Transcript v 2

EENG 2009
Part 5. Nodal Analysis
5.1
5.2
5.3
5.4
Nodes and Node Voltages
Nodal Analysis By Example
Supernodes
Dependent Sources
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5.1 Nodes and Node Voltages
Nodes
Recall that nodes are the connected segments of conductor
that remain when we remove the circuit elements (which at
present for us are either resistors or sources).
Example 1
Identify the nodes for the given circuit.
16 
9A
8
12 
3A
Solution:
Redraw the circuit with the circuit elements removed and
note the connected segments of conductors. We see that
there are 3 nodes for this circuit:
A single
node!
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Node Voltages
A node voltage associated with a given node
is defined to be the voltage difference between
the given node and a reference node, which
has been chosen from among the nodes. For a
circuit with N essential nodes, there are N–1
node voltages.
Once the set of node voltages is determined,
all the other voltages and currents can be
obtained in a straightforward manner.
Example 2
Identify a reference node and corresponding
node voltages for the given circuit (whose
nodes we found in the previous example).
16 
9A
8
12 
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3A
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Example 2 Solution:
Using the results of Example 1, we draw the
circuit with the nodes emphasized:
16 
9A
8
12 
3A
Next choose the bottom node as the reference node,
and designate the node voltages for the other two
nodes as v1 and v2:
v1
v2
16 
node 1
9A
8
node 2
12 
3A
Reference node, ground,
earth, “sea level”
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Measurement of Branch and Node Voltages
voltmeter
16 V
+ –
v1
16 
The voltage being
measured is a
branch voltage.
v2
1
9A
8
2
12 
voltmeter
64 V
+ –
3A
voltmeter
Node
Voltages!
48 V
+ –
The voltage across the 16- branch is not a node voltage. It
is a branch voltage, and is actually the difference between the
two node voltages v1 and v2.
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Branch Voltages In Terms of Node Voltages
Example 3
Find the relationships among the branch voltages
and the node voltages.
v1 + v12 – v
2
+
2
1
+
v13
–
3
v23
–
Solution:
There are three branch voltages, v13 , v12 , and v23, and two
node voltages, v1 and v2 (not counting v3) .
The branch voltages v13 and v23 are clearly equal to the
node voltages v1 and v2:
v13 = v1
v23 = v2
The branch voltage v12 is a combination of the node
voltages v1 and v2.
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Solution (cont.):
We can write the branch voltage v12 in terms of the
node voltages v13 and v23 by applying KVL, as
follows:
+
v1 +
v12 –
v2
2
+
1
v13
–
v23
–
3
v12 = v13 – v23 = v1 – v2
Note the correlation with the order of the subscripts:
The subscripts of v12 are 1 & 2 and the subscripts
of the two node voltages being subtracted are 1 & 2.
In general,
vj
node j
vk
node k
vjk = vj – vk
EENG 2009 FS 2006 Part
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Memorize
me!
152
Branch Currents In Terms of Node Voltages
Example 4
Find the relationships among the branch current
and the node voltages. Note that neither voltagepolarity markings nor a current reference arrow is
shown on the diagram!
R
v1
node 1
v2
node 2
reference node
Solution:
The branch current flowing from node 1 to node 2 is:
i12 = ( v1 – v2 ) / R
The branch current flowing from node 2 to node 1 is:
i21 = ( v2 – v1 ) / R
In general:
vj
node j
R
vk
node k
ijk = ( vj – vk ) / R
ikj = ( vk – vj ) / R
Note that to avoid
confusion, ijk, ikj, and their
corresponding reference
directions are not shown
on the diagram.
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5.2 Nodal Analysis By Example
The rationale for nodal analysis is that once the node
voltages are determined, all the other voltages and
currents can be obtained in a simple manner.
The reference node is chosen by the circuit analyst.
In electronic circuits, we frequently choose the node
to which lots of branches are connected. In power
systems, we usually choose “ground” or “earth.”
Look for the associated symbol on the circuit:
Basic Procedure For Nodal Analysis:
1. Identify the nodes and the node voltages.
2. Write KCL at the proper nodes. (Here’s
where we use the branch-current / node-voltage
relationship we just developed.)
3. Solve for the node voltages.
4. Use the node voltages to solve for any other
required quantities.
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Example 1
Find the values of the node voltages.
