07-nodal analysis - · pdf filenodal analysis with voltage sources • properties of a...
TRANSCRIPT
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07-Nodal Analysis Text: 3.1 – 3.4
ECEGR 210
Electric Circuits I
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Overview
• Introduction
• Nodal Analysis
• Nodal Analysis with Voltage Sources
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Introduction
• Basic Circuit Laws
Ohm’s Law
Kirchhoff’s Voltage Law (KVL)
Kirchhoff’s Current Law (KCL)
• Now we seek to analyze any linear circuit systematically using
Nodal analysis
Mesh analysis
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Introduction
• Express circuit as a system of linear equations
• Solve linear equations
Substitution
Gaussian elimination
Cramer’s Rule
Numerical methods
Others
• Remember: for a unique solution there must be an equal number of independent equations as unknowns
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11 12 13 1 1
21 22 23 2 2
31 32 33 3 3
y y y x b
y y y x b
y y y x b
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Introduction
• Let X be a vector of variables to be solved for
• Nodal Analysis:
X: node voltages
Y: conductances
b: currents
• Mesh Analysis:
X: loop currents
Y: resistances
b: voltages
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11 12 13 1 1
21 22 23 2 2
31 32 33 3 3
y y y x b
y y y x b
y y y x b
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Nodal Analysis
Steps to solve a circuit with N nodes
1. Assign a reference node
2. Assign a variable to the voltages at all nodes wrt the reference node (N-1 variables)
3. Apply KCL to each of the N-1 non-reference nodes to generate N-1 equations
4. Use Ohm’s Law to express currents as functions of node voltages
5. Solve resulting simultaneous equations
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Nodal Analysis
• Let I1, I2, R1, R2, R3 be known (given)
• Solve for the voltages at each node
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I1
I2
R1 R3
R2
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Step 1: Assign Reference Node
• Recall: all voltage is a relative measure
• Reference node is also known as ground
In power systems it is actually the ground (earth)
In electronics it could be the metallic chassis
Or arbitrary
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I1
I2
R1 R3
R2
(usually the “bottom” of circuit)
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Step 2: Assign Variables to Nodes
• There are two remaining nodes (variables)
• Assign voltages and polarities wrt reference node
• No need to write voltage across R2
Does not add an independent equation (no new information)
Can solve later as V1 – V2
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I1
I2
R1 R3
R2
+
- V1
+
- V2
x x
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Step 3: Apply KCL
• KCL gives one equation per node (two equations)
• I1 = I2 + i1 + i2
• I2 = i3 – i2
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I1
I2
i1 i3
i2
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Step 4: Apply Ohm’s Law
• Need to write i1, i2, i3 in terms of the node voltages
• Remember sign convention
• i1 = (V1 - 0)/R1
• i2 = (V1 – V2)/R2
• i3 = (V2 - 0)/R3
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I1
I2
i1 i3
i2
x x V2 V1
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Step 5: Solve Equations
• Recap
two variables (V1, V2)
two KCL equations
• KCL equations use intermediate variables i1, i2, i3
I1 = I2 + i1 + i2
I2 = i3 – i2
• Use substitution to express KCL in terms of voltages
i1 = (V1 - 0)/R1
i2 = (V1 – V2)/R2
i3 = (V2 - 0)/R3
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Step 5: Solve Equations
• Via substitution:
I1 = I2 + V1/R1 + (V1 – V2)/R2
I2 = V2/R3 – (V1 – V2)/R2
• Can we solve this system of equations?
• Yes!
