Nodal analysis with a supernode: 22-Elec-A1 Circuits, December 2013, Q1

Question 1: Nodal analysis with a supernode [12 + 8]

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 (bottom rail as reference/ground):

Element	Value	Connection
Source $E_1$	10 V	+ terminal to node S, − to ground
$R_a$	5 Ω	node S → node 1
$R_b$	2 Ω	node 1 → ground
Source $E_2$	5 V	node 1 → node 2, + at node 2
$R_c$	3 Ω	node 2 → ground
$R_d$	4 Ω	node 2 → ground

Find. $V_1$, $V_2$, and the powers $P_{5\Omega}$, $P_{4\Omega}$.

Figure 1 — DC network for Q1. Node S is held at +10 V by the source; the 5 V source sits directly between nodes 1 and 2, so those two nodes form a supernode.

Approach. The 5 V source connects two non-reference nodes, so nodes 1 and 2 are treated as a supernode: one KCL equation around the pair plus the source constraint $V_2-V_1=5$.

Source constraint. The 5 V source has its + terminal at node 2, so $$V_2-V_1=5\ \text{V}.$$
Supernode KCL. Sum all currents leaving the {1,2} supernode (node S is fixed at 10 V): $$\frac{V_1-10}{5}+\frac{V_1}{2}+\frac{V_2}{3}+\frac{V_2}{4}=0.$$
Substitute and clear fractions. Put $V_2=V_1+5$ and multiply by the LCD 60: $$12(V_1-10)+30V_1+20(V_1+5)+15(V_1+5)=0\;\Rightarrow\;77V_1+55=0.$$
Solve the node voltages. $$V_1=-\tfrac{55}{77}=\boxed{-0.714\ \text{V}},\qquad V_2=V_1+5=\tfrac{30}{7}=\boxed{4.286\ \text{V}}.$$
Power in the 5 Ω. Its voltage is $V_S-V_1=10-(-0.714)=10.714\ \text{V}$, so $$P_{5\Omega}=\frac{(10.714)^2}{5}=\boxed{22.96\ \text{W}}.$$
Power in the 4 Ω. It sits across node 2: $$P_{4\Omega}=\frac{V_2^{\,2}}{4}=\frac{(4.286)^2}{4}=\boxed{4.59\ \text{W}}.$$

Quantity	Result
Node 1 voltage	$V_1=-0.714\ \text{V}\ (=-5/7)$
Node 2 voltage	$V_2=4.286\ \text{V}\ (=30/7)$
Power in 5 Ω	$\boxed{P_{5\Omega}=22.96\ \text{W}}$
Power in 4 Ω	$\boxed{P_{4\Omega}=4.59\ \text{W}}$

Concept & Theory

Topic. Nodal analysis with a voltage source bridging two ungrounded nodes (supernode).

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

Why this works. The branch current through an ideal voltage source is unknown, so it cannot appear in an individual node equation. Enclosing both terminals in a Gaussian surface makes that internal current cancel; the missing equation is supplied by the fixed potential difference the source imposes. A negative $V_1$ simply means node 1 sits below ground — the 5 V source lifts node 2 back above it.

Common pitfall. Writing separate KCL at node 1 and node 2 while ignoring the source current, or dropping the sign on 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: Nodal analysis with a supernode

Question 1: Nodal analysis with a supernode [12 + 8]