R2
is1
R1
R3
is2
Solution:
Step 1. There are 3 nodes. Choose the bottom
node as the reference node, and designate the
other two nodes as v1 and v2.
v2
v1
R2
node 2
node 1
is1
R1
R3
is2
node 3
With 3 nodes and one of them as the reference
node, there are two nodal equations needed.
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Solution (cont.):
Step 2. Write KCL at nodes 1 and 2.
v1
v2
R
2
node 1
is1
R1
node 2
R3
is2
KCL at node 1 (Summing the currents flowing
out of node 1):
v1 / R1 + (v1 – v2) / R2 – is1 = 0
At node 2:
(v2 – v1) / R2 + v2 / R3 + is2 = 0
These are the equations to solve for v1 and v2.
We can put them in orderly form (and discover a
shortcut to writing them), as we show next.
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Development of the Algorithmic Method for
Writing Nodal- Analysis Equations
Consider the previous circuit and its nodal
equations:
v1
is1
R1
R2
v2
is2
R3
v1 / R1 + (v1 – v2) / R2 – is1 = 0
(v2 – v1) / R2 + v2 / R3 + is2 = 0
Collect terms in v1 and v2 and put the
independent-source terms on the RHS:
(1 / R1 + 1 / R2) v1 – 1/ R2 v2
= is1
– 1/ R2 v1 + (1 / R2 + 1 / R3) v2 = – is2
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Now, using conductances:
(G1 + G2) v1 – G2 v2
= is1
– G2 v1 + (G2 + G3) v2 = – is2
v1
R2
v2
is1
R1
R3
(1)
(2)
is2
Observe that the following algorithm is valid:
In (1), the coefficient of v1 =
 (conductances connected to node 1)
In (1), the coefficient of v2 =
–  (conductances between nodes 1 & 2)
In (2), the coefficient of v2 =
 (conductances connected to node 2)
In (2), the coefficient of v1 =
–  (conductances between nodes 1 & 2)
In (1) and (2) the RHS =
(current sources entering the node)
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Example 2
Find v1 and v2.
16 
+
9A
+
8 3A
V1
V2
-
Solution:
16 
v1
node 1
9A
12 
v2
Electrically the
same as the
node 2
node 2
8 3A
12 
KCL @ node 1:
v1 / 8 + (v1 – v2) / 16 – 9 = 0
KCL @ node 2:
(v2 – v1) / 16 + v2 /12 – 3 = 0
Collecting terms:
(1/8 + 1/16) v1 – 1/16 v2
–1/16 v1 + (1/16 + 1/12) v2
= 9
= 3
The solution is v1 = 64 V, v2 = 48 V.
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Solution (cont.) Note that these equations could
have been written directly by using the algorithm!
v1
9A
8
16 
v2
3A
12 
(1/8 + 1/16)v1 - 1/16 v2 = 9
–1/16 v1 + (1/16 + 1/12) v2 = 3
The TI-85 & TI-86 keystrokes for solving this equation are:
2nd SIMULT
Number = 2 ENTER
a1,1
=
8 –1 + 16 –1 ENTER
a1,2
=
– 16 –1 ENTER
b1
= 9 ENTER
a2,1
=
– 16 –1 ENTER
a2,2
=
16 –1 + 12 –1 ENTER
b2
=
3 ENTER
SOLVE
x1
= 64.000
x2
= 48.000
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Example 3.
Find the voltage differences across the sources.
4
5A
4
6A
2
3A
Solution:
4
3A
node 2
v2
v1
node 1
5A
6A
4
2
KCL @ nodes 1 and 2:
(v1 – v2) / 4 + 5 + v1 / 2 – 3 = 0
(v2 – v1) / 4 – 5 – 6 + v2 / 4 = 0
Solution: v1 = – 4 V, v2 = 20 V
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Example 3. (cont.)
We still have to find the voltage differences across
the three sources, as follows.*
v2 = 20 V
5A
4
v1 = – 4 V
3A
+
v3A
–
+
v5A
–6 A
2
+
v6A
–
4
v3A = v1 = – 4 V
v6A = v2 = 20 V
v5A = v2 – v1 = 24 V
* At this point it is up to us to choose the polarity
markings for the sources, as none were specified.
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12 
Example 4.
Find the current i and the
power absorbed by the 6 
resistor.
