The node voltage method
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Equivalent resistance
•
Voltage / current dividers
•
Source transformations
•
Node voltages
•
Mesh currents
•
Superposition
Not every circuit lends itself to “short-cut” methods. Sometimes we
need a formal approach that does not rely on seeing a trick that can be
used. The node-voltage is the first (and maybe most used) of our three
formal methods.
The node-voltage method allows for the calculation of the voltages at
each node of the circuit, relative to a reference node. Once the node
voltages are known, all currents in the circuit can be determined easily.
The method leads to a set of simultaneous that must be solved. Bigger
circuits will have more nodes and require more equations (more math).
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node-voltage method – 1
Recall:
IS
iVS1
IS2
R1
a
VS1 +
–
iR1
b
R2
R3
iR2
iR3
c
iVS2
+ VS2
–
d
Using KCL at each node:
a:
b:
c:
d:
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4 equations, but 5 unknowns (iVS1, iVS2,
iR1, iR2, and iR3).
Need more info. Or a better method.
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node-voltage method – 2
The node voltage method
Look at an example.
R1
VS
10 V
+
–
10 !
R2
5!
IS 1 A
Step 1 - Identify all of the nodes in the circuit.
a
VS
+
–
R1
b
R2
IS
c
This small circuit has three nodes. In principle, three unknown
voltages, but we will try to reduce this.
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node-voltage method – 3
Step 2 - Choose one node to be the reference, usually called ground.
Since voltage is a relative quantity (only voltage differences
matter), we can choose one point in the circuit to be V = 0. The
voltages at the other nodes will be with respect this ground node.
a
VS
+
–
R1
b
R2
IS
vc = 0
In general, we can choose any node as the reference, but some
choices are better than others. Typically, we should choose a
node that is connected to the positive or negative terminal of a
voltage source.
For this circuit, that implies node a or node c. This time, we
choose node c, so we can now say vc = 0.
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node-voltage method – 4
Step 3 - Identify any other nodes for which the voltages (with respect
to ground) are known.
va = V S
VS
+
–
R1
b
R2
IS
vc = 0
For this circuit, the voltage source tells us that node a is VS higher
in voltage than the ground node. Therefore va = VS.
Since the voltage at a is now known, we have reduced the
number of unknowns down to one – the voltage at node b.
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node-voltage method – 5
Step 4 - Look for other ways (like resistor reductions) that could be
used to reduce the number of unknown voltages further.
It is not necessary to calculate the voltage at every node. If we
can eliminate non-essential nodes, we should do so.
In this example, we are already down to one node – not much more
we can do.
Step 5 - Assign voltage variables to each of the remaining unknown nodes.
va = VS
VS
+
–
R1
vb
R2
IS
For this example, there
is only one unknown
node voltage.
vc = 0
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node-voltage method – 6
Step 6 - Assign currents to all the branches connected to each of the
unknown nodes. You are at liberty to choose whatever directions
you want for the current arrows.
R1
va = VS
VS
+
–
iR1
IS
vb
R2
iR2
IS
vc = 0
Step 7 - Use KCL to write equations balancing the currents at each
unknown node.
iR1 + IS = iR2
Step 8 - Use Ohm’s law to express the resistor currents in terms of the
node voltages on either side of the resistor. Pay attention to polarity!
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node-voltage method – 7
Step 9 - Substitute the Ohm’s law expressions for the resistor currents
into the KCL equations to form the node-voltage equations.
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The circuit analysis is done. The rest is just math.
Step 10 - Solve the equation(s).
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=
=
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=
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+
+(
+
)(
)
= .
Finally, the resistor voltages and currents (using Ohm’s law) can be calculated.
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= .
= .
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node-voltage method – 8
node-voltage method – summary
1. Identify all of the nodes in the circuit.
2. Choose one node to be ground.
3. Identify nodes for which the voltage is known due to sources.
4. Use resistor reductions to eliminate any other non-essential nodes.
5. Assign variables for the voltages at the remaining unknown nodes.
6. Assign currents to all of the branches connected to the nodes.
7. Write KCL equations balancing the currents at each of the nodes.
8. Use Ohm’s law to express resistor currents in terms of the (unknown)
node voltages on either side of the resistor. (Be sure to get the correct
polarity!)
9. Substitute the resistor currents into the KCL equations to form the
node-voltage equations. (Set of equations relating the unknown node
voltages.)
10.Do the math to solve the equations and determine the node voltages.
Determine currents, powers, etc., if needed.
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node-voltage method – 9
Example
Same circuit, but choose a different ground.
1. Identify the nodes.
a
VS
10 V
+
–
R1
b
10 !
R2
5!
IS 1 A
c
2. Choose one to be ground. This time, choose node a.
va = 0
VS
+
–
R1
b
R2
IS
c
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node-voltage method – 10
3. Identify nodes for which the voltage is known.
Clearly, va – vc = VS. Therefore, vc = – VS.
va = 0
VS
R1
+
–
b
R2
IS
vc = –VS
5&6. Assign variables for the unknown voltages. Assign currents for
each branch connected to the nodes.
va = 0
VS
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–
R1
iR1
IS
vb
R2
iR2
vc = –VS
IS
node-voltage method – 11
7. Write KCL equations balancing the currents at each node.
va = 0
VS
+
–
R1
IS
vb
iR1
R2
IS
iR2
vc = –VS
iR1 + IS = iR2
8. Use Ohm’s law to express the resistor currents in terms of the
node voltages.
