In Chapters 1 & 2, we introduced several tools in circuit analysis:

• Ohm's Law
• Kirchhoff's laws
• Circuit reduction

Circuit reduction, it should be noted, is not fundamentally different from direct application of Ohm's and Kirchhoff's laws - it is simply a convenient re-statement of these laws for specific combinations of circuit elements.

In Chapter 1, we saw that direct application of Ohm's law and Kirchhoff's laws to a specific circuit using the exhaustive method often results in a large number of unknowns - even if the circuit is relatively simple. A correspondingly large number of equations must be solved to determine these unknowns. Circuit reduction allows us, in come cases, to simplify the circuit to reduce the number of unknowns in the system. Unfortunately, not all circuits are reducible and even analysis of circuits that are reducible depends upon the engineer “noticing” certain resistance combinations and combining them appropriately.

In cases where circuit reduction is not feasible, approaches are still available to reduce the total number of unknowns in the system. Nodal analysis and mesh analysis are two of these. Nodal and mesh analysis approaches still rely upon application of Ohm's law and Kirchhoff's laws - we are just applying these laws in a very specific way in order to simplify the analysis of the circuit. One attractive aspect of nodal and mesh analysis is that the approaches are relatively rigorous - we are assured of identifying a reduced set of variables, if we apply the analysis rules correctly. Nodal and mesh analysis are also more general than circuit reduction methods - virtually any circuit can be analyzed using nodal or mesh analysis.

Since nodal and mesh analysis approaches are fairly closely related, section 3.1 introduces the basic ideas and terminology associated with both approaches. Section 3.2 provides details of nodal analysis, and mesh analysis is presented in section 3.3.

• Use nodal analysis techniques to analyze electrical circuits
• 
  Use mesh analysis techniques to analyze electrical circuits

As noted in the introduction, both nodal and mesh analysis involve identification of a “minimum” number of unknowns, which completely describe the circuit behavior. That is, the unknowns themselves may not directly provide the parameter of interest, but any voltage or current in the circuit can be determined from these unknowns. In nodal analysis, the unknowns are the node voltages. In mesh analysis, the unknowns are the mesh currents. We introduce the concept of these unknowns via an example below.

Consider the circuit shown in Fig. 3.1(a). The circuit nodes are labeled in Fig. 3.1(a), for later convenience. The circuit is not readily analyzed by circuit reduction methods. If the exhaustive approach toward applying KCL and KVL is taken, the circuit has 10 unknowns (the voltages and currents of each of the five resistors), as shown in Fig. 3.1(b). Ten circuit equations must be written to solve for the ten unknowns. Nodal analysis and mesh analysis provide approaches for defining a reduced number of unknowns and solving for these unknowns. If desired, any other desired circuit parameters can subsequently be determined from the reduced set of unknowns.

In nodal analysis, the unknowns will be node voltages. Node voltages, in this context, are the independent voltages in the circuit. It will be seen later that the circuit of Fig. 3.1 contains only two independent voltages - the voltages at nodes b and c¹⁾. Only two equations need be written and solved to determine these voltages! Any other circuit parameters can be determined from these two voltages.

In nodal analysis, Kirchhoff's current law is written at each independent voltage node; Ohm's law is used to write the currents in terms of the node voltages in the circuit.

In mesh analysis, the unknowns will be mesh currents. Mesh currents are defined only for planar circuits; planar circuits are circuits which can be drawn in a single plane such that no elements overlap one another. When a circuit is drawn in a single plane, the circuit will be divided into a number of distinct areas; the boundary of each area is a mesh of the circuit. A mesh current is the current flowing around a mesh of the circuit. The circuit of Fig. 3.1 has three meshes:

1. The mesh bounded by $V_s$, node a, and node d
2. The mesh bounded by node a, node c, and node b
3. The mesh bounded by node b, node c, and node d

These three meshes are illustrated schematically in Fig. 3.2. Thus, in a mesh analysis of the circuit of Fig. 3.1, three equations must be solved in three unknowns (the mesh currents). Any other desired circuit parameters can be determined from the mesh currents.

In mesh analysis, Kirchhoff's voltage law is written around each mesh loop; Ohm's law is used to write the voltage in terms of the mesh currents in the circuit. Since KVL is written around closed loops in the circuit, mesh analysis is sometimes known as loop analysis.

• In nodal analysis:
  • Unknowns in the analysis are called the node voltages
  • Node voltages are the voltages at the independent nodes in the circuit
  • Two nodes connected by a voltage source are not independent. The voltage source constrains the voltages at the nodes relative to one another. A node which is not independent is also called dependent.
• In mesh analysis:
  • Unknowns in the analysis are called mesh currents.
  • Mesh currents are defined as flowing through the circuit elements which form the perimeter of the circuit meshes. A mesh is any enclosed, non-overlapping region in the circuit (when the circuit schematic is drawn on a piece of paper).

