Encyclopedia of Electrical Engineering

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.
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,
$$i_1 + i_4 = i_2 + i_3$$
or
$${v_1 - v_2 \over 2} + {v_1 - v_3 \over 4} = {v_2 - 0 \over 8}+ {v_3 - 0 \over 6}$$
$$12v_1 - 12v_2+ 6v_1 - 6v_3 = 3v_2 + 4v_3$$
$$18v_1 - 15v_2 - 10v_3 = 0 \tag{1}$$
**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.
**Solution: **

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 $$2 = {v_1 - 0 \over 2} + {v_2 - 0 \over 4}+ 7$$ $$8 = 2v_1 + v_2 + 28$$ or $$v_2 = -20 - 2v_1 \tag{1.1}$$ 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 $$v_2 = v_1 + 2 = -20 - 2v_1$$ or $$3v_1 = -22 \Rightarrow v_1 = -7.333 V$$ and $$v_2 = v_1 +2 = -5.333 V$$ Note that the 10Ω resistor does not make any difference because it is connected across the supernode.

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 $$2 = {v_1 - 0 \over 2} + {v_2 - 0 \over 4}+ 7$$ $$8 = 2v_1 + v_2 + 28$$ or $$v_2 = -20 - 2v_1 \tag{1.1}$$ 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 $$v_2 = v_1 + 2 = -20 - 2v_1$$ or $$3v_1 = -22 \Rightarrow v_1 = -7.333 V$$ and $$v_2 = v_1 +2 = -5.333 V$$ Note that the 10Ω resistor does not make any difference because it is connected across the supernode.

Nodal Analysis
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