Post on 17-Jan-2016
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ECE 221Electric Circuit Analysis I
Chapter 10Circuit Analysis 3 Ways
Herbert G. Mayer, PSUStatus 10/21/2015
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Syllabus
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
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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
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Problem 1
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Circuit for Sample Problem 1
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Solve Solve Problem 1Problem 1
Via KCL, KVLVia KCL, KVL
Using Arithmetic SubstitutionUsing Arithmetic Substitution
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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 = 0
(3)’ i3 = (R2*i2)/R3 + v2/v3
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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
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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
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Solve Solve Problem 1Problem 1
Via KCL, KVLVia KCL, KVL
Using Cramer’s RuleUsing Cramer’s Rule
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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
| 1 -1 -1 | | 0 |Δ = | R1 0 R3 |, R = | v1 |
| 0 R2 -R3 | |-v2 |
| 1 -1 -1 |Δ = |100 0 300 |
| 0 200 -300 |
| 1 -1 1 |S = | -1 1 -1 |
| 1 -1 1 |
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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| 0 -1 -1 |
N(i1) = N1 = | 10 0 300 ||-20 200 -300|
N1 = -10 | -1 -1 | -20|-1 -1|| 200 -300| | 0 300|
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) = N2 = | 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) = N3 = | 100 0 10 |
| 0 200 -20 |
N3 = 1 | 0 10 | -100 | -1 0 || 200 -20 | | 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 Solve Problem 1Problem 1
Using NoVoMoUsing NoVoMo
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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 = 30 + 3*20/2
V300*( 1 + 3 + 2/3 ) = 60
Students Compute V300
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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 * 3/2 = 30 + 3*20/2
V300*( 1 + 3 + 3/2 ) = 60
V300 = 60 * 2 / 11
V300 = 10.9090 V
Students Compute i3
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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 * 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
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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
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
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Solve Solve Problem 1Problem 1
Using MeCuMoUsing MeCuMo
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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 = ib*(R2+R3) - R3*ia
-v2 = 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
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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
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Problem 2Problem 2
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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
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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 + 4 + 3 = 0
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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’)
(3’)-(2’)
-500*i2 - 600*i2 = -7 -9 = -16
i2*1,100 = 16
i2 = 16 / 1,100
i2 = 14.54 mA
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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
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Solve Solve Problem 2Problem 2
Via KCL, KVLVia KCL, KVL
Using Cramer’s RuleUsing Cramer’s Rule
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Solve Problem 2 Using Cramer’s Rule
i1 = i2 + i3
i1*100 + i2*200 -3 = 0
-i2*200 + i3*300 +4 +3 = 0
Normalized:
i1 - i2 - i3 = 0
100*i1 + 200*i2 + 0 = 3
0 - 200*i2 + 300*i3 = -7
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Cramer’s Characteristic Determinant
Normalize i1, i2, i3 positions
| -1 1 1 | | 0 |D = | 100 200 0 |, R = | 3 |
| 0 -200 300 | | -7|
| 1 -1 1 |S = | -1 1 -1 |
| 1 -1 1 |
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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 1 1 |N(i1) = N1 = | 3 200 0 |
| -7 -200 300 |
N1 = 0 - 3| 1 1 | -7 | 1 1 ||-200 300 | |200 0 |
Students Compute N1,
Given Δ = -110 k
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Numerator Determinant N1, and i1| 0 1 1 |
N(i1) = N1 = | 3 200 0 || -7 -200 300 |
N1 = 0 - 3| 1 1 | -7 | 1 1 ||-200 300 | |200 0 |
N1 = -3*(300+200) -7*(-200) =
N1 = -1,500 + 1,400
N1 = -10
Now Students Compute i1
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Numerator Determinant N1, and i1| 0 1 1 |
N(i1) = N1 = | 3 200 0 || -7 -200 300 |
N1 = 0 - 3| 1 1 | -7 | 1 1 ||-200 300 | |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
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Numerator Determinant N2, and i2
| -1 0 1 |N(i2) = N2 = |100 3 0 |
| 0 -7 300 |
N2 = -1 | 3 0 | -100 | 0 1 | + 0| -7 300 | | -7 300|
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
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Solve Solve Problem 2Problem 2
Using NoVoMoUsing NoVoMo
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Solve Problem 2 by Node Voltage Method
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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
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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
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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
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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
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Solve Solve Problem 2Problem 2
Using MeCuMoUsing MeCuMo
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Solve Problem 2 by Mesh Current Method
Again we analyze 2 meshes, with fictitious currents ia and ib
Circuit is repeated below for convenience
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Mesh Current In Sample Problem 2
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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
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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
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Solve Problem 2 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 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