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22-Elec-A1 Circuits · May 2013

Question 2 of 6: Node-voltage analysis

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper: National Examinations — 07-Elec-A1 Circuits, May 2013. Closed-book, 3-hour paper; a Laplace-transform table is supplied. Six questions (Q1 has two parts); any five of equal value constitute a complete paper. Full worked solutions to all six questions are given below.

Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — nodal/mesh analysis, Thevenin and maximum-power transfer, first-order transients, AC power and power-factor correction, Laplace-domain circuit analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of the s-domain and initial-condition models.

every network shown here was reconstructed element-by-element from the original drawing. Reference-node and source-polarity choices are stated with each solution so a reader can reproduce every sign.

Question 2: Node-voltage analysis [8 + 4 + 4]

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.

ElementConnection
6 Ω$V_1 \leftrightarrow V_3$ (top branch)
4 Ω$V_1 \leftrightarrow V_2$
10 Ω$V_2 \leftrightarrow V_3$
2 Ω$V_1 \leftrightarrow$ reference
$I_a=5\ \text{A}$injected into $V_2$
$15\ \Omega + V_s=10\ \text{V}$$V_3 \to$ reference (series)

Find. The three node voltages $V_1$, $V_2$, $V_3$.

6 Ω 4 Ω 10 Ω V₁ V₂ V₃ 2 Ω Iₐ = 5 A 15 Ω + − Vₛ = 10 V
Figure 3 — Three-node network for Q2. The bottom rail is the reference; the 15 Ω and 10 V source form a grounded series branch from node 3.
Check. The 10 V source is taken with its + terminal toward the 15 Ω (upper) so the branch current out of node 3 is $(V_3-10)/15$. This is the polarity drawn in the figure; reversing it only changes the sign of the 10 V term.

Approach. Write one KCL equation per non-reference node, expressing every branch current through Ohm's law; the 5 A source and the grounded series (15 Ω, 10 V) branch enter as known injections.

  1. KCL at each node. Currents leaving each node sum to the injected current: $$\text{(1)}\ \frac{V_1}{2}+\frac{V_1-V_2}{4}+\frac{V_1-V_3}{6}=0,$$ $$\text{(2)}\ \frac{V_2-V_1}{4}+\frac{V_2-V_3}{10}=5,$$ $$\text{(3)}\ \frac{V_3-V_1}{6}+\frac{V_3-V_2}{10}+\frac{V_3-10}{15}=0.$$
  2. Clear fractions. Multiplying through gives the linear system $$11V_1-3V_2-2V_3=0,\quad -5V_1+7V_2-2V_3=100,\quad -5V_1-3V_2+10V_3=20.$$
  3. Solve the system. Gaussian elimination (or Cramer's rule) yields $$\boxed{V_1=9.43\ \text{V},\quad V_2=25.09\ \text{V},\quad V_3=14.25\ \text{V}.}$$
  4. Check. Substituting back, node 2 gives $\dfrac{25.09-9.43}{4}+\dfrac{25.09-14.25}{10}=3.91+1.08=5.0\ \text{A}$ — matching the 5 A source, confirming the solution.
NodeVoltage
$V_1$$9.43\ \text{V}$
$V_2$$25.09\ \text{V}$
$V_3$$14.25\ \text{V}$