1. The circuit below has three nodes, A, B, and C. Which two nodes are dependent? Why?

2. Identify meshes in the circuit below.

As noted in section 3.1, in nodal analysis we will define a set of node voltages and use Ohm's law to write Kirchhoff's current law in terms of these voltages. The resulting set of equations can be solved to determine the node voltages; and other circuit parameters (e.g. currents) can be determined from these voltages.

The steps used in nodal analysis are provided below. The steps are illustrated in terms of the circuit of Fig. 3.3.

One node will be arbitrarily as a reference node or datum node. The voltages of all other nodes in the circuit will be defined to be relative to the voltage of this node. Thus, for convenience, it will be assumed that the reference node voltage is zero volts. It should be emphasized that this definition is arbitrary - since voltages are actually potential differences, choosing the reference voltage as zero is primarily a convenience.

For our example circuit, we will choose node d as our reference node and define the voltage at this node to be 0V, as shown in Fig. 3.4.

We now define the voltages at the independent nodes. These voltages will be the unknowns in our circuit equations. In order to define independent nodes:

• “Short-circuit” all voltage sources
• “Open-circuit” all current sources

After removal of the sources, the remaining nodes (with the exception of the reference node) are defined as independent nodes (the nodes which were removed in this process are dependent nodes. The voltages at these nodes are sometimes said to be constrained). Label the voltages at these nodes - they are the unknowns for which we will solve.

For our example circuit of Fig. 1, removal of the voltage source (replacing it with a short circuit) results in nodes remaining only at nodes b and c. This is illustrated in Fig. 3.5.

With the independent voltages defined as in Step 2, replace the sources and define the voltages at the dependent nodes in terms of the independent voltages and the known voltage difference.

For our example, the voltage at node a can be written as a known voltage $V_s$ above the reference voltage, as shown in Fig. 3.6.

Define currents and write Kirchhoff's current law at all independent nodes. Currents for our example are shown in Fig. 3.7 below. The defined currents include the assumed direction of positive current - this defines the sign convention for our currents. To avoid confusion, these currents are defined consistently with those shown in Fig. 3.1(a). The resulting equations are (assuming that currents leaving the node are defined as positive):

Node b:

$$-i_1+i_3+i_4=0 (Eq. 3.1)$$

Node c:

$$-i_2-i_3+i_5=0 (Eq. 3.2)$$

The currents defined in Step 4 can be written in terms of the node voltages defined previously. For example, from Fig. 3.7: $i_1= \frac{V_S-V-b}{R_1}$, $i_3=\frac{V_b-V_c}{R_3}$, and $i_4=\frac{V_b-0}{R_4}$, sop equation (3.1) can be written as:

$$-\frac{V_s-V_b}{R_1} + \frac{V_b-V_c}{R_3} + \frac{V_b-0}{R_4} = 0 $$

So the KCL equation for node b becomes:

$$ \left( \frac{1}{R_1} + \frac{1}{R_3} + \frac{1}{R_4} \right) V_b - \frac{1}{R_3} V_c = \frac{1}{R_1}V_S (Eq. 3.3)$$

Likewise, the KCL equation for node c can be written as:

$$-\frac{1}{R_3} V_b + \left( \frac{1}{R_3} + \frac{1}{R_2} + \frac{1}{R_5} \right) V_c = \frac{1}{R_2} V_S (Eq. 3.4)$$

If the circuit being analyzed contains only independent sources, and the sign convention used in KCL equations is the same as used above (currents leaving nodes are assumed positive), the equations written at each node will have the following form:

• The term multiplying the voltage at that node will be the sum of the conductances connected to that node. For the example above, the term multiplying $V_b$ in the equation for node b is $\frac{1}{R_1} + \frac{1}{R_3} + \frac{1}{R_4}$ while the term multiplying $V_c$ nin the equation for node c is $\frac{1}{R_3} + \frac{1}{R_2} + \frac{1}{R_5}$.
• The term multiplying the voltages adjacent to the node will be the negative of the conductance connecting the two nodes. For the example above, the term multiplying $V_c$ in the equation for node b is $-\frac{1}{R_3}$, and the term multiplying $V_b$ in the equation for node c is $-\frac{1}{R_3}$.

If the circuit contains dependent sources, or a different sign convention is used when writing the KCL equations, the resulting equations will not necessarily have the above form.

Step 5 will always result in N equations in N unknowns, where N is the number of independent nodes identified in Step 2. These equations can be solved for the independent voltages. Any other desired circuit parameters can be determined from these voltages.

The example below illustrates the above approach.

