2.3.1 -Y ð-TÞ Conversion Formula
2.4 Node Analysis
2.4.1 Circuits Having No Dependent Sources
2.4.1.1 Node Analysis of Circuits Having No Voltage Sources
The above-mentioned method can be used to analyze circuits that are excited by independent current source(s) only. Let us take a look at the following example.
(Example 2.2) A Circuit Excited by Independent Current Sources Only
For the circuit of Figure 2.7(a), the top three nodes are labeled 1,2, and 3 and the bottom node 0 as the reference node. Then the formula (2.10) is applied to write a set of node equations as
y11 y12 y13
Figure 2.7(b) shows the solution for the node voltages and the resulting branch currents.
Note. See Appendix B for more details about matrix inversion.
2.4.1.2 Node Analysis of Circuits Having Voltage Sources
It is not possible to write a KCL equation for a node connected directly to a voltage source(s), because the current through a voltage source cannot be expressed in terms of node voltages. There are two ways of handling circuits having a voltage source(s), the supernode method and the source transformation method. The supernode method regards the group of nonreference nodes connected only via the voltage source(s) as a supernode and applies KCL individually to the nonreference nodes as well as the supernode(s). One thing to note is that if a voltage source is directly connected to the reference node, the node voltage at the other end node of the voltage source is already known so that KCL does not have to be applied to that node, to say nothing of the reference node. Consequently, the number of node
Figure 2.7 Circuit with independent current sources and its solution for Example 2.2
equations or unknown node voltage variables decreases by the number of voltage sources that are connected to the reference node. The source transformation method associates every voltage source with an element in series, transforms it into the equivalent current source by using the technique introduced in Section 1.5.2 as needed, and applies the formula (2.10) to set up the node equation(s). Another thing to note is that if a portion of the circuit involving the interested voltages/currents has been modified via source transformation, it should be restored back to the original connection after solving the node equation(s).
It is desired to have not only the capability of applying the supernode method and the source transformation method but also an eye for judging which one takes less trouble. Note that the supernode method requires individual application of KCL to each (nonreference) node, while it does not need a source transformation(s).
(Example 2.3) A Circuit Having Independent Voltage Sources with Series Elements
Find the current, iR3, through the 3S resistor in the circuit of Figure 2.8(a) by using the node analysis.
Since each of the voltage sources is connected to the reference node at one end, KCL does not have to be applied to the other end nodes 1 and 3 of the voltage sources. Thus KCL is applied to node 2 only, yielding the node equation, which is solved as
iR1þ iR2þ iR3¼ 4; 1ðv2 v1Þ þ 2ðv2 0Þ þ 3ðv2 v3Þ ¼ 4v1¼5; v!3¼16v2¼ 6; v2¼ 1 V ðE2:3:1Þ Technically, it is possible to proceed hypothetically to write the node equations in matrix–vector form as if there were no voltage source:
This naturally yields the same result:
5 þ 6v2þ 3 ¼ 4; 6v2¼ 6; v2¼ 1 V ðE2:3:3Þ
Now, the source transformation method will be tried. Since both of the two voltage sources have series resistors, they can easily be transformed into the equivalent current sources as depicted in Figure 2.8(b). Since the resulting circuit has two nodes, we apply KCL only to one node, i.e. the top node 2 or just use the formula (2.10), to write the node equation and solve it to get the same result as iR1þ iR2þ iR3¼ 5 þ 4 3; 1v2þ 2v2þ 3v2¼ 6; 6v2¼ 6; v2¼ 1 V ðE2:3:4Þ Whichever method is used, we find the current, iR3, through the 3S resistor in the circuit of Figure 2.8(a), based on the node voltages:
iR3¼ 3 ðv2 v3Þ ¼ 3½1 ð1Þ ¼ 6 A ðE2:3:5Þ
Figure 2.8 Circuit with current and voltage sources for Example 2.3
2.4 Node Analysis 43
(Question) If you find the current, iR3, through the 3S resistor in the circuit of Figure 2.8(b), do you get the same result? If not, find out what is wrong with it.
(Example 2.4) A Circuit Having an Independent Voltage Source with No Series Element
Find the voltage, vR3, across the 3S resistor in the circuit of Figure 2.9(a) by using the node analysis.
Since the circuit has a voltage source between two nonreference nodes 1 and 2, the supernode method and the source transformation method can be tried.
(a) Supernode Method
Referring to Figure 2.9(a), we regard the group of two nodes 1 and 2 connected via the 1 V voltage source as a supernode, say 12, and apply KCL to supernode 12 and node 3 to write the node equations as
G1v1þ G2ðv1 v3Þ þ G4v2¼ v1þ 2ðv1 v3Þ þ 4 v2¼ 14
G2ðv3 v1Þ þ G3v3¼ 2ðv3 v1Þ þ 3 v3¼ 14 ðE2:4:1Þ Since there are only two equations with three unknowns, one more equation is needed, which will be provided by the matchmaker, the 1V voltage source, between the two nodes 1 and 2:
v1 v2¼ 1; v2¼ v1þ 1 ðE2:4:2Þ
We substitute this equation into Equation (E2.4.1) and solve it as 7 2
2 5
v1
v3
¼ 18 14
; v1
v3
¼ 7 2
2 5
1
18 14
¼ 1
7 5 ð2Þð2Þ 5 2 2 7
18 14
¼ 2 2
ðE2:4:3Þ
Figure 2.9 Circuit and its equivalents for Example 2.4
Figure 2.9(d) shows the voltage and current solutions as a whole, where vR3¼ v3¼ 2 V.
(b) Source Transformation Method
To transform the 1V voltage source into the equivalent current source, it should first be connected with an element in series. For this purpose, it is duplicated to make two copies in parallel, as shown in Figures 2.9(b1) or (c1). Then we associate each of them with the 14 A current source and the 4S resistor, respectively, to transform the latter into a 1 V 4 S ¼ 4 A current source in parallel with the 4 S resistor as depicted in Figure 2.9(b2), while the former voltage source connected in series with the 14 A current source has disappeared having lost its role (see Section 1.5.1.5). Then we use the formula (2.10) to set up the node equation for the equivalent circuit in Figure 2.9(b2) and solve it to get the same result as Equation (E2.4.3):
1þ 2 þ 4 2
Alternatively, we can associate each of the duplicated 1V voltage sources with the 1S resistor and the 2S resistor, respectively, to transform the latter into a 1 V 2S ¼ 2A current source in parallel with the 2S resistor and the former into a 1 V 1 S ¼ 1A current source in parallel with the 1S resistor, as depicted in Figure 2.9(c2). Then we use the formula (2.10) to set up the node equation for the equivalent circuit in Figure 2.9(c2) and solve it as
1þ 2 þ 4 2