DC nodal analysis with a supernode: 22-Elec-A1 Circuits, May 2014, Q1

Question 1: DC nodal analysis with a supernode [8 + 6 + 6]

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.

Given. A DC network referenced to the bottom rail (ground):

Element	Value	Connection
Source	20 V	+ at node $V_3$, − to ground
$R$ (top)	4 Ω	$V_3 \leftrightarrow V_2$
$R$	6 Ω	$V_2 \leftrightarrow$ ground
$R$	9 Ω	$V_1 \leftrightarrow$ ground
$R$	5 Ω	$V_2 \leftrightarrow V_1$ (branch 1)
Source	10 V	$V_2 \leftrightarrow V_1$ (branch 2), + at $V_1$; current $I_2$ flows $V_2\to V_1$

Find. The node equations, $V_1$, $V_2$, $V_3$, and the branch current $I_2$.

Figure 1 — DC network. The 20 V source fixes $V_3$; the 10 V source sits directly between nodes 1 and 2 (in parallel with the 5 Ω), so those two nodes form a supernode.

Approach. The 20 V source ties $V_3$ to ground, so $V_3$ is known; the 10 V source bridges two ungrounded nodes, so $V_1$ and $V_2$ are handled as a supernode (one KCL enclosing both, plus the source constraint).

Source node. The 20 V source connects $V_3$ to ground, so directly $$V_3=\boxed{20\ \text{V}}.$$
Supernode constraint. The 10 V source has its + terminal at node 1: $$V_1-V_2=10\ \text{V}.$$
Supernode KCL (nodes 1&2). Sum the currents leaving the enclosed pair through the external branches — the 9 Ω at node 1, the 6 Ω at node 2, and the 4 Ω toward $V_3=20$ V (the 5 Ω and the 10 V source are internal to the supernode): $$\frac{V_1}{9}+\frac{V_2}{6}+\frac{V_2-20}{4}=0.$$
Solve. Substitute $V_1=V_2+10$ and multiply by the LCD 36: $$4(V_2+10)+6V_2+9(V_2-20)=0\;\Rightarrow\;19V_2-140=0,$$ giving $$V_2=\frac{140}{19}=\boxed{7.37\ \text{V}},\qquad V_1=V_2+10=\frac{330}{19}=\boxed{17.37\ \text{V}}.$$
Branch current $I_2$. Apply KCL at node 1 alone: the current entering through the 10 V branch ($I_2$) must leave through the 9 Ω (to ground) and the 5 Ω (to node 2): $$I_2=\frac{V_1}{9}+\frac{V_1-V_2}{5}=\frac{17.37}{9}+\frac{10}{5}=1.93+2.00=\boxed{3.93\ \text{A}}.$$

Quantity	Result
$V_3$	$20\ \text{V}$ (fixed by source)
$V_2$	$7.37\ \text{V}\ (=140/19)$
$V_1$	$17.37\ \text{V}\ (=330/19)$
Branch current $I_2$ ($V_2\to V_1$)	$\boxed{3.93\ \text{A}}$

Concept & Theory

Topic. Nodal analysis with a grounded source (known node) and a floating voltage source (supernode).

Key relations. KCL $\sum I_{out}=0$; Ohm’s law $I=\Delta V/R$; supernode = one KCL enclosing both terminals of a floating source plus the constraint $V_1-V_2=E$.

Why this works. The current through an ideal voltage source is unknown, so it cannot appear in a single node’s KCL. Enclosing both terminals in one Gaussian surface makes that internal current cancel; the missing equation is restored by the fixed potential difference the source imposes. Note the 5 Ω carries a fixed $10/5=2$ A because it sits directly across the 10 V source.

Common pitfall. Trying to write separate KCL at nodes 1 and 2 (the source current is unknown), or dropping the sign of the constraint — the + terminal defines which node is higher.

Source. Sadiku, Fundamentals of Electric Circuits §3.3–3.4 (nodal analysis, supernodes).

Question 1 of 6: DC nodal analysis with a supernode

Question 1: DC nodal analysis with a supernode [8 + 6 + 6]