Question 2 of 6: Mesh analysis with a current source in the shared branch
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — 07-Elec-A1 Circuits, December 2014. Closed-book, 3-hour paper; a Laplace-transform table and Y–Δ formulas are supplied in the appendix. Six questions of equal value; any five constitute a complete paper. Full worked solutions to all six questions are given below so the set can be used for study regardless of which five a candidate chose.
Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — series/parallel reduction and the Wheatstone bridge, mesh/supermesh and nodal analysis, first-order transients, AC steady-state, Thévenin’s theorem and maximum-power transfer, and 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 source models.
each network below is redrawn element-by-element from the original drawing. The Wheatstone bridge of Q1 is shown in the equivalent rectangular form (electrically identical to the diamond on the paper). Every reference node, mesh direction and source polarity used is stated with its solution so a reader can reproduce every sign. Phasor answers are quoted in magnitude∡angle form with angles in degrees; the source of Q5 is rms.
Question 2: Mesh analysis with a current source in the shared branch [8 + 8 + 4]
Given. Two clockwise meshes sharing a central branch that contains a $5\,\Omega$ resistor in series with a $10\text{ A}$ current source (arrow up):
Element
Value
Location
Source
$30$ V (+ up)
left branch
$R$
$10\,\Omega$
top of mesh 1
$R$
$4\,\Omega$
top of mesh 2
$R$
$2\,\Omega$
right branch
Source
$20$ V ($-$ up, + down)
right branch
Shared branch
$5\,\Omega$ + $10$ A (up)
between the meshes
Find. The mesh equations; $I_1$ and $I_2$; and $P_{5\Omega}$.
[Figure not reproduced: Figure-2 (redrawn). Both mesh currents are clockwise. The $10\text{ A}$ source sits in the branch shared by the two meshes, so it fixes $I_2-I_1$ and the two meshes are combined into one supermesh for the KVL equation. See the official exam paper.]
Approach. A current source in the shared branch fixes the difference of the two mesh currents, so it supplies the constraint equation; the second equation is one KVL taken around the supermesh (the outer loop, skipping the current-source branch).
Current-source constraint. The source drives $10\text{ A}$ upward in the shared branch. With both mesh currents clockwise, the upward branch current is $I_2-I_1$, so $$I_2-I_1=10\ \text{A}.$$
Supermesh KVL (i). Traverse the outer loop clockwise, summing voltage drops (a source is a rise from $-$ to $+$). Passing up through the $30$ V ($-\!\to\!+$, a rise), across the $10\,\Omega$ and $4\,\Omega$ (drops $10I_1,\,4I_2$), down through the $2\,\Omega$ (drop $2I_2$), then through the $20$ V whose polarity is $-$ top/$+$ bottom (also a rise): $$-30+10I_1+4I_2+2I_2-20=0\;\Rightarrow\;10I_1+6I_2=50.$$
Solve the pair (ii). Substitute $I_2=I_1+10$: $$10I_1+6(I_1+10)=50\;\Rightarrow\;16I_1=-10,$$ so $$I_1=\boxed{-0.625\ \text{A}},\qquad I_2=I_1+10=\boxed{9.375\ \text{A}}.$$ The negative $I_1$ means mesh 1 actually circulates counter-clockwise.
Independent check (nodal). Referencing the bottom rail with the $30$ V fixing the top-left node at $30$ V, nodal analysis gives $V_{TM}=36.25$ V, $V_{TR}=-1.25$ V, whence the $10\,\Omega$ current $(30-36.25)/10=-0.625$ A$=I_1$ and the $4\,\Omega$ current $(36.25+1.25)/4=9.375$ A$=I_2$ — agreeing exactly.
Power in the $5\,\Omega$ (iii). The $5\,\Omega$ is in series with the $10\text{ A}$ source, so it carries the full source current, $|I_2-I_1|=10\text{ A}$: $$P_{5\Omega}=I^2R=10^2\times5=\boxed{500\ \text{W}}.$$