5A
6
i
4
18 
6
Solution:
v2 12 
node 2
6
5A
i
4
v3
node 3
node 1
v1
18 
6
KCL @ 1: (v1 – v3)/4 + (v1 – v2)/12 + v1/6 = 0
KCL @ 2: (v2 – v1)/12 + (v2 – v3)/6 – 5
= 0
KCL @ 3: v3/18 + (v3 – v1)/4 + (v3 – v2)/6
= 0
Solving for the node voltages gives:
v1 = 21 V, v2 = 45 V, v3 = 27 V
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Solution (cont.):
To calculate i we apply Ohm’s Law (which requires
calculating the branch voltage across the 4 
i 4
resistor):
v3
v1
i = ( v3 – v1) / 4
= (27 – 21) / 4
= 3/2 A*
* Now we know that the true current thru the 4  resistor really is
flowing from left-to-right. At node 1 we wrote the expression for the
current flowing from right to left, and at node 3 from left-to-right.
To find the power absorbed by the 6  resistor:
v2 = 45 V
6
12 
p6 = (v2 – v3)2 / 6
4
5A
v1
v3 = 27 V
18 
6
EENG 2009 FS 2006 Part
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= (45 – 27)2 / 6
= 54 W
164
Example 5
+
Find va
1 mA
2 k
va 1 k
3 k
+
–
–
6V
Solution:
(Note that in this circuit
the 6 V voltage source
has one of its nodes
connected to the
reference node. This
makes solution easier!)
va is the only 1 mA
unknown node
voltage, so write
KCL at node a:
Node voltage
already known!
node A
va
+
va 1 k
–
2 k
3 k
6V
+
–
6V
va/1000 + va/3000 + (va – 6)/2000 – 10–3 = 0
(6 + 2 + 3) va = 18 + 6
va = 2.18 V
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Solution (cont.):
va
+
1 mA
va
i1k
1 k
2 k
3 k
6V
+
–
–
6V
In this circuit, the independent current source
produces mA, the independent voltage source
produces volts, and the resistors have values in k.
Currents in the circuit will be in the mA range. For
example,
i1k = 2.18 V / 1 k
= 2.18 mA.
In these situations it is convenient when writing
KCL to express the currents in mA and write the
equation as follows:
va / 1 + va/ 3 + (va – 6) / 2 – 1 = 0
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5.3 Supernodes
A supernode is a set of nodes connected to each
other by voltage sources, but not to the reference
node by a path of voltage sources.
Example 1
Identify the supernode.
Solution:
supernode
10 V
+ –
v1
3A
13 
5
v2
4
14 A
Note that the supernode includes the component(s)
in parallel with the voltage source. (It includes all
of the circuit elements connected between the two
nodes with the node voltages v1 and v2.)
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Solution Procedure for Circuits with Supernodes
Step 1. Write KCL equation for a surface
enclosing the supernode.
Step 2. Write a KVL equation relating the nodal
voltages in the supernode (a constraint equation).
Step 3. Solve the equations.
Example 2.
Find v1 and v2.
Solution:
3A
10 V
supernode
+ –
v1
13 
5
v2
4
14 A
1. KCL @ supernode: 3 + v1/5 + v2/4 – 14 = 0
2. Constraint equation: v1 – v2 = 10
3. Re-writing these two equations:
4 v1 + 5 v2 = 220
– v1 + v2 = – 10
Solving gives: v1 = 30 V, v2 = 20 V
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Example 3
12 k
Find v1
6V
+ –
6 k
v1
3 k
14 mA
2 mA
Solution:
In this circuit there are two regular nodes and one
supernode:
supernode
regular node
v3 12 k v2 6 V v1
+ –
6 k
14 mA
3 k
2 mA
regular node
The solution procedure in this case is:
Step 1. Write the KCL equation for the
supernode and the KCL equation for the regular
node that was not chosen as the supernode.
Step 2. Write the constraint equation.
Step 3. Solve.
