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22-Elec-A1 Circuits · May 2013

Question 3 of 6: Mesh-current 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 3: Mesh-current analysis [8 + 3 + 3 + 3 + 3]

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. Two 10 V sources stack on the left. Mesh 1 (top-left) is bounded by the upper 10 V, the 1 Ω, the 4 Ω and the 2 Ω. Mesh 2 (bottom-left) is bounded by the lower 10 V, the 2 Ω, the 5 Ω and the 3 Ω. Mesh 3 (right) is bounded by the 4 Ω, the 6 Ω and the 5 Ω. The 2 Ω is shared by meshes 1 and 2.

Find. The three mesh currents and the current $I$ through the 2 Ω resistor.

1 Ω + − 10 V + − 10 V 2 Ω I 4 Ω 5 Ω 3 Ω 6 Ω I₁ I₂ I₃
Figure 4 — Three-mesh network for Q3. All mesh currents $I_1,I_2,I_3$ are taken clockwise; $I$ is the current in the shared 2 Ω branch (arrow pointing from the centre node toward the source side).

Approach. Assign all three mesh currents clockwise and apply KVL around each loop; shared resistors carry the difference of the two adjacent mesh currents.

  1. Mesh 1 (KVL). Traversing clockwise through the 1 Ω, 4 Ω (shared with $I_3$) and 2 Ω (shared with $I_2$), driven by the upper 10 V: $$1\,I_1+4(I_1-I_3)+2(I_1-I_2)=10\ \Rightarrow\ 7I_1-2I_2-4I_3=10.$$
  2. Mesh 2 (KVL). Through the 2 Ω, 5 Ω (shared with $I_3$) and 3 Ω, driven by the lower 10 V: $$2(I_2-I_1)+5(I_2-I_3)+3I_2=10\ \Rightarrow\ -2I_1+10I_2-5I_3=10.$$
  3. Mesh 3 (KVL). Source-free loop through the 6 Ω, 5 Ω and 4 Ω: $$6I_3+5(I_3-I_2)+4(I_3-I_1)=0\ \Rightarrow\ -4I_1-5I_2+15I_3=0.$$
  4. Solve the 3×3 system. $$\boxed{I_1=3.04\ \text{A},\quad I_2=2.42\ \text{A},\quad I_3=1.62\ \text{A}.}$$
  5. Current through the 2 Ω. It is shared by meshes 1 and 2; in the arrow's direction $$I=I_1-I_2=3.04-2.42=\boxed{0.63\ \text{A}}.$$
QuantityResult
Mesh current $I_1$$3.04\ \text{A}$
Mesh current $I_2$$2.42\ \text{A}$
Mesh current $I_3$$1.62\ \text{A}$
Current in 2 Ω$\boxed{I=0.63\ \text{A}}$