Two independent equations, two unknowns
The rest is just math
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Step 5: Solve Equations
• In matrix form
I1 = I2 + V1/R1 + (V1 – V2)/R2
I2 = V2/R3 – (V1 – V2)/R2
• Many methods of solving linear equations
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1 2 2 1 1 2
2 2
2 3 2
1 1 1
R R R V I I
V I1 1 1
R R R
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Example
• Let
I1 = 10A
I2 = 5A
R1 = 6Ω
R2 = 4Ω
R3 = 2Ω
• Find all voltages
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I1
I2
i1 i3
i2
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Example
• Equations:
I1 = I2 + V1/R1 + (V1 – V2)/R2
I2 = V2/R3 – (V1 – V2)/R2
• Using circuit values:
10 = 5 + V1/6 + (V1 – V2)/4
5 = V2/2 – (V1 – V2)/4
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Example
• Starting with the second equation
5x4 = [V2/2 – (V1 – V2)/4]x4
Yields
20 = 2V2 – (V1 – V2) = 3V2 – V1
Now the first equation:
10x12 = [5 + V1/6 + (V1 – V2)/4]x12
120=60 + 2V1+ 3V1 – 3V2
60 = 5V1 – 3V2
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Example
20 = -V1 + 3V2 60 = 5V1 – 3V2
Solve using elimination
• 20x5 = [-V1 + 3V2]x5 = 100 = -5V1 + 15V2
60 = 5V1 – 3V2
160 = 0 +12V2
Yields:
• V2 = 13.333V
• V1 = 20V
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Example
• Solving in matrix form:
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1 2 2 1 1 2
2 2
2 3 2
1
2
1 1 1
R R R V I I
V I1 1 1
R R R
1 1 1V 56 4 4
V 51 1 1
4 2 4
Can use Matlab, Gaussian elimination Cramer’s Rule, matrix inversion
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Solution by Matrix Inversion
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1
2
1 1 1V 56 4 4
V 51 1 1
4 2 4
1
2
1 1
2
V 5
V 5
V 5 20
V 5 13.3
G
G
G (matrix)
For small matrices use “inv()” command in Matlab
1 3 1
1 1.67
Gwhere
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Nodal Analysis with Voltage Sources
• Nodal analysis requires knowledge of current through elements
• Recall that the current through a voltage source is not readily known
• Example:
Current through resistor is I = (Va – Vb)/R
But what is the current through a voltage source? How is the “R” in the equation to be interpreted?
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+
- V
+
-
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Nodal Analysis with Voltage Sources
• If the voltage source is connected to the reference node, then it can be easily included in nodal analysis
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I1
I2
i1
i2
+
- V3
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Nodal Analysis with Voltage Sources
• Only one node with unknown voltage
• I1 = I2 + V1/R1 + (V1 – V3)/R2
• Compare this to earlier example (2 eqns, 2 unknowns)
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I1
I2
+
- V3
+
- V1
R2
R1
x
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Nodal Analysis with Voltage Sources
• If voltage source not connected to reference:
Move the reference so it is! (only works if there is one voltage source)
Or
Use Supernode concept
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Nodal Analysis with Voltage Sources
• Supernode: a closed surface containing a voltage source and its two nodes AND any elements in parallel with it
• Current into supernode = current out of supernode
• Also applies to dependent voltage sources
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+ -
Supernodes
I I
node
+ -
I I
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Nodal Analysis with Voltage Sources
• Properties of a supernode
Voltage source inside the supernode provides a constraint equation needed to solve for node voltages
A supernode has no voltage of its own
There is only one KCL equation for the supernode, but two unknown node voltages
Additional equation is found from application of KVL
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Nodal Analysis with Voltage Sources
• Find the node voltages
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+ -
2V
10W
2A 7A 4W 2W
assume reference is here
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Nodal Analysis with Voltage Sources
• Make a supernode out of the voltage source
• Write KCL for supernode
2 = i1 + i2 + 7 (only one equation since it is a supernode)
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+ -
2V
10W
2A 7A 4W 2W
assume reference is here
i1 i2
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Nodal Analysis with Voltage Sources
• Write i1, i2 in terms of node voltage
• i1 = (v1 – 0)/2
• i2 = (v2 – 0)/4
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+ -
2V
10W
2A 7A 4W 2W
assume reference is here
i1 i2
v1 v2
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Nodal Analysis with Voltage Sources
• Substitute into supernode KCL
• 2 = i1 + i2 + 7 (supernode KCL)
2 = 0.5v1 + 0.25v2 + 7 (after substitution)
8 = 2v1 + v2 +28 (multiplying by 4)
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+ -
2V
10W
2A 7A 4W 2W i1 i2
v1 v2
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Nodal Analysis with Voltage Sources
• Apply KVL around loop
-v1 + 2 + v2 = 0 (second equation)
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+ -
2V
10W
2A 7A 4W 2W i1 i2
v1 v2
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Nodal Analysis with Voltage Sources
• How many independent equations?
