Hello again,
Ok, let me see if i can get you started.
They call it Nodal Analysis because we pay strict attention to each node in the circuit. A node is a junction of two or more components, so this circuit has 6 nodes. We can call one node ground however and declare it to be at 0v, so that reduces this circuit to 5 nodes plus ground. The next thing we would do is assign nodal voltages to each node of the circuit, calling them v1, v2, v3,..., etc., up to the number of nodes. That gives us 5 voltage nodes and the voltage is assumed to be referenced to 0v (the ground node). Note that we will call IO2 the ground node, so it is assumed to be equal to 0v.
What we do next is interesting, as we more or less assume that we already know all of those voltages v1 through v5, and begin calculating the current through each element, 'knowing' those voltages. Of course the current through a resistor is simply I=E/R, we can use that to calculate the currents with each E being the difference in voltage between the two nodes that the component connects between. This means each E comes out to for example v2-v1, which is simply the subtraction of two node voltages.
We also assign currents I1 though however many branches we have. For example, we would call the current though R1 to be I1 and that current would be equal to:
I1=(v1-v2)/R1
Notice that the current arrow would follow conventional current flow and pointing from left to right and that is why we subtract v1-v2 and not v2-v1 above.
We can call the current though R3 to be I3, and that current would be:
I3=(v2-v3)/R3
and following that plan we can calculate the expressions for all the currents through all of the resistors.
Once we have all these expressions, we follow another rule that says that the current into any node is equal to the current leaving that node, or stated differently, the algebraic sum of currents entering a node is zero, so we start to sum currents for each node. For example, the sum of currents for node 2 (junction of R1 and R3) is:
I1-I3-I6=0
Note that we subtracted I3 and I6 because we are assuming that the top of each associated resistor is positive with respect to the bottom.
We continue in this manner until we have done every node, and what we end up with is a set of simultaneous equations in all of these unknown currents, but since we 'know' what each current is we substitute each current with what we calculated previously using the node voltages. What we end up with in the end is a set of simultaneous equations in 5 unknown voltages v1, v2, etc., but we can immediately knock that down to 4 unknowns because we know that v1 is equal to 24 volts. Thus, we end up with 4 equations in 4 unknowns and we then use the mathematics of simultaneous equations to solve for all 4 voltages. This gives us the voltage of every node in the circuit referenced to ground, so now we can go back and really calculate all of the currents by using the sum of currents into the nodes that we used previously, but this time we get the actual value of each current rather than an algebraic expression.