Class 5
ASSIGNMENTS DUE •
ELECTRIC CIRCUITS ECSE-2010 Spring 2003
Today (Thursday): •
• • • •
Class 5
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)
CONTROLLED SOURCES
REVIEW
+ v1 −
• Controlled/Dependent Sources: • VCVS, VCCS, CCVS, CCCS • Used to Model Electronic Devices • Make Circuits Interesting (and Harder)
• Equivalent Resistance:
2Ω
− 10 V
+
• Replace any Load Network with Req • Load Network = R’s, Controlled Sources, No Independent Sources
3v1 Volts
Voltage Controlled Voltage Source (VCVS)
CONTROLLED SOURCES
CONTROLLED SOURCES i1
+ v1 −
2Ω
2Ω
10 V
Voltage Controlled Current Source (VCCS)
4v1 Amps
10 V
4Ω
5i1 Amps
Current Controlled Current Source (CCCS)
1
CONTROLLED SOURCES
EQUIVALENT RESISTANCE
i
i
i 2Ω
− 10 V
6i Volts
+
+
+
v −
v − [Controlled Sources, R's No Independent Sources]
Current Controlled Voltage Source (CCVS)
Req with Controlled Sources
REVIEW • 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
vt
Source(s)]
Connect v t , define i t
Use KCL, KVL to express v t in terms of i t
R eq =
vt it
WHERE ARE WE? • Circuit Elements: • • • • •
v i
• Req with Controlled Sources:
it
[with Controlled
R eq =
R eq
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
2
NODE EQUATIONS
EXAMPLE
• Systematic Technique for Solving ANY Linear Circuit:
v1
2Ω
1Ω
v2
• Will Always Work! • Not Always the Easiest Technique
• Will Also Learn About Mesh Equations:
3A
2Ω
10 V
• Can Use Either Technique; But Cannot Mix
• Very Powerful Techniques!! • Will Use For Rest of Course
NODE EQUATIONS • See Example Ckt: • Node Equation Procedure:
EXAMPLE 10 V
vv11
2Ω
1Ω
v2 v2
• Label Unknown Node Voltages (v1, v2, …) 2Ω
10 V
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
EXAMPLE 10 V
vv11
2Ω i1
10 V
1Ω i2
2Ω
• Sum of Currents Out = 0 • Express i’s in terms of Node Voltages
i3
v2 v2 i4 i5 3A
0V KCL: i1 + i 2 + i3 = 0
KCL: i 4 + i5 = 0; i5 = −3 A
3
NODE EQUATIONS • Example:
KCL at v1:
NODE EQUATIONS • Example:
(v1 − 10) (v1 − 0) (v1 − v2 ) + + =0 2 2 1
Node v1 =>
Node v2 => − v1 + v2 = 3
=> 2v1 − v2 = 5 KCL at v2 :
2v1 − v2 = 5
(v2 − v1 ) + ( − 3) = 0 1
Add: =>
v1 = 8 V
=>
v2 = 11 V
=> ( − v1 + v2 ) = 3
ACTIVITY 5-1
NODE EQUATIONS
60 V
• 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
5Ω
10 Ω
R
20 Ω 12 A
4Ω
8A
40 V
ACTIVITY 5-1
ACTIVITY 5-1 v1 − 60
• 6 Nodes - 2 Voltage Sources - 1 (Ref): => 3 Unknown Node Voltages:
60 V 10 Ω
• v1, v2, v3
5Ω
v1 R
v 2 20 Ω
v3
40 V
12 A 8A
4Ω
40 V
0V
4
ACTIVITY 5-1
ACTIVITY 5-1 • 6 Nodes - 2 Voltage Sources - 1 (Ref): => 3 Unknown Node Voltages:
v1 − 60
At Node v 2 : At Node v3 :
v1 − 60 − v3 v1 − 40 v1 − v 2 + + =0 10 5 R v 2 − v1 v − 40 v 2 + 12 + 2 + =0 R 20 4 v3 − (v1 − 60) + (−12) + 8 = 0 10
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 v − 2 1 3 − 1 0 10 10 [ V ] = [ Is ] [G ] volts amps siemons
∑i
out
=0
R
v 2 20 Ω
v3
40 V
12 A 8A
4Ω
40 V
0V
ACTIVITY 5-1
ACTIVITY 5-1 At Node v1:
5Ω
v1
node
10 Ω
• 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
60 V
1 1 1 1 1 + + ) + v 2 (− ) + v3 (− ) = 6 + 8 = 14 10 5 R R 10 1 1 1 1 + ) + v3 (0) = 2 − 12 = −10 v1 (− ) + v 2 ( + R R 20 4 1 1 v1 (− ) + v 2 (0) + v3 ( ) = 12 − 6 − 8 = −2 10 10
v1 (
3 Equations; Solve for v1 , v 2 , v3 Maple, MATLAB, Calculator, Cramer's Rule, etc
ACTIVITY 5-1 Matrix Equation:
[ G ] [ V ] = [ Is ] Want to Solve for [ V ] Let's Use MAPLE
5
ACTIVITY 5-2
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
• 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
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);
• 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
ACTIVITY 5-3a
10 kΩ
• KCL at Node v: 30 V
6 kΩ
8V
v
(v + 8) V +
30 V
5 mA
v
2 kΩ
−
• • • •
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!
