Encyclopedia of Electrical Engineering

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.
**Fig. 1: **Node-voltage
definition and notation.
In nodal analysis, we are interested in finding the node voltages. To define a set of node
voltages, we first select a reference node.
#### Node Voltage

Fig.no.1 shows the notation used to define node-voltage variables. In this figure
the reference node is indicated by the ground symbol and the node voltages are identified by a voltage symbol next to all the other nodes. This notation means that the
positive reference mark for the node voltage is located at the node in question,
whereas the negative mark is at the reference node. Obviously, any circuit with $N$
nodes involves $N-1$ node voltages since one node, the reference node, is known.
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 :

**Example 1: **Calculate the node voltages in the circuit shown in Fig. 2(a).
**Fig. 2: **For Example 1(a) original
circuit, (b) circuit for analysis.
**Solution: **

Consider Fig. 2(b), where the circuit in Fig. 2(a) has been prepared for nodal analysis. 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. The reference node is selected, and the node voltages $v_1$ and $v_2$ are now to be determined. At node 1, applying KCL and Ohm's law gives $$i_1 = i_2 + i_3$$ $$5 = {v_1 - v_2 \over 4}+ {v_1 - 0 \over 2}$$ Multiplying each term in the last equation by 4, we obtain $$20 = v_1 - v_2 + 2v_1$$ or $$3v_1 - v_2 = 20 \tag{1}$$ At node 2, we do the same thing and get $$i_2 + i_4 = i_1 + i_5$$ $${v_1 - v_2 \over 4} + 10 = 5 + {v_2 - 0 \over 6}$$ Multiplying each term by 12 results in $$3v_1 - 3v_2 + 120 = 60 + 2v_2$$ or $$-3v_1 + 5v_2 = 60 \tag{2}$$ Using the elimination technique, we add Eqs. (1) and (2). $$4v_2 = 80 \Rightarrow v_2 = 20 V$$ Substituting v2 = 20 in Eq. (1) gives $$3v_1 - 20 = 20 \Rightarrow 3v_1 = 40$$ $$v_1 = 13.33 V$$
##### Nodal Analysis Related Questions

- Select a node as the reference node. Assign voltages $v_1, v_2, . . . , v_{n-1}$ to the remaining $n - 1$ nodes. The voltages are referenced with respect to the reference node.
- Apply KCL to each of the $n - 1$ nonreference nodes. Use Ohm's
law to express the branch currents 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. We can express this principle as $$i = {v_{\text{higher}} - v_{\text{lower}} \over R}$$ - Solve the resulting simultaneous equations to obtain the unknown node voltages.

Consider Fig. 2(b), where the circuit in Fig. 2(a) has been prepared for nodal analysis. 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. The reference node is selected, and the node voltages $v_1$ and $v_2$ are now to be determined. At node 1, applying KCL and Ohm's law gives $$i_1 = i_2 + i_3$$ $$5 = {v_1 - v_2 \over 4}+ {v_1 - 0 \over 2}$$ Multiplying each term in the last equation by 4, we obtain $$20 = v_1 - v_2 + 2v_1$$ or $$3v_1 - v_2 = 20 \tag{1}$$ At node 2, we do the same thing and get $$i_2 + i_4 = i_1 + i_5$$ $${v_1 - v_2 \over 4} + 10 = 5 + {v_2 - 0 \over 6}$$ Multiplying each term by 12 results in $$3v_1 - 3v_2 + 120 = 60 + 2v_2$$ or $$-3v_1 + 5v_2 = 60 \tag{2}$$ Using the elimination technique, we add Eqs. (1) and (2). $$4v_2 = 80 \Rightarrow v_2 = 20 V$$ Substituting v2 = 20 in Eq. (1) gives $$3v_1 - 20 = 20 \Rightarrow 3v_1 = 40$$ $$v_1 = 13.33 V$$