=
=
(
)
9. Substitute resistor current equations into the KCL equations to
form the node-voltage equations.
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node-voltage method – 12
10. Solve it.
+
=
+
=
+
=
=
+
+
+
=
(
)(
)
+
=
.
At first glance, this seems wrong – vb here is different from its
value in the previous calculation.
But remember that only voltage differences matter. In choosing a
different reference node, all other node voltages will be shifted
accordingly.
vR1 = VS – vb = 3.33 V and vR2 = vb – 0 = 6.67 V, in both cases. The
corresponding currents will be the same, as well.
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node-voltage method – 13
another example
VS1 +
–
10 V
R1
R3
1 k!
5 k!
R2
3 k!
IS
5 mA
+
–
R3
c
VS2
20 V
1. Identify the nodes.
a
VS1 +
–
R1
b
R2
IS
+
–
VS2
d
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node-voltage method – 14
2. Choose one to be ground. Since there are voltage sources
connected to d, that would seem to be a good choice.
3. Identify nodes for which the voltage is known. With d as ground,
then we see that va = VS1 and vc = VS2.
va = VS1
VS1 +
–
R1
R3
b
R2
IS
vc = VS2
+
–
VS2
Only one node left, so only one
unknown voltage to be found.
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node-voltage method – 15
5&6. Assign a variable for the unknown voltage. Assign currents for
each branch connected to the node.
va = VS1
R1
VS1 +
–
iR1
R3
vb
R2
iR2
IS
iR3
vc = VS2
+
–
VS2
7. Write KCL equations balancing the currents at the node.
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8. Express the resistor currents in terms of the node voltage.
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node-voltage method – 16
9. Substitute resistor currents into the KCL equation to form the
node-voltage equation.
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+
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10. Solve it.
+
+
(
)+(
=
( .
=
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+
+
=
)+(
=
+
)(
+
)=
+
+
)
.
node-voltage method – 17
Example
In the circuit below, we might like to find the currents of the
resistors. The resistor that bridges across the top makes the short-cut
methods unusable.
R3 30 !
R1
10 !
R2
R4
20 !
40 !
VS +
–
100 V
1. Identify the nodes.
x
R1
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R5
50 !
R3
R2
y
VS +
–
g
R4
z
R5
node-voltage method – 18
2. Choose one to be ground. Due to the single voltage source, the
bottom node seems to be a likely choice.
3. Identify nodes for which the voltage is known. With g as ground,
then we see that vy = VS.
R3
x
R1
R2
vy = VS
R4
z
R5
VS +
–
vg = 0
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node-voltage method – 19
5&6. Assign a variable for the unknown voltage. Assign currents for
each branch connected to the node.
R3
R2
vx
R1
iR2
iR1
iR3
VS +
–
R4
iR4
vz
2 nodes this time!
R5
iR5
7. Write KCL equations balancing the currents at the node.
x:
=
+
z:
+
=
8. Express the resistor currents in terms of the node voltages.
=
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=
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node-voltage method – 20
9. Substitute resistor currents
into the KCL equations to form
the node-voltage equations.
x:
=
z:
+
+
=
10. Solve the equations
=
+
+
+
+
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=
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=
3.667vx – 0.667vz = 100 V
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-1.333vx +3.133vz = 100 V
vx = 35.85 V and vz = 47.175 V
node-voltage method – 21
Final example
(a big one)
Again, we might like to know how much power is being generated
and dissipated in the various elements for the circuit below. Our
short-cut methods are useless here, but the node-voltage works in
exactly the same manner as the previous examples. The math is a bit
more tedious for a bigger circuit – there will be 3 equations in 3
unknowns in this case – but the method is still straight forward.
R3 50 !
R1 10 !
VS
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+ 15 V
–
R4 20 !
R2 20 ! IS1
0.25 A
R5 30 !
R6
40 !
IS2
0.5 A
node-voltage method – 22
1. Identify the nodes.
2. Choose one to be ground. (The bottom one is good, again.)
3. Identify known node voltages. (The left-most node is obviously at VS.)
4. Assign variable names to the remaining, unknown nodes.
5. Assign resistor currents in each branch.
R3
va = V S
a
VS +
–
vb R4 iR4
vc R5
b
c
R1
iR1
iR3
R2
iR2
vd
iR5
IS1
e
d
R6
iR6
IS2
ve = 0
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node-voltage method – 23
iR3
R3
vb R4 iR4
R1
VS
iR1
+
–
R2
vc R5
iR5
IS1
iR2
vd
R6
IS2
iR6
7. Write KCL equations balancing the currents at the node.
=
+
+
=
+
=
+
8. Express the resistor currents in terms of the node voltages.
9. Substitute resistor currents into the KCL equations.
=
+
+
+
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node-voltage method – 24
10. Solve the equations
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+ .
+ .
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Two web-site solvers:
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vb = 9.554 V
vc = 8.217 V"
vd = 13.709 V
http://math.bd.psu.edu/~jpp4/finitemath/3x3solver.html
http://www.1728.org/unknwn3.htm
node-voltage method – 25