Nodal Analysis



Nodal analysis provides a general procedure for analyzing circuits using node voltages as the circuit variables. Choosing node voltages instead of element voltages as circuit variables is convenient and reduces the number of equations one must solve simultaneously.
To simplify matters, we shall assume in this section that circuits do not contain voltage sources. Circuits that contain voltage sources will be analyzed in the next section.
In nodal analysis, we are interested in finding the node voltages. Given a circuit with n nodes without voltage sources, the nodal analysis of the circuit involves taking the following three steps.

Steps to Determine Node Voltages:



  1. Select a node as the reference node. Assign voltages v1, v2,. . . ,vn-1 to the remaining n − 1 nodes. The voltages are referenced with respect to the reference node.

  2. Apply KCL to each of the n − 1 nonreference nodes. Use Ohm’s law to express the branch currents in terms of node voltages

  3. Solve the resulting simultaneous equations to obtain the unknown node voltages.

We shall now explain and apply these three steps.
The first step in nodal analysis is selecting a node as the reference or datum node. The reference node is commonly called the ground since it is assumed to have zero potential. A reference node is indicated b y any of the three symbols in Fig. 3.1. The type of ground in Fig. 3.1(c) is called a chassis ground and is used in de vices where the case, enclosure, or chassis acts as a reference point for all circuits. When the potential of the earth is used as reference, we use the earth ground in Fig. 3.1(a) or (b). We shall always use the symbol in Fig. 3.1(b).

Figure 3.1 Common symbols for indicating a reference node, (a) common ground, (b) ground, (c) chassis ground.

Once we have selected a reference node, we assign voltage designations to nonreference nodes. Consider , for example, the circuit in Fig. 3.2(a). Node 0 is the reference node ( v = 0), while nodes 1 and 2 are assigned voltages v1 and v2, respectively. Keep in mind that the node voltages are defined with respect to the reference node. As illustrated in Fig. 3.2(a), each node voltage is the voltage rise from the reference node to the corresponding nonreference node or simply the voltage of that node with respect to the reference node

Figure 3.2 Typical circuit for nodal analysis

As the second step, we apply KCL to each nonreference node in the circuit. To avoid putting too much information on the same circuit, the circuit in Fig. 3.2(a) is redrawn in Fig. 3.2(b), where we no w add i1, i2, and i3 as the currents through resistors R1, R2, and R3, respectively. At node 1, applying KCL gives

At node 2,

We now apply Ohm’s law to express the unknown currents i1, i2, and i3 in terms of node voltages. The key idea to bear in mind is that, since resistance is a passive element, by the passive sign convention, current must always flow from a higher potential to a lower potential.

Current flows from a higher potential to a lower potential in a resistor

We can express this principle as :




Substituting Eq. (3.4) in Eqs. (3.1) and (3.2) results, respectively, in


In terms of the conductances, Eqs. (3.5) and (3.6) become



Example :


Calculate the node voltages in the circuit shown in Figure :


Solution:




Notice how the currents are selected for the application of KCL. Except for the branches with current sources, the labeling of the currents is arbitrary but consistent. (By consistent, we mean that if, for example, we assume that i2 enters the 4-Ω resistor from the left-hand side, i2 must leave the resistor from the right-hand side.) The reference node is selected, and the node voltages v1 and v2 are now to be determined. At node 1, applying KCL and Ohm’s law gives


Multiplying each term in the last equation by 4, we obtain : 




At node 2, we do the same thing and get :


Multiplying each term by 12 results in : 


Now we have two simultaneous Eqs. (3.1.1) and (3.1.2). We can solve the equations using any method and obtain the values of v1 and v2.

Using the elimination technique, we add Eqs. (3.1.1) and (3.1.2)

Substituting v2 = 20 in Eq. (3.1.1) gives :