Question 2 of 6: Mesh equations and the 20 V source current

Question 2: Mesh equations and the 20 V source current [12 + 8]

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 $I_1\ldots I_4$. A $3\,\Omega$ sits on the top of mesh 1; a $4\text{ A}$ independent source (arrow to the right) is shared by meshes 1 and 2; a $5\,\Omega$ is the left leg of mesh 2; a dependent source $2I$ (arrow down) is shared by meshes 2 and 3; a $2\,\Omega$ tops mesh 3; a $4\,\Omega$ is shared by meshes 3 and 4; a $5\,\Omega$ tops mesh 4; and the $20\text{ V}$ source ($+$ up) is the right leg of mesh 4, carrying the current $I$ (arrow up).

Find. (a) the mesh equations; (b) $I$, the upward current in the $20\text{ V}$ branch.

[Figure not reproduced: Figure 2 — redrawn. Two current sources ($4\text{ A}$ and the controlled $2I$) each straddle a mesh pair, so meshes 1–2–3 merge into one supermesh; mesh 4 is ordinary. The control current $I$ is the branch current of the $20\text{ V}$ source. See the official exam paper.]

Approach. Both current sources sit between meshes, so write their branch constraints, merge meshes 1–2–3 into a supermesh for one KVL, add an ordinary KVL for mesh 4, and close the system with the control law $I=-I_4$ (the source arrow $I$ opposes the clockwise $I_4$).

1. Control-variable link. The $20\text{ V}$ branch is the right leg of mesh 4; clockwise $I_4$ runs down it while the labelled $I$ runs up, so $$I=-I_4.$$
2. Current-source constraints (the two “a” equations from the sources). The $4\text{ A}$ source (rightward) sets the mesh-1/mesh-2 difference, and the $2I$ source (downward) sets the mesh-2/mesh-3 difference: $$I_2-I_1=4,\qquad I_2-I_3=2I=-2I_4.$$
3. Supermesh KVL around meshes 1–2–3. Travelling the outer loop (through the $3\,\Omega$ of $I_1$, the $2\,\Omega$ and shared $4\,\Omega$ of $I_3$, and the $5\,\Omega$ of $I_2$) and summing $IR$ drops: $$3I_1+5I_2+6I_3-4I_4=0,$$ where the $6I_3$ is $2\,\Omega+4\,\Omega$ carried by $I_3$ and $-4I_4$ is the mesh-4 share of the $4\,\Omega$.
4. Ordinary KVL for mesh 4. Around mesh 4 (top $5\,\Omega$, the $20\text{ V}$ rise/drop, and the shared $4\,\Omega$): $$5I_4+20+4(I_4-I_3)=0\;\Rightarrow\;9I_4-4I_3=-20.$$ These four boxed relations are the requested mesh equations.
5. Eliminate. Substitute $I_1=I_2-4$ and $I_3=I_2+2I_4$ into the supermesh KVL: $$3(I_2-4)+5I_2+6(I_2+2I_4)-4I_4=0\;\Rightarrow\;7I_2+4I_4=6.$$ From mesh 4, $-4(I_2+2I_4)+9I_4=-20\Rightarrow I_4=4I_2-20$.
6. Solve. $7I_2+4(4I_2-20)=6\Rightarrow23I_2=86\Rightarrow I_2=3.739\text{ A}$, so $I_4=4(3.739)-20=-5.043\text{ A}$. Then $$I=-I_4=\boxed{5.04\text{ A}}.$$ For completeness $I_1=-0.261\text{ A}$, $I_3=-6.348\text{ A}$; check $I_2-I_3=3.739+6.348=10.09=2I$ ✓.