AC mesh analysis with current sources: 22-Elec-A1 Circuits, December 2013, Q3

Question 3: AC mesh analysis with current sources [10 + 5 + 5]

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 2×2 mesh grid with four clockwise mesh currents $I_1$ (bottom-left), $I_2$ (bottom-right), $I_3$ (top-left), $I_4$ (top-right):

Branch	Element
Left, upper	$-j5\ \Omega$ (capacitor)
Left, lower	$15\angle0^{\circ}$ V source
Centre vertical, upper	$5\angle10^{\circ}$ A source (↑)
Centre horizontal, left	$4\ \Omega$ (carries $V_o$)
Centre horizontal, right	$j10\ \Omega$ (inductor)
Centre vertical, lower	$-j4\ \Omega$ (capacitor)
Right, upper	$5\ \Omega$
Right, lower	$2\angle12^{\circ}$ A source (↑)

Find. The mesh equations, $I_1\ldots I_4$, and $V_o=4(I_3-I_1)$ (+ terminal on the right of the 4 Ω).

[Figure not reproduced: Figure 3 — Redrawn 2×2 mesh grid. Two current sources sit on shared/outer branches, so they fix current relations rather than needing KVL; $V_o$ is measured across the central 4 Ω. See the official exam paper.]

Approach. Each current source pins a mesh-current relation. The $2\angle12^{\circ}$ A source is on the outer right branch (mesh $I_2$ only), so it fixes $I_2$ outright; the $5\angle10^{\circ}$ A source is shared by $I_3,I_4$, giving a constraint plus one supermesh KVL. A KVL around mesh $I_1$ closes the system.

Current-source constraints. The outer $2\angle12^{\circ}$ A source (arrow up) opposes clockwise $I_2$, so $$I_2=-2\angle12^{\circ}=1.956\angle{-}168^{\circ}\ \text{A}.$$ The shared $5\angle10^{\circ}$ A source (arrow up) carries $I_4-I_3$ upward: $$I_4-I_3=5\angle10^{\circ}\ \text{A}.$$
KVL, mesh $I_1$. Around the bottom-left loop (4 Ω shared with $I_3$, $-j4$ shared with $I_2$, then the 15 V rise): $$4(I_1-I_3)-j4(I_1-I_2)=15\angle0^{\circ}.$$
Supermesh KVL, $I_3\cup I_4$. Around the outer boundary of the two top meshes (skipping the shared source), with the $-j5$, $4\,\Omega$, $5\,\Omega$ and $j10$ drops: $$5I_4+j10(I_4-I_2)+4(I_3-I_1)-j5\,I_3=0.$$
Reduce to two unknowns. Substituting $I_4=I_3+5\angle10^{\circ}$ and the known $I_2$ gives the linear pair $$\begin{aligned}(4-j4)I_1-4I_3&=15-j4I_2,\\-4I_1+(9+j5)I_3&=-(5+j10)\,(5\angle10^{\circ})+j10I_2.\end{aligned}$$
Solve. Solving the $2\times2$ complex system: $$I_1=\boxed{3.69\angle{-}55.9^{\circ}\ \text{A}},\quad I_3=\boxed{8.30\angle{-}121.4^{\circ}\ \text{A}},\quad I_4=I_3+5\angle10^{\circ}=\boxed{6.25\angle{-}84.5^{\circ}\ \text{A}}.$$
Voltage across the 4 Ω. With the + terminal on the right, $$V_o=4(I_3-I_1)=\boxed{30.2\angle{-}147.8^{\circ}\ \text{V}}.$$

Quantity	Result
$I_1$	$3.69\angle{-}55.9^{\circ}$ A
$I_2$	$2\angle{-}168^{\circ}$ A (fixed by source)
$I_3$	$8.30\angle{-}121.4^{\circ}$ A
$I_4$	$6.25\angle{-}84.5^{\circ}$ A
$V_o$ across 4 Ω	$\boxed{30.2\angle{-}147.8^{\circ}\ \text{V}}$

Concept & Theory

Topic. Mesh (loop) analysis of an AC network containing current sources — constraint equations and the supermesh.

Key relations. A current source on the outer branch of one mesh fixes that mesh current; a source shared by two meshes fixes their difference and merges them into a supermesh, whose KVL uses only the passive drops on the outer boundary.

Why this works. The voltage across a current source is unknown, so it cannot enter a normal KVL loop. Replacing the two individual KVLs by (constraint + supermesh KVL) supplies exactly the same number of independent equations without that unknown voltage. Phasor impedances $j\omega L$ and $1/j\omega C$ let KVL be written algebraically in the complex plane.

Common pitfall. Getting the sign of a source–mesh relation wrong (compare the source arrow to the clockwise sense), or forgetting that the shared 4 Ω and $j10$ each carry a difference of two mesh currents.

Source. Sadiku §3.5–3.6 (mesh analysis, supermesh) and §10 (sinusoidal steady-state / phasors).

Question 3 of 6: AC mesh analysis with current sources

Question 3: AC mesh analysis with current sources [10 + 5 + 5]