0V
1 Unknown Node Voltage
Find v using Node Equations
6
ACTIVITY 5-3b 1A
8Ω
• 5 Nodes - 2 Voltage Sources - 1 (Ref)
7Ω
• 2 Unknown Node Voltages, v1, v2 ic
10 Ω
24 V
24 V
v1
3Ω ib
ACTIVITY 5-3b
2 Unknown Node Voltages
90 V
v2
15 Ω 6Ω 0V
• Write a KCL at each Unknown Node • Sum of Currents Out = 0
2Ω
4Ω
ia
Show that v1 = 24 V; v 2 = −6 V Find i a , i b , i c
ACTIVITY 5-3b KCL at v1 : v1 − 24 v1 − v 2 v1 v1 − (v 2 + 90) + + + =0 3 15 6 10 KCL at v 2 : v2 (v + 90) − v1 v 2 − v1 v 2 − 24 + 2 + + −1 = 0 2+4 10 15 7+8
ACTIVITY 5-3b
ACTIVITY 5-3b 1 1 1 1 1 1 Node v1: v1 ( + + + ) + v 2 (− − ) = 8 + 9 = 17 3 15 6 10 15 10 Node v 2 : v1 (−
1 1 1 1 1 1 24 − ) + 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
With v1 = 24 V; v2 = −6 V v2 = −1 A 2+4 24 − v1 24 − v2 => i b = + + 1 = 3 A 3 8+7 24 − v1 => ic = i b − = 3A 3 => ia =
7
ELECTRIC CIRCUITS ECSE-2010 Spring 2003
Today (Thursday): •
• • • •
Class 5
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)
CONTROLLED SOURCES
REVIEW
+ v1 −
• Controlled/Dependent Sources: • VCVS, VCCS, CCVS, CCCS • Used to Model Electronic Devices • Make Circuits Interesting (and Harder)
• Equivalent Resistance:
2Ω
− 10 V
+
• Replace any Load Network with Req • Load Network = R’s, Controlled Sources, No Independent Sources
3v1 Volts
Voltage Controlled Voltage Source (VCVS)
CONTROLLED SOURCES
CONTROLLED SOURCES i1
+ v1 −
2Ω
2Ω
10 V
Voltage Controlled Current Source (VCCS)
4v1 Amps
10 V
4Ω
5i1 Amps
Current Controlled Current Source (CCCS)
1
CONTROLLED SOURCES
EQUIVALENT RESISTANCE
i
i
i 2Ω
− 10 V
6i Volts
+
+
+
v −
v − [Controlled Sources, R's No Independent Sources]
Current Controlled Voltage Source (CCVS)
Req with Controlled Sources
REVIEW • 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
vt
Source(s)]
Connect v t , define i t
Use KCL, KVL to express v t in terms of i t
R eq =
vt it
WHERE ARE WE? • Circuit Elements: • • • • •
v i
• Req with Controlled Sources:
it
[with Controlled
R eq =
R eq
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
2
NODE EQUATIONS
EXAMPLE
• Systematic Technique for Solving ANY Linear Circuit:
v1
2Ω
1Ω
v2
• Will Always Work! • Not Always the Easiest Technique
• Will Also Learn About Mesh Equations:
3A
2Ω
10 V
• Can Use Either Technique; But Cannot Mix
• Very Powerful Techniques!! • Will Use For Rest of Course
NODE EQUATIONS • See Example Ckt: • Node Equation Procedure:
EXAMPLE 10 V
vv11
2Ω
1Ω
v2 v2
• Label Unknown Node Voltages (v1, v2, …) 2Ω
10 V
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
EXAMPLE 10 V
vv11
2Ω i1
10 V
1Ω i2
2Ω
• Sum of Currents Out = 0 • Express i’s in terms of Node Voltages
i3
v2 v2 i4 i5 3A
0V KCL: i1 + i 2 + i3 = 0
KCL: i 4 + i5 = 0; i5 = −3 A