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Solution (cont.)
v3
6 k
12 k
v2 6 V
+ –
14 mA
v1
3 k
2 mA
Step 1: KCL @ the supernode:
(v2 – v3) / (12x103 ) + v1 / (3x103 ) + 2x10-3 = 0
Step 1 (cont.): KCL @ the non-reference node:
v3 / (6x103 ) + (v3 – v2) / (12x103 ) – 14x10-3 = 0
Step 2: Constraint equation:
v2 – v1 = 6
Step 3: Solving the 3 equations gives:
v1 = 6 V
v2 = 12 V
v3 = 60 V
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Example 4
Find vo
30 V
20 V +
–
+
–
2 k
4 k
+
vo
–
4 k
+
vo
–
5 k
Solution:
vo
30 V
2 k
+
–
20 V +
–
v1
5 k
v2
Step 1: KCL @ the supernode:
v1 / 2x103 + v2 / 5x103 + vo / 4x103 = 0
Step 2: Constraint equations:
vo – v1 = 30
vo – v2 = 20
Step 3: Solving the 3 equations gives:
vo = 20 V
v1 = – 10 V
v2 = 0 V
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5.4 Dependent Sources
Dependent sources require constraint equations.
Example 1.
Find v1, v2,
and v3.
2
7A
1
v1
+
v1
–
1
v2
2
4 vx
4
1/2  v
3
– vx + 3
3/2 A
Solution:
KCL @ node 1: v1 / 2 + (v1–v2) / 1 – 4vx + 7 = 0
KCL @ node 2: (v2–v1) / 1 + v2 / 4 + (v2–v3) /(1/2) = 0
KCL @ node 3: (v3–v2) /(1/2) – 3/2 – 7 = 0
constraint equation: vx = v3 – v2
Four equations in four unknowns. The solution is:
v1 = 24 V, v2 = 30.25 V,
v2 = 26 V, vx = 4.25 V
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Example 2.
Find i1 and i2.
2 i1
v2
v1
15 A
i1
2
3
i2
4
Solution:
KCL @ node 1: v1 / 2 + 2 i1 + (v1–v2) / 3 = 15
KCL @ node 2: (v2–v1) / 3 + v2 / 4 – 2 i1 = 0
constraint:
v1 / 2 = i1
Solving the 3 simultaneous equations in v1, v2, i1,
and then using the relationship i2 = v2 / 4 gives:
i1 = 7 A
i2 = 8 A
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1
Example 3.
Find vx.
– vx +
1/4 
1/2 
6A
+ –
3 vx
30 V
1/2 
1/3 
Solution:
First, redraw to emphasize the nodes. Then choose
a reference node and label the node voltages. Then
write the KCL and constraint equations.
v2
Note that for the
chosen reference
node there is no
supernode present,
but one of the node v1
voltages is now
known to be –30 V.
1/4 
1/2 
1
6A
+ –
3 vx
1/2 
EENG 2009 FS 2006 Part
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– 30 V
30 V
1/3 
174
Solution (cont.):
v2
node 2
1/4 
1/2
1
node 1
6A
+ –
v1
3 vx
1/2 
–30 V
30 V
1/3 
node 1: – vx / 1 + v1 / (1/2) + 3 vx = 0
node 2:
vx / 1 + v2 / (1/2) + (v2+30)/(1/4) – 6 = 0
constraint: v2 – v1 = vx
Solving:
v1 = –12 V
v2 = 18 v
vx = 6 v
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Example 5. (Contains
dependent sources and a
supernode, too!)
Find v and v1.
6A
+
1
+
v
–
v1
4 i1
6V
–
– +
+ –
1
2
1.5 v1
4
i1
Solution:
First, identify
the nodes and
supernode.
There is one
supernode and +v
two regular
v
nodes (including –
the reference
node).
6A
vc
node c
+
1
6V
+ –
1
va
4 i1
– +
4
i1
2
v1
–
vb
supernode
1.5 v1
reference node
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Solution (cont.):
6A
vc
1
6V
supernode
v
+
v
–
+ –
va
1
Could also
write –i1 here
4 i1
– +
4
i1
node c
+
2
v1
v–b
1.5 v1
reference node
KCL @ supernode:
v / 1 + va / 4 – 1.5v1 + (vb –vc) / 1 – 6 = 0
KCL @ node c:
(vc –vb) / 1 + vc / 2 + 6 = 0
Constraint equations:
v – va = 6
vb – va = 4 i1
i1 = – v / 1
vc – vb = v1
Solving:
v = – 2 V, v1 = – 4 V
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6 unknowns:
v
v1
va
vb
vc
i1
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Example 6.
6i
Find v
2
i
– +
2
+ v –
4A
8V
+ –
2
+
–
4V
Solution:
Try working this out on your own. If you select
the reference node carefully you can save yourself
some effort.
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