Nodal Analysis with Voltage Sources

Up to this point we have analyzed circuits containing only resistors and independent current sources. Applying KCL in such circuits is simplified because the sum of currents at a node only involves the output of current sources or resistor currents expressed in terms of the node voltages.
Adding voltage sources to circuits modifies node analysis procedures because the current through a voltage source is not directly related to the voltage across it. While initially it may appear that voltage sources complicate the situation, they actually simplify node analysis by reducing the number of equations required.
Consider the following two possibilities in node analysis with voltage sources.
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. 1], for example,
$$v_1 = 10 V$$
Thus our analysis is somewhat simplified by this knowledge of the voltage at this node.
Fig. 1: A circuit with a supernode.
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.

What is a supernode?

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. 1], nodes 2 and 3 form a supernode. 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 knowing 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 supernode in [Fig. 1],
$$i_1 + i_4 = i_2 + i_3$$
$${v_1 - v_2 \over 2} + {v_1 - v_3 \over 4} = {v_2 - 0 \over 8}+ {v_3 - 0 \over 6}$$
$$12v_1 - 12v_2+ 6v_1 - 6v_3 = 3v_2 + 4v_3$$
$$18v_1 - 15v_2 - 10v_3 = 0 \tag{1}$$
Fig. 2: Applying KVL to a supernode.
To apply Kirchhoff's voltage law to the supernode in [Fig. 1], we redraw the circuit as shown in [Fig. 2]. Going around the loop in the clockwise direction gives
$$-v_2 + 5 + v_3 = 0$$
$$ \Rightarrow v_2 - v_3 = 5 \tag{2}$$
Example 1: For the circuit shown in [Fig. 3], find the node voltages.
Fig. 3: Applying: (a) KCL to the supernode, (b) KVL to the loop.
The supernode contains the 2V source, nodes 1 and 2, and the $10Ω$ resistor. Applying KCL to the supernode as shown in [Fig. 4(a)] gives
$$2 = i_1 + i_2 + 7$$
Expressing $i_1$ and $i_2$ in terms of the node voltages
$$2 = {v_1 - 0 \over 2} + {v_2 - 0 \over 4}+ 7$$
$$8 = 2v_1 + v_2 + 28$$
$$v_2 = -20 - 2v_1 \tag{1.1}$$
To get the relationship between $v_1$ and $v_2$, we apply KVL to the circuit in[ Fig. 3(b)]. Going around the loop, we obtain
$$-v1 - 2 + v_2 = 0$$
$$v_2 = v_1 + 2 \tag{1.2}$$
From Eqs. (1.1) and (1.2), we write
$$v_2 = v_1 + 2 = -20 - 2v_1$$
$$3v_1 = -22 \Rightarrow v_1 = -7.333 V$$
$$v_2 = v_1 +2 = -5.333 V$$
Note that the $10Ω$ resistor does not make any difference because it is connected across the supernode.

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