Mesh analysis with a dependent source: 22-Elec-A1 Circuits, May 2017, Q2

Question 2: Mesh analysis with a dependent source [8 + 8 + 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. Four clockwise mesh currents $I_1,I_2,I_3,I_4$. A $2\text{ A}$ independent source is shared by meshes $1$ and $2$; a dependent source $4I_o$ (arrow up) is shared by meshes $1$ and $3$; $I_o$ is the current in the $1\,\Omega$ resistor (top of mesh $4$), drawn pointing left. Resistors: $5\,\Omega$ (only in mesh $1$), $4\,\Omega$ (mesh $2$), $3\,\Omega$ (mesh $3$), $6\,\Omega$ (shared by meshes $3$ and $4$), $1\,\Omega$ (mesh $4$); a $15\text{ V}$ source (+ up) closes mesh $4$.

Find. the four mesh currents and $P_{6\Omega}$.

[Figure not reproduced: Figure 2 — redrawn, all four mesh currents clockwise. The two current sources are internal branches, so the meshes they join are merged into one super-mesh for the KVL walk. Because $I_1$ links to both $I_2$ (via $2\text{ A}$) and $I_3$ (via $4I_o$), meshes $1$–$2$–$3$ form a . See the official exam paper.]

Approach. Take each current source as a mesh-current constraint, write one super-mesh KVL around the combined loop $1$–$2$–$3$ and one ordinary KVL for mesh $4$, express $I_o$ through $I_4$, then solve the $4\times4$ set; the $6\,\Omega$ carries $I_3-I_4$.

(a) Source constraints. The $2\text{ A}$ source fixes the difference of the meshes on either side of it, and the dependent source fixes the other pair: $$I_2-I_1=2,\qquad I_3-I_1=4I_o.$$ The controlling current is the $1\,\Omega$ current $I_o$; drawn opposite to the clockwise $I_4$, so $$I_o=-I_4.$$
(a) Super-mesh KVL (meshes 1–2–3). Walking the outer boundary clockwise and summing $IR$ drops (each resistor carries its own mesh current; the $6\,\Omega$ on the boundary carries $I_3-I_4$): $$5I_1+4I_2+3I_3+6\,(I_3-I_4)=0.$$
(a) Mesh-4 KVL. Around mesh $4$ (through the $6\,\Omega$, the $1\,\Omega$ and the $15\text{ V}$ rise): $$6\,(I_4-I_3)+1\cdot I_4+15=0.$$ These four boxed relations are the requested mesh equations.
(b) Reduce and solve. Substituting $I_o=-I_4$ gives $I_3-I_1=-4I_4$. With $I_2=I_1+2$, the super-mesh KVL becomes $5I_1+4(I_1+2)+3I_3+6I_3-6I_4=0\Rightarrow 9I_1+9I_3-6I_4=-8$, and mesh 4 gives $-6I_3+7I_4=-15$. Together with $I_3=I_1-4I_4$ the linear solve yields $$\boxed{I_1=-2.87\text{ A},\ \ I_2=-0.87\text{ A},\ \ I_3=1.29\text{ A},\ \ I_4=-1.04\text{ A}.}$$ Hence $I_o=-I_4=1.04\text{ A}$ (consistent: $I_3-I_1=4.16=4I_o$).
(c) Power in the $6\,\Omega$. Its current is $I_3-I_4=1.29-(-1.04)=2.33\text{ A}$, so $$\boxed{P_{6\Omega}=(I_3-I_4)^2\,(6)=(2.33)^2(6)=32.5\text{ W}.}$$

Quantity	Value
$I_1$	$-2.87\text{ A}$
$I_2$	$-0.87\text{ A}$
$I_3$	$1.29\text{ A}$
$I_4$	$-1.04\text{ A}$
Control current $I_o$	$1.04\text{ A}$
Power in $6\,\Omega$	$32.5\text{ W}$

Topic: Mesh (loop) analysis with independent and current-controlled dependent current sources — the super-mesh technique.

Key relations: a current source shared by two meshes gives $I_{\text{near}}-I_{\text{far}}=\pm I_{src}$; KVL $\sum V=0$ around a super-mesh omits the shared source branch; a shared resistor carries the difference of the two mesh currents.

Why this works: A current source has an unknown voltage, so you cannot write KVL through it directly. Instead you use it twice: once as a constraint equation (its value equals the mesh-current difference) and once by routing the KVL loop around it (the super-mesh). Here two sources chain meshes $1$, $2$ and $3$ into a single super-mesh, so only one KVL is written for the trio; the dependent source adds no new unknown because its value is tied to $I_o=-I_4$. Counting: four unknowns, two source constraints, two KVL equations — exactly determined.

Common pitfall: mis-signing $I_o$ (it is drawn opposite the clockwise $I_4$, so $I_o=-I_4$, not $+I_4$), or writing KVL straight through a current source. Negative mesh currents are fine — they just mean the physical current runs counter-clockwise.

Source: Sadiku, Fundamentals of Electric Circuits, §3.5 (mesh analysis) and §3.6 (mesh analysis with current sources / super-mesh).

Question 2 of 6: Mesh analysis with a dependent source

Question 2: Mesh analysis with a dependent source [8 + 8 + 4]