Question 1 of 6: Mesh analysis of a four-mesh DC network

Question 1: Mesh analysis of a four-mesh DC network [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. A DC network with four meshes. Taking the bottom rail as reference and all four mesh currents clockwise:

Find. The four clockwise mesh currents and the output voltage $V_o$ across the right-hand $4\,\Omega$ resistor.

[Figure not reproduced: Figure 1 — four-mesh DC network, redrawn. All mesh currents are taken clockwise; the $5\,\text{A}$ source sits in the branch common to $I_2$ and $I_3$, so those two meshes form a supermesh. See the official exam paper.]

Approach. Apply KVL to meshes $I_1$ and $I_4$; because the $5\,\text{A}$ source is shared between $I_2$ and $I_3$ it fixes their difference and the pair is handled as a supermesh, giving four equations for four currents.

1. Mesh $I_1$ (KVL, clockwise). The $3\,\text{V}$ source aids the loop; the two $4\,\Omega$ arms are shared with $I_4$ (top) and $I_2$ (right):$$4(I_1-I_4)+4(I_1-I_2)-3=0\;\Rightarrow\;\boxed{\,8I_1-4I_2-4I_4=3\,}$$
2. Mesh $I_4$ (top loop). Self-resistance $4+2+4=10\,\Omega$; the $8\,\text{V}$ rise is crossed from $+$ to $-$ as the loop returns along the middle rail:$$4I_4+2(I_4-I_2)+8+4(I_4-I_1)=0\;\Rightarrow\;-4I_1-2I_2+10I_4=-8$$
3. Supermesh around $I_2$ and $I_3$. Traversing the outer boundary (skipping the shared current source):$$4(I_2-I_1)-8+2(I_2-I_4)+4I_3=0\;\Rightarrow\;-4I_1+6I_2+4I_3-2I_4=8$$
4. Current-source constraint. The $5\,\text{A}$ source drives current up out of the $I_2$/$I_3$ branch:$$\boxed{\,I_3-I_2=5\,}$$
5. Solve the $4\times4$ system. Substituting and eliminating gives$$I_1=-1.75\text{ A},\quad I_2=-2.29\text{ A},\quad I_3=2.71\text{ A},\quad I_4=-1.96\text{ A}.$$ The negative signs simply mean those loop currents physically circulate counter-clockwise.
6. Output voltage. $V_o$ is the drop across the right-hand $4\,\Omega$ resistor, which carries $I_3$:$$V_o=4\,I_3=4(2.708)=\boxed{\,10.83\text{ V}\,}$$ An independent nodal check (reference = bottom rail, $V_A=3$, supernode $V_C=V_B+8$) gives node D at $65/6=10.83\text{ V}$, confirming $V_o$.