8 = 2v1 + v2 +28
-v1 + 2 + v2 = 0
• How many unknowns?
v1, v2
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Solve!
+ -
2V
10W
2A 7A 4W 2W i1 i2
v1 v2
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Nodal Analysis with Voltage Sources
• Solving…
8 = 2v1 + v1 - 2 +28
v1 = -5.333V
v2 = -7.333V
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+ -
2V
10W
2A 7A 4W 2W i1 i2
v1 v2
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Example
• Find the node voltages in the circuit shown
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+ -
10V
2W
4W
2W
8W 5A
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Example
• 3 unknown voltages: V1, V2, V3
• Need three independent equations (that do not introduce new variables)
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+ -
10V
2W
4W
2W
8W
V1 V2 V3
5A
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Example
• Identify the supernode
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+ -
10V
2W
4W
2W
8W
V1 V2 V3
5A
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Example
• Write KCL for the supernode
i5 = i1+i4+i3 (supernode)
• Write KCL for the other node
5+i4 = i5 (node 2)
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+ -
10V
2W
4W
2W
8W
V1 V2 V3
5A i1 i3
i4 i5
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Example
• Express currents in terms of voltages
i1 = (V1 – 0)/4
i3 = (V3 – 0)/8
i4 = (V1 – V2)/2
i5 = (V2 – V3)/2
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+ -
10V
2W
4W
2W
8W
V1 V2 V3
5A i1 i3
i4 i5
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Example
• Have two equations
KCL for node 2, supernode
• Need one more independent equation
• Do KVL around top loop
10 = 2i4 + 2i5
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+ -
10V
2W
4W
2W
8W
V1 V2 V3
5A i1 i3
i4 i5
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Example
• 5 + i4 = i5 (node 2)
• i6 = i1 + i5 + i3 (supernode)
i1 = (V1 – 0)/4 = 0.25V1
i3 = (V3 – 0)/8 = 0.125V3
i4 = (V1 – V2)/2 = 0.5V1 – 0.5V2
i5 = (V2 – V3)/2 = 0.5V2 – 0.5V3
5 + 0.5V1 – 0.5V2 = 0.5V2 – 0.5V3 (solving node 2 eqn)
10 + V1 – V2 = V2 – V3
0.5V2 - 0.5V3 = 0.25V1 + 0.5V1 – 0.5V2 + 0.125V3 (supernode eqn)
4V2 - 4V3 = 2V4 + 4V1 – 4V2 + V3
10 = 2i4 + 2i5 (KVL of top loop)
10 = V1 – V2 + V2 – V3
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Example
10 + V1 – V2 = V2 + V3 (node 2)
10 = -V1 + 2V2 + V3 (after rearranging)
4V2 - 4V3 = 2V1 + 4V1 – 4V2 - V3 (supernode)
0 = 6V1 -8V2 + 3V3 (after rearranging)
10 = V1 – V2 + V2 – V3 (KVL of top loop)
10 = V1 + 0V2– V3 (after rearranging)
Solving
V1 = 12.22V
V2 = 10V
V3= 2.22V
Dr. Louie 45
1
2
3
1 2 1 10
6 8 3 0
1 0 1 10
V
V
V
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Example
• Find Vx
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2A
+
-
Vx 20W 10W 0.2Vx
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Example
• One node, write equation from KCL
2 =I1 + I2 + 0.2Vx
• From Ohm’s Law
I1 = Vx/10
I2 = Vx/20
• Solving:
2 = 0.1Vx + 0.05Vx + 0.2Vx
Vx = 5.71V
Dr. Louie 47
2A
+
-
Vx
20W 10W 0.2Vx I1 I2