3
NODE EQUATIONS • Example:
KCL at v1:
NODE EQUATIONS • Example:
(v1 − 10) (v1 − 0) (v1 − v2 ) + + =0 2 2 1
Node v1 =>
Node v2 => − v1 + v2 = 3
=> 2v1 − v2 = 5 KCL at v2 :
2v1 − v2 = 5
(v2 − v1 ) + ( − 3) = 0 1
Add: =>
v1 = 8 V
=>
v2 = 11 V
=> ( − v1 + v2 ) = 3
ACTIVITY 5-1
NODE EQUATIONS
60 V
• 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
5Ω
10 Ω
R
20 Ω 12 A
4Ω
8A
40 V
ACTIVITY 5-1
ACTIVITY 5-1 v1 − 60
• 6 Nodes - 2 Voltage Sources - 1 (Ref): => 3 Unknown Node Voltages:
60 V 10 Ω
• v1, v2, v3
5Ω
v1 R
v 2 20 Ω
v3
40 V
12 A 8A
4Ω
40 V
0V
4
ACTIVITY 5-1
ACTIVITY 5-1 • 6 Nodes - 2 Voltage Sources - 1 (Ref): => 3 Unknown Node Voltages:
v1 − 60
At Node v 2 : At Node v3 :
v1 − 60 − v3 v1 − 40 v1 − v 2 + + =0 10 5 R v 2 − v1 v − 40 v 2 + 12 + 2 + =0 R 20 4 v3 − (v1 − 60) + (−12) + 8 = 0 10
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 v − 2 1 3 − 1 0 10 10 [ V ] = [ Is ] [G ] volts amps siemons
∑i
out
=0
R
v 2 20 Ω
v3
40 V
12 A 8A
4Ω
40 V
0V
ACTIVITY 5-1
ACTIVITY 5-1 At Node v1:
5Ω
v1
node
10 Ω
• 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
60 V
1 1 1 1 1 + + ) + v 2 (− ) + v3 (− ) = 6 + 8 = 14 10 5 R R 10 1 1 1 1 + ) + v3 (0) = 2 − 12 = −10 v1 (− ) + v 2 ( + R R 20 4 1 1 v1 (− ) + v 2 (0) + v3 ( ) = 12 − 6 − 8 = −2 10 10
v1 (
3 Equations; Solve for v1 , v 2 , v3 Maple, MATLAB, Calculator, Cramer's Rule, etc
ACTIVITY 5-1 Matrix Equation:
[ G ] [ V ] = [ Is ] Want to Solve for [ V ] Let's Use MAPLE
5
ACTIVITY 5-2
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
• 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
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);
• 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
ACTIVITY 5-3a
10 kΩ
• KCL at Node v: 30 V
6 kΩ
8V
v
(v + 8) V +
30 V
5 mA
v
2 kΩ
−
• • • •
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!
0V
1 Unknown Node Voltage
Find v using Node Equations
6
ACTIVITY 5-3b 1A
8Ω
• 5 Nodes - 2 Voltage Sources - 1 (Ref)
7Ω
• 2 Unknown Node Voltages, v1, v2 ic
10 Ω
24 V
24 V
v1
3Ω ib
ACTIVITY 5-3b
2 Unknown Node Voltages
90 V
v2
15 Ω 6Ω 0V
• Write a KCL at each Unknown Node • Sum of Currents Out = 0
2Ω
4Ω
ia
Show that v1 = 24 V; v 2 = −6 V Find i a , i b , i c
ACTIVITY 5-3b KCL at v1 : v1 − 24 v1 − v 2 v1 v1 − (v 2 + 90) + + + =0 3 15 6 10 KCL at v 2 : v2 (v + 90) − v1 v 2 − v1 v 2 − 24 + 2 + + −1 = 0 2+4 10 15 7+8
ACTIVITY 5-3b
ACTIVITY 5-3b 1 1 1 1 1 1 Node v1: v1 ( + + + ) + v 2 (− − ) = 8 + 9 = 17 3 15 6 10 15 10 Node v 2 : v1 (−
1 1 1 1 1 1 24 − ) + 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
With v1 = 24 V; v2 = −6 V v2 = −1 A 2+4 24 − v1 24 − v2 => i b = + + 1 = 3 A 3 8+7 24 − v1 => ic = i b − = 3A 3 => ia =
7
