Node‑voltage analysis of a four‑node network: 22-Elec-A1 Circuits, December 2019, Q2

Question 2: Node‑voltage analysis of a four‑node network [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 not reproduced: Figure‑2 — redrawn four‑node network. A 15 V source ties $V_1$ to the reference; three 5 A sources inject at $V_1,V_3,V_4$. Branch resistors as labelled; $V_x$ is defined across the 5 Ω link between $V_3$ and $V_4$. See the official exam paper.]

Given. Four non‑reference nodes over a common ground. A 15 V dc source connects the reference to $V_1$ (so $V_1$ is fixed at 15 V), and independent 5 A sources feed $V_1$, $V_3$ and $V_4$. Resistive branches: $V_1\!-\!V_2=5\,\Omega$, $V_1\!-\!V_3=6\,\Omega$, $V_2\!-\!V_3=2\,\Omega$, $V_3\!-\!V_4=5\,\Omega$; shunts $V_2\!\to\!0=5\,\Omega$, $V_3\!\to\!0=6\,\Omega$, and at $V_4$ a $2\,\Omega\|6\,\Omega=1.5\,\Omega$ leg.

Branch	Value	Branch	Value
$V_1\!-\!V_2$	$5\,\Omega$	$V_2\!\to\!0$	$5\,\Omega$
$V_1\!-\!V_3$	$6\,\Omega$	$V_3\!\to\!0$	$6\,\Omega$
$V_2\!-\!V_3$	$2\,\Omega$	$V_4\!\to\!0$	$2\\|6=1.5\,\Omega$
$V_3\!-\!V_4$	$5\,\Omega$	Sources	$15$ V @ $V_1$; $5$ A @ $V_1,V_3,V_4$

Find. The four node‑voltage equations and the solved voltages $V_1\!-\!V_4$ (and $V_x=V_3-V_4$).

Check — source placement. In the figure the 15 V source reads as tying $V_1$ directly to ground, which fixes $V_1=15$ V and makes the 5 A at $V_1$ flow through the source. The three KCL equations for $V_2,V_3,V_4$ below are unaffected by that reading; if your copy shows the 15 V floating between two nodes, replace the $V_1$ line with the corresponding supernode constraint.

Approach. Take the bottom rail as reference. The grounded 15 V source sets $V_1$ outright; write KCL (currents leaving = injected) at $V_2,V_3,V_4$ and solve the 3×3 system.

Node‑voltage equations. With the reference at the bottom rail:$$V_1=15\ \text{V}\quad(\text{set by the grounded source}),$$$$\text{(}V_2\text{)}:\ \frac{V_2-V_1}{5}+\frac{V_2-V_3}{2}+\frac{V_2}{5}=0,$$$$\text{(}V_3\text{)}:\ \frac{V_3-V_1}{6}+\frac{V_3-V_2}{2}+\frac{V_3-V_4}{5}+\frac{V_3}{6}=5,$$$$\text{(}V_4\text{)}:\ \frac{V_4-V_3}{5}+\frac{V_4}{1.5}=5.$$
Clear fractions. Substituting $V_1=15$ and multiplying through gives the linear system $$9V_2-5V_3=30,\qquad -15V_2+31V_3-6V_4=225,\qquad -3V_3+13V_4=75.$$
Solve. Back‑substitution (or matrix inversion) yields $$\boxed{V_2=11.42\ \text{V},\quad V_3=14.55\ \text{V},\quad V_4=9.13\ \text{V}.}$$
Auxiliary voltage. $V_x=V_3-V_4=14.55-9.13=\boxed{5.42\ \text{V}.}$ All three KCL residuals evaluate to zero at these voltages.

Node	Voltage
$V_1$	$15.00\,\text{V (fixed)}$
$V_2$	$11.42\,\text{V}$
$V_3$	$14.55\,\text{V}$
$V_4$	$9.13\,\text{V}$
$V_x=V_3-V_4$	$5.42\,\text{V}$

Concept & Theory

Topic: nodal analysis with independent voltage and current sources.

Key relations: KCL in node‑voltage form $\sum_k \dfrac{V_i-V_k}{R_{ik}} = I_{\text{inj},i}$; a voltage source from a node to the reference fixes that node’s voltage directly.

Why this works: each independent current source appears as a known injection on the right‑hand side, and each resistor contributes conductance terms symmetric between its two nodes. A ground‑referenced ideal voltage source removes one unknown (here $V_1$), shrinking the system to a 3×3 solve.

Common pitfall: sign of the current‑source injection (into the node is $+$ on the right side) and forgetting to combine the parallel $2\,\Omega\|6\,\Omega$ at $V_4$ into $1.5\,\Omega$ before writing KCL.

Source: Hayt, Engineering Circuit Analysis, 9th ed., §4.1–4.4.

Question 2 of 6: Node‑voltage analysis of a four‑node network

Question 2: Node‑voltage analysis of a four‑node network [10 + 10]