Find the voltage V for the circuit shown below:

Steps 1, 2, and 3:Choosing the reference voltage as shown below, identifying voltages at dependent nodes, and defining voltages $V_A$ and $V_B$ at the independent nodes results in the circuit schematic shown below:

Steps 4 and 5: Writing KCL at nodes A and B and converting currents to voltages using Ohm's law results in the following two equations:

Node A:

$$\frac{V_A - 6}{2} + \frac{V_A - V_B}{2} + \frac{V_A - V_B}{1} = 0 \Rightarrow \left( \frac{1}{2} + \frac{1}{2} + \frac{1}{1} \right) V_A - V_B = 3 \Rightarrow 2V_A - V_B = 3$$

Node B:

$$\frac{V_B-V_A}{1} + \frac{V_B-0}{0.5} + 16 = 0 \Rightarrow \left( \frac{1}{1} + \frac{1}{0.5} \right) V_B - V_A = -16 \Rightarrow 3V_B - V_A = 16 $$

Step 6: Solving the above equations results in $V_A=5V$ and $V_B=7V$. The voltage $V$ is:

$$V=V_A - V_B = -2V$$

Several comments should be made relative to the above example: 1. Steps 4 and 5 (applying KCL at each independent node and using Ohm's law to write these equations in terms of voltages) have been combined into a single step. This approach is fairly common, and can provide a significant savings in time.

2. There may be a perceived inconsistency between the two node equations, in the assumption of positive current direction in the 1Ω resistor. In the equation for node A, the current is apparently assumed to be positive from node A to node B, as shown below:

This leads to the corresponding term in the equation for node A becoming: $\frac{V_A - V_B}{1}$. In the equation for node B, however, the positive current direction appears to be from node B to node A, as shown below:

This definition leads to the corresponding term in the equation for node B becoming: $\frac{V_B - V_A}{1}$/

The above inconsistency in sign is, however, insignificant. Suppose that we had assumed (consistently with the equation for node A) that the direction of positive current for the node B equation is from Node A to B. Then, the corresponding term in the equation for node B would have been: $-\frac{V_A - V_B}{1}$ (note that a negative sign has been applied to this term to accommodate our assumption that currents flowing into nodes are negative). This is equal to $\frac{V_B - V_A}{1}$, which is exactly what our original result was.

3. The current source appears directly in the nodal equations.

When we write nodal equations in these chapters, we will generally assume that any unknown currents are flowing away from the node for which we are writing the equation, regardless of any previous assumptions we have made for the direction of that current. The signs will work out, as long as we are consistent in our sign convention between assumed voltage polarity and current direction and our sign convention relative to positive currents flowing out of nodes.

The sign applied to currents induced by current sources must be consistent with the current direction assigned by the source.

In the previous examples, we identified dependent nodes and determined constrained voltages. Kirchhoff's current law was then only written at independent nodes. Many readers find this somewhat confusing, especially if the dependent voltages are not relative tot he reference voltage. We will thus discuss these steps in more detail here in the context of an example, introducing the concept of a supernode in the process.

Example: for the circuit below, determine the voltage difference, V, across the 2mA source.

Choose reference node (somewhat arbitrarily) as shown below; label the reference node voltage, $V_R$, as zero volts.

Short circuit voltage sources, open circuit current sources as shown below and identify independent nodes/voltages. For our example, this result in only one independent voltage, labeled as $V_A$ below.

The known voltages are written in terms of node voltages identified above. There is some ambiguity in this step. For example, either of the representations below will work equally well - either side of the voltage source can be chosen as the node voltage, and the voltage on the other side of the source written in terms of this node voltage. Make sure, however, that the correct polarity of the voltage source is preserved. In our example, the left side of the source has a potential that is three volts higher than the potential of the right side of the source. This fact is represented correctly by both of the choices below.

It is this step that sometimes causes confusion among readers, particularly when voltage sources are present in the circuit. Conceptually, it is possible to think of two nodes connected by an ideal voltage source as forming a single supernode (some authors use the term generalized node rather than supernode). A node is rigorously defined as having a single, unique voltage. However, although the two nodes connected by a voltage source do not share the same voltage, they are not entirely independent - the two voltages are constrained by one another. This allows us to simplify the analysis somewhat.

For our example, we will arbitrarily choose the circuit to the left above to illustrate this approach. The supernode is chosen to include the voltage source and both nodes to which it is connected, as shown below. We define two currents leaving the supernode, $i_1$ and $i_2$, as shown. KCL, applied at the supernode, results in:

$$-2mA + i_1 + i_2 = 0$$

As before, currents leaving the node are assumed to be positive. This approach allows us to account for the current flowing through the voltage source without ever explicitly solving for it.

For the single KCL equation written above, this results in:

$$-2mA + \frac{V_A - 0}{3k \Omega} + \frac{(V_A - 3) - 0}{6k \Omega} = 0$$

Solution of the equation above results in $V_A = 5V$. Thus, the voltage difference across the current source is $V=5V$.

