Use nodal analysis to find $V_o$ given that $V_s = 5 V$, $I_s = 8 A$, $R_1 = 2 \Omega$, $R_2 = 8 \Omega$, $R_3 = 10 \Omega$, $R_4 = 6 \Omega$, and $R_5 = 3 \Omega$.
The best choice for the reference node is the bottom node, since that node has the most elements connected to it and it has one terminal of the voltage source connected to it.
Use the convention from Hayt, Kemmerly and Durbin that currents enter the node from current sources and leave the node through resistors.

The nodal analysis equation at node 1:
$V_1$ + $V_2$ + $V_3$ + $V_4$ =

The nodal analysis equation at node 2:
$V_1$ + $V_2$ + $V_3$ + $V_4$ =

The nodal analysis equation at node 3:
$V_1$ + $V_2$ + $V_3$ + $V_4$ =

The nodal analysis equation at node 4:
$V_1$ + $V_2$ + $V_3$ + $V_4$ =


$V_1 =$ $V$
$V_2 =$ $V$
$V_3 =$ $V$
$V_4 =$ $V$
$V_o =$ $V$