A tricky node-voltage situation - Iowa State...
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EE 201 super nodes – 1 A tricky node-voltage situation The node-method will always work – you can always generate enough equations to determine all of the node voltages. The method we have outlined well in almost all cases, but there is one situation where things can be a bit sticky. Whenever there is a voltage source that does not have one of its terminals connected to ground, there will be an unknown current that gets added into the mix. 10 V 1 A 10 ! 5 ! To see the source of the problem, let’s consider the 2-source, 2-resistor circuit one more time. + – R 1 R 2 V S I S a b c + – R 1 R 2 V S I S b v b = 0 v a v c As rookies, we might think that it is OK to choose node b as the reference. That leaves nodes a and c to be determined with respect to b.
Transcript of A tricky node-voltage situation - Iowa State...
EE 201 super nodes – 1
A tricky node-voltage situationThe node-method will always work – you can always generate enough equations to determine all of the node voltages. The method we have outlined well in almost all cases, but there is one situation where things can be a bit sticky. Whenever there is a voltage source that does not have one of its terminals connected to ground, there will be an unknown current that gets added into the mix.
10 V 1 A
10 !
5 !
To see the source of the problem, let’s consider the 2-source, 2-resistor circuit one more time.
+–
R1
R2VS IS
a b
c
+–
R1
R2VS IS
bvb = 0va
vc
As rookies, we might think that it is OK to choose node b as the reference. That leaves nodes a and c to be determined with respect to b.
EE 201 super nodes – 2
We proceed in the usual fashion, defining the currents and setting up KCL equations. We note that since we don’t know va directly, we need to write a KCL there, and that this will involve iVS.
+–
R1
R2VS IS
bvb = 0va
vc
iR1iR2
iVS
Writing the KCL equations:
+–
R1
R2VS IS
bvb = 0va
vc
iR1iR2
iVSa: iVS = iR1
c: iR2 = iVS + IS
EE 201 super nodes – 3
Using Ohm’s law to turn these into node-voltage equations:
�vcR2
= iVS + ISiVS =vaR1
And now we see the essence of the difficulty here – in using Ohm’s law to convert the KCL equations to node-voltage equation, we can’t do anything with the voltage-source current. We don’t know its value and we can’t apply Ohm’s law to it. We are left with three unknowns (va, vc and iVS), but only two equations.
+–
R1
R2VS IS
bvb = 0va
vc
iR1iR2
iVS
Two equations, three unknowns – clearly we need another equation. The “ungrounded” voltage source, which is causing the mathematical difficulty here, also gives us the way out. The source relates the voltages at nodes a and c, giving us a third equation. We might call this the auxiliary equation.
va = vc + VS
EE 201 super nodes – 4
�vcR2
= iVS + ISiVS =vaR1 va = vc + VS
Once again, we have reached a point where we have extracted everything we need in find all the properties of the circuit – the rest is just math.
�vcR2
=vaR1
+ IS
Then substitute in for va using the right-hand equation at the top, giving a single equation that can be solved for vc.
�vcR2
=vc + VS
R1+ IS
As an exercise in algebra, we can proceed in any number of ways. Once approach would be to substitute the left equation for iVS into the middle equation, giving
EE 201 super nodes – 5
�vcR1R2
= vc + VS + R1IS
�vc�1+
R1R2
�= VS + R1IS
vc = �VS + R1IS1+ R1
R2
Plug in the numbers: vc = �10 V+ (10Ω) (1 A)
1+ 10Ω5Ω
= �6.67 V
va = vc + VS = –6.67 V + 10 V = 3.33 V.
iR2 =vR2R2
=6.67 V5Ω = 1.33 AiR1 =
vR1R1
=3.33 V10Ω = 0.333 A
And we have the exact same results as seen twice before:
vR1 = va – vb = 3.33 V – 0 = 3.33 V. vR1 = vb – vc = 0 V – (–6.67 V) = +6.67 V.
Writing out the remaining algebra,
EE 201 super nodes – 6
+–
+–
Consider the circuit below.
We see that there are four nodes. The problem comes in picking one to be ground. Either 2 or 4 would be a reasonable candidate. We’ll flip a coin and pick 4 as ground. Then node 1 is clearly at VS1 = 15 V.
1 2 3
4
VS1
VS2
IS
R1
R2 R31 k!
15 V2 k! 3 k!
5 V
2 mA
Another example
EE 201 super nodes – 7
Label the two nodes and the currents.
Again, we see the same sort of difficulty. The current through the second source is not known, and we cannot use Ohm’s law to relate it to the node voltages on either side. So, in addition to the two unknown node voltages, we have a third unknown, iVS2.
+–
+–va vb
iR1 iVS2iR2 iR3VS1
VS2
IS
R1
R2 R3
EE 201 super nodes – 8
Use KCL to balance currents
iR1 = iR2 + iVS2
+–
+–va vb
iR1 iVS2iR2 iR3VS1
VS2
IS
R1
R2 R3
iVS2 + IS = iR3
VS1 � va
R1=
va
R2+ iV S2
iV S2 + IS =vb
R3Three unknowns.
Can eliminate iVS2:VS1 � va
R1=
va
R2+
vb
R3� IS
EE 201 super nodes – 9
+–
+–va vb
iR1 iVS2iR2 iR3VS1
VS2
IS
R1
R2 R3
Use information from the second source: vb - va = VS2
Substitute into the other equation:
VS1 � va
R1=
va
R2+
va + VS2
R3� IS
Solve for va:
va
1 +
R3
R1+
R2
R2
�=
R3
R1VS1 � VS2 + R3IS
va = 8.36 V
vb = 13.36 V
EE 201 super nodes – 10
+–
+–va vb
iR1 iR2 iR3VS1
VS2
IS
R1
R2 R3
iR1 + IS = iR2 + iR3
VS1 � va
R1+ IS =
va
R2+
va + VS2
R3
vb - va = VS2
Right where we were before without having to mess with iVS2.
An alternative approach that gets to the same point in one less step is the super node. Enclose the other voltage source, along with the two nodes connected in an imaginary container – call it a super node. KCL applies to the super node – what goes in must come out. But note the super node is not all at one voltage.
Write a KCL equation equation for the currents crossing the boundary of the super node:
Relate resistor currents to the voltages:
Relate va and vb using VS2:
