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ELECTRIC CIRCUITS
ECSE-2010
Spring 2003
Class 5
ASSIGNMENTS DUE
•
Today (Thursday):
•
•
Activities 5-1, 5-2, 5-3 (In Class)
Next Monday:
•
•
•
•
HW #2 Due
Experiment #1 Report Due
Activities 6-1, 6-2, 6-3, 6-4 (In Class)
Next Tuesday/Wednesday:
•
•
Will do Experiment #2 In Class (EP-2)
Activities 7-1, 7-2, (In Class)
REVIEW
• Controlled/Dependent Sources:
• VCVS, VCCS, CCVS, CCCS
• Used to Model Electronic Devices
• Make Circuits Interesting (and Harder)
• Equivalent Resistance:
• Replace any Load Network with Req
• Load Network = R’s, Controlled Sources,
No Independent Sources
CONTROLLED SOURCES
 v1 
2
10 V


Voltage Controlled Voltage Source (VCVS)
3v1 Volts
CONTROLLED SOURCES
 v1 
2
10 V
Voltage Controlled Current Source (VCCS)
4v1 Amps
CONTROLLED SOURCES
i1
2
10 V
4
Current Controlled Current Source (CCCS)
5i1 Amps
CONTROLLED SOURCES
i
2
10 V


Current Controlled Voltage Source (CCVS)
6i Volts
EQUIVALENT RESISTANCE
i
i


v

v

[Controlled Sources, R's
No Independent Sources]
R eq
R eq
v

i
Req with Controlled Sources
it
[with Controlled
Source(s)]
Connect vt , define i t
Use KCL, KVL to
express v t in terms of i t
vt
vt
R eq 
it
REVIEW
• Req with Controlled Sources:
• Connect Test Voltage vt
• Define it Using Active Convention
• Find it in terms of vt (Use KCL, KVL,
Ohm’s Law, etc.)
• Req = vt / it
WHERE ARE WE?
• Circuit Elements:
•
•
•
•
•
Ideal, Independent i and v sources
Ideal, Controlled i and v sources
Resistors
Potentiometers
Have Learned How to Define all i’s and v’s
using Active/Passive Convention
• Have Begun to do Circuit Analysis = Solving
for all i’s and v’s
WHERE ARE WE?
• Circuit Analysis Techniques:
•
•
•
•
•
•
Ohm’s Law for R’s
KCL and KVL
Simplify Circuits using Series/Parallel R’s
Replace Load Network with Req
Sometimes Hard to Know Where to Start
Will Now Develop Systematic Techniques
that will Always Work
• Node Equations Today
• Mesh Equations on Monday
NODE EQUATIONS
• Systematic Technique for Solving
ANY Linear Circuit:
• Will Always Work!
• Not Always the Easiest Technique
• Will Also Learn About Mesh
Equations:
• Can Use Either Technique; But Cannot Mix
• Very Powerful Techniques!!
• Will Use For Rest of Course
EXAMPLE
v1
2
10 V
2
1
v2
3A
NODE EQUATIONS
• See Example Ckt:
• Node Equation Procedure:
• Label Unknown Node Voltages (v1, v2, …)
EXAMPLE
10 V
10 V
1
vv11
2
2
v2 v2
3A
0V
NODE EQUATIONS
• Node Equation Procedure:
• Label Unknown Node Voltages (v1, v2, …)
• # of Unknown Node Voltages = # of Nodes # of Voltage Sources - 1 (Reference)
• Example: 4 Nodes - 1 Voltage Source - 1 = 2
Unknown Node Voltages; v1, v2
• Write a KCL at Each Unknown Node Voltage
• Sum of Currents Out = 0
• Express i’s in terms of Node Voltages
EXAMPLE
10 V
i1
10 V
1
vv11
2
i2
2
i3
v2 v2
i4
i5
3A
0V
KCL: i1  i 2  i3  0
KCL: i 4  i5  0; i5  3 A
NODE EQUATIONS
• Example:
(v1  10) (v1  0) (v1  v 2 )
KCL at v1:


0
2
2
1
 2v1  v2  5
(v 2  v1 )
KCL at v 2 :
 (  3)  0
1
 (  v1  v2 )  3
NODE EQUATIONS
• Example:
Node v1 
2v1  v 2  5
Node v 2   v1  v 2  3
Add: 
v1  8 V

v2  11 V
NODE EQUATIONS
• Writing a KCL at Each Unknown
Node Voltage will Always Provide #
of Linear Equations = # Unknowns:
• Can Always Solve for v1, v2, ….
• Node Equations will ALWAYS work
• Can take it to the bank
ACTIVITY 5-1
60 V
10 
5
R
20 
12 A
8A
4
40 V
ACTIVITY 5-1
• 6 Nodes - 2 Voltage Sources - 1 (Ref):
=> 3 Unknown Node Voltages:
• v1, v2, v3
ACTIVITY 5-1
v1  60
60 V
10 
5
v1
R
v 2 20 
v3
40 V
12 A
8A
4
0V
40 V
ACTIVITY 5-1
• 6 Nodes - 2 Voltage Sources - 1 (Ref):
=> 3 Unknown Node Voltages:
• v1, v2, v3
• Write a KCL at each Unknown Node
Voltage, relating Currents to the
Node Voltages using Ohm’s Law
• Usually best to sum Currents OUT of
the Node
 iout  0
ACTIVITY 5-1
v1  60
60 V
5
v1
i
out
node
10 
R
v 2 20 
v3
40 V
12 A
8A
4
0V
40 V
0
ACTIVITY 5-1
v1  60  v3 v1  40 v1  v 2
At Node v1:


0
10
5
R
v 2  v1
v 2  40 v 2
At Node v 2 :
 12 

0
R
20
4
v3  (v1  60)
At Node v3:
 ( 12)  8  0
10
ACTIVITY 5-1
1 1 1
1
1
v1 (   )  v 2 ( )  v3 ( )  6  8  14
10 5 R
R
10
1
1 1 1
v1 ( )  v 2 ( 
 )  v3 (0)  2  12  10
R
R 20 4
1
1
v1 ( )  v 2 (0)  v3 ( )  12  6  8  2
10
10
3 Equations; Solve for v1 , v 2 , v3
Maple, MATLAB, Calculator, Cramer's Rule, etc
ACTIVITY 5-1
In Matrix Form:
1
1
1 1 1

 
10  5  R
R
10  v   14 


1
1 1 1




 1

 +
0  v 2    10 


R
R 20 4

  v3    2 
1 
  1
0
 10
10 
V

I



s
G
 
volts amps
siemons
ACTIVITY 5-1
Matrix Equation:
 G   V    Is 
Want to Solve for  V 
Let's Use MAPLE
ACTIVITY 5-2
• For Large Matrices, Use MAPLE
• Usually for n > 2
•
•
•
•
Must Include with (linalg):
Two Ways to Solve using MAPLE:
See MAPLE Scripts
Must Be Careful with Syntax
ACTIVITY 5-2
• MAPLE SCRIPT – Classic Method:
>with (linalg)
>G:=R->matrix(3,3,[1/5+1/10+1/R,
-1/R,-1/10,-1/R,1/20+1/4+1/R,0,
-1/10,0,1/10]):
>G(R);
>Is:=matrix(3,1,[14,-10,-2]);
ACTIVITY 5-2
>v:=R->evalm(inverse(G(R))&*Is):
#&* means matrix multiplication
>v(R);
>v(1);
>ia:=R->(v(R)[2,1]-2):
>ia(R);
>ia(1);
ACTIVITY 5-2
• MAPLE SCRIPT – Gaussjord Method:
>with (linalg)
>G:=R->matrix(3,4,[1/5+1/10+1/R,
-1/R,-1/10,14,-1/R,1/20+1/4+1/R,0,
-10,-1/10,0,1/10,-2]):
>G(R);
>B:=R->guassjord(G(R)):
>B(R);
#leads to same result
ACTIVITY 5-3a
10 k
30 V
8V
6 k
v
(v  8) V

2 k
30 V
5 mA
v

0V
1 Unknown Node Voltage
Find v using Node Equations
ACTIVITY 5-3a
• KCL at Node v:
•
•
•
•
Sum of Currents Out of Node = 0
(v-30)/6 + (v+8)/2 + (v+8-30)/10 - 5 = 0
v (1/6 + 1/2 + 1/10) = 30/6 – 8/2 + 22/10 + 5
=> (23/30) v = 8.2 => v = 10.7 V
• Writing a KCL at each Unknown
Node always works!
ACTIVITY 5-3b
1A
8
2 Unknown Node Voltages
7
ic
10 
24 V
24 V
v1
3
ib
90 V
v2
15 
6
0V
2
4
ia
Show that v1  24 V; v 2  6 V
Find ia , i b , i c
ACTIVITY 5-3b
• 5 Nodes - 2 Voltage Sources - 1 (Ref)
• 2 Unknown Node Voltages, v1, v2
• Write a KCL at each Unknown Node
• Sum of Currents Out = 0
ACTIVITY 5-3b
KCL at v1 :
v1  24 v1  v2 v1 v1  (v 2  90)

 
0
3
15
6
10
KCL at v2 :
v2
(v2  90)  v1 v 2  v1 v 2  24



1  0
24
10
15
7 8
ACTIVITY 5-3b
1 1 1 1
1 1
Node v1: v1 (    )  v 2 (  )  8  9  17
3 15 6 10
15 10
1 1
1 1 1 1
24
Node v 2 : v1 (  )  v 2 (    )  9 
1
15 10
6 10 15 15
15
ACTIVITY 5-3b
Multiply by 30:
v1 (20)  v 2 (5)  510
v1 (5)  v 2 (12)  192
G:  matrix(2,3,[20,  5,510,  5,12, 192]);
V:  gaussjord(G);
 v1  24 V
 v 2  6 V
ACTIVITY 5-3b
With v 1  24 V; v 2  6 V
v2
 i a 
 1 A
24
24  v1  24  v 2 
 i b 

 1  3 A
3
 87

24  v1
 i c  i b 
 3A
3