
0 |,
R =
300 |
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
N(i1):
| 0
| 3
| -7
1
200
-200
N1 = 0 - 3|
1
1 |
|-200 300 |
1 |
0 |
300 |
-7 | 1
|200
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
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1|
0|
Numerator Determinant N2, and i2
N(i2):
N2 =
-1
| -1
0
| 100 3
|
0 -7
| 3
| -7
1 |
0 |
300 |
0 | -100 | 0
300 |
| -7
1 | + 0 =
300|
N2 = -(900) - 100* (7)= -1,600
i2 = -1,600 / -110,000
i2 = 14.54 mA
From this follows i3 = i1-i2
i3 = -13.63 mA
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Node-Voltage With Sample Problem2
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Solve Problem2 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
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Solve Problem2 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
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Solve Problem2 by Mesh-Current Method
 Again we analyze 2 meshes, with fictitious
currents ia and ib
 The circuit is repeated here for convenience
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Mesh-Current With Sample Problem2
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Solve Problem2 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
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Solve Problem2 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
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Solve Problem2 by Mesh-Current Method
With i2 = i1 – i3, it follows:
i2 = 14.54 mA
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Which Method is easiest?
• It seems the Mesh-Current Method is
simplest for these problems
• With the smallest number of equations
• And less chances for students’ sign
confusion, as we follow the same sign (or:
direction) through the whole mesh
• But for a large number of unknowns
Cramer’s Rule is the only way to compute
accurately
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