Node-voltage analysis with a floating source (supernode): 22-Elec-A1 Circuits, Undated paper, Q2

Question 2: Node-voltage analysis with a floating source (supernode) [10 + 10]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Figure 2 — three non-reference nodes. $8\,\Omega$ across $V_1$–$V_3$; $3\,\Omega$ across $V_1$–$V_2$; a $5\,\Omega$+$20\text{ V}$ branch from $V_1$ to ground; $6\,\Omega$ from $V_2$; $9\,\Omega$ from $V_3$; and a $10\text{ V}$ source directly between $V_2$ and $V_3$ ($+$ toward $V_3$).

Given. Reference = bottom rail. The $10\text{ V}$ source sits between two non-reference nodes $V_2$ and $V_3$ (a floating source) with its $+$ terminal at $V_3$, so $V_2$ and $V_3$ form a supernode with constraint $V_3-V_2=10$. The $V_1$ branch to ground is a $5\,\Omega$ in series with a $20\text{ V}$ source ($+$ toward $V_1$), so its current leaving $V_1$ is $(V_1-20)/5$.

Find. $V_1,\,V_2,\,V_3$.

Approach. Write KCL at $V_1$ and one KCL around the $V_2$–$V_3$ supernode, close the system with the source constraint, and solve the $3\times3$ set.

Supernode constraint. The ideal $10\text{ V}$ source fixes $$V_3-V_2=10.$$
KCL at $V_1$ (sum of currents leaving $=0$): $$\frac{V_1-V_3}{8}+\frac{V_1-V_2}{3}+\frac{V_1-20}{5}=0.$$
KCL around the supernode $(V_2,V_3)$ — add every current leaving the two nodes to ground and to $V_1$: $$\frac{V_2-V_1}{3}+\frac{V_2}{6}+\frac{V_3-V_1}{8}+\frac{V_3}{9}=0.$$
Substitute $V_3=V_2+10$ and collect. The two equations reduce to$$0.6583\,V_1-0.4583\,V_2=5.25,\qquad -0.4583\,V_1+0.7361\,V_2=-2.361.$$
Solve. The linear pair gives $\boxed{V_1=10.13\text{ V}}$, $\boxed{V_2=3.10\text{ V}}$, and hence $V_3=V_2+10=\boxed{13.10\text{ V}}$. Back-substitution makes both KCL residuals zero, confirming the solution.

Node	Voltage
$V_1$	$10.13\text{ V}$
$V_2$	$3.10\text{ V}$
$V_3$	$13.10\text{ V}$

Concept & Theory

Topic: nodal analysis with a source between two non-reference nodes (the supernode technique).

Key relations: KCL $\sum i_{\text{leaving}}=0$; a resistor current $(V_a-V_b)/R$; supernode = the two enclosed nodes share one KCL equation plus the source voltage constraint.

Why this works: an ideal voltage source between two nodes fixes their difference but not the branch current through it, so its current cannot be written in terms of node voltages. Enclosing both nodes in a supernode lets that unknown current cancel (it enters one node and leaves the other), and the source equation restores the lost degree of freedom.

Common pitfall: sign of the series-source branch at $V_1$. With the source’s $+$ toward $V_1$, the mid-branch behaves like a $20\text{ V}$ node behind $5\,\Omega$, giving $(V_1-20)/5$; dropping the $20$ (treating it as a plain $5\,\Omega$ to ground) is the usual error. Source: Sadiku, 7th ed., §3.3 (supernodes).

Question 2 of 6: Node-voltage analysis with a floating source (supernode)

Question 2: Node-voltage analysis with a floating source (supernode) [10 + 10]