super-node


We now consider how voltage sources affect nodal analysis. We use the circuit in Fig. 3.7 for illustration. Consider the following two possibilities.

Figure 3.7 A circuit with a super-node

CASE 1: If a voltage source is connected between the reference node and a nonreference node, we simply set the voltage at the nonreference node equal to the voltage of the voltage source. In Fig. 3.7, for example,

Thus, our analysis is somewhat simplified by this knowledge of the voltage at this node.

CASE 2: If the voltage source (dependent or independent) is connected between two nonreference nodes, the two nonreference nodes form a generalized node or supernode; we apply both KCL and KVL to determine the node voltages.

A supernode is formed by enclosing a (dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it.

In Fig. 3.7, nodes 2 and 3 form a supernode. (W e could have more than two nodes forming a single supernode. F or example, see the circuit in Fig. 3.14.) We analyze a circuit with supernodes using the same three steps mentioned in the previous section except that the supernodes are treated differently. Why? Because an essential component of nodal analysis is applying KCL, which requires kno wing the current through each element. There is no way of knowing the current through a voltage source in advance. However, KCL must be satisfied at a supernode like any other node. Hence, at the super node in Fig. 3.7,

To apply Kirchhoff’s voltage law to the supernode in Fig. 3.7, we redraw the circuit as shown in Fig. 3.8.
Figure 3.8 Applying KVL to a supernode.

Going around the loop in the clockwise direction gives :

From Eqs. (3.10), (3.11b), and (3.12), we obtain the node voltages. Note the following properties of a supernode:
  1. The voltage source inside the supernode provides a constraint equation needed to solve for the node voltages.
  2. A supernode has no voltage of its own
  3. A supernode requires the application of both KCL and KVL

Example

For the circuit shown in Figure, find the node voltages.

Solution


The supernode contains the 2-V source, nodes 1 and 2, and the 10- Ω resistor. Applying KCL to the supernode as shown in Figure : 



Expressing i1 and i2 in terms of the node voltages :


To get the relationship between v1 and v2, we apply KVL to the circuit in Figure : 


Going around the loop, we obtain :


From Eqs. (3.3.1) and (3.3.2), we write :


and v2 = v1 + 2 = −5.333 V. Note that the 10-Ω resistor does not make any difference because it is connected across the supernode.