The use of supernodes can be convenient, but is not a necessity. An alternate approach, for those who do not wish to identify supernodes, is to restrain separate nodes on either side of the voltage source and then write a constraint equation relating these voltages. Thus, in cases where the reader does not recognize a supernode, the analysis can proceed correctly. We now revisit the previous example, but use constraint equations rather than the previous supernode technique.

In this approach, Steps 2 and 3 (identification of independent nodes) are not necessary. One simply writes Kirchhoff's current law at all nodes and then writes constraints equations for the voltage sources. A disadvantage of this approach is that currents through voltage sources must be accounted for explicitly; this result in a greater number of unknowns (and equations to be solved) than the supernode technique.

Example (revisited): for the circuit below, determine the voltage difference, V, across the 2mA source.

Choice of a reference voltage proceeds as previously. However, now we will not concern ourselves too much with identification of independent nodes. Instead, we will just make sure we account for voltages and currents everywhere in the circuit. For our circuit, this results in the node voltages and currents shown below. Notice that we have now identified two unknown voltages ($V_A$ and $V_B$) and three unknown currents, one of which ($i_3$) is the current through the voltage source.

Now we write KCL at each of the identified nodes, making sure to account for the current through the voltage source. This results in the following equations (assuming currents leaving the node are positive):

Node A: $-2mA+i_1+i_3=0$

Node B: $-i_3+i_2=0$

Using Ohm's law to convert the currents $i_1$ and $i_2$ to voltages results in:

Node A: $-2mA+ \frac{V_A - 0}{3k \Omega} + i_3 = 0$

Node B: $-i_3 + \frac{V_B - 0}{6k \Omega} = 0$

Notice that we cannot, by inspection, determine anything about the current $i_3$ from the voltages; the voltage-current relationship for an ideal source is not known.

The two equations above have three unknowns - we cannot solve for the node voltages from them without a third equation. This third equation is the constraint equation due to the presence of the voltage source. For our circuit, the voltage source causes a direct relationship between $V_A$ and $V_B$:

$$V_B=V_A-3$$

These thee equations (the two KCL equations, written in terms of the node voltages and the constraint equation) constitute three equations in three unknowns. Solving these for the node voltage $V_A$ results in $V_A=5V$, so the voltage across the current source is $V=5V$.

The example below uses the concept of a supernode to write the governing KCL equations. In the example below, Steps 1, 2, and 3 have been condensed into a single process, as have Steps 4 and 5. It is suggested that the reader re-do the example below using constraint equations. Note again that the current sources appear directly in the KCL equations.

For the circuit below, find the power generated or absorbed by the 2V source and the power generated or absorbed by the 2A source.

Steps 1, 2, and 3: we choose our reference node (arbitrarily) as shown below. Shorting voltage sources and open-circuiting current sources identifies three independent node voltages (labeled below as $V_A$, $V_B$, and $V_C$) and one dependent node, with voltage labeled below as $V_A-2$.

Steps 4 and 5: Writing KCL at nodes A, B, and C and converting the currents to voltages using Ohm's law results in the equations below. Note that we have (essentially) assumed that all unknown currents at a node are flowing out of the node, consistent with our node 2 for example 1 above.

Node A:

$$2A+ \frac{V_A - 0V}{4\Omega} + \frac{(V_A - 2V) - V_B}{4\Omega} + \frac{(V_A - 2V} - V_C}{8\Omega} = 0 \Rightarrow 5V_A - 2V_B - V_C = -10$$

Node B:

$$\frac{V_B - 0V}{3\Omega} + \frac{(V_A - 2V)}{4\Omega} + 3A = 0 \Rightarrow 7V_B - 3V_A = 42$$

Node C:

$$\frac{V_C - (V_A - 2)}{8\Omega} - 3A = 0 \Rightarrow V_C - V_A = 22$$

Step 6: Solving the above results in $V_A = 0V$, $V_B = -6V$, and $V_C=22V$. Thus, the voltage difference across the 2A source is zero volts, and the 2A source delivers no power. KCL at node A indicates that the current through the 2V source is 2A, and the 2V source generates 4W.

In the presence of dependent sources, nodal analysis proceeds approximately as outlined above. The main difference is the presence of additional equations describing the dependent source. As before, we will discuss the treatment of dependent sources in the context of examples.

Write the nodal equations for the circuit below. The dependent source is a voltage controlled voltage source. $I_S$ is an independent current source.

As always, the choice of reference node is arbitrary. To determine independent voltages, dependent voltage sources are short-circuited in the same way as independent voltage sources. Thus, the circuit below has two independent nodes; the dependent voltage source and the nodes on either side of it form a supernode. The reference voltage, independent voltages, supernode, and resulting dependent voltage are shown below.

We now, as previously, write KCL for each independent node, taking into account the dependent voltage resulting from the presence of the supernode: