simulate this circuit – Schematic created using CircuitLab
See the schematic above. This is a simplified schematic from a more complex circuit divided using nodal analysis, but my question is only directed toward this part.
Could you explain me clearly why the equilibrum equation for node Va (nodal analysis) is (note the "s" is used as Laplace transform, i.e. derivative or integral) :
$$
V_{a} \cdot (\frac{1}{R_{4}} + \frac{1}{R_{8}} + \frac{1}{L_{1}s}) – V_{b} \cdot \frac{1}{L_{1}s} – V_{c} \cdot \frac{1}{R_{4}} = \frac{v_{s}}{R_{8}}
$$
The part confusing me is related to \$ R_{8} \$, I guess because it's placed after the voltage source, between it and the ground. I would guess \$v_{s} = V_{a} \$ right? That would mean technically from the above equation (taken from the solutions book) that \$ R_{8} \$ would not appear in the (simplified) equation as \$ V_{a}/R_{8} \$ and \$ v_{s}/R_{8} \$ would are on both sides of the equation?
What I was doing intuitively is:
$$
\frac{v_{s} – 0}{R_{8}} + \frac{V_{a} – V_{c}}{R_{4}} + \frac{V_{a} – V_{b}}{L_{1}s} = 0
$$
By the way the current directions \$(i_4, i_8, i_1)\$ are given as-is in the book's exercise's schematic. I only tried to follow them in the above equation, seems like I'm all wrong…
How would you approach this using nodal analysis?
Side note, if \$ R_{8} \$ had been between \$V_{a}\$ and \$v_{s}\$ instead, I would have gotten the same answer as the book. Yet, since \$ R_8 \$ is next to the ground, I don't think I could throw \$ \frac{V_{a} – v_{s}}{R_{8}} \$, no?
Thanks
Best Answer
Label the node where R8 connect to the vs source as
Vd.Now \$V_d = V_a - v_s\$.
So you can calculate \$I_8=\frac{V_d}{R_8}=\frac{V_a-v_s}{R_8}\$.
If you rearrange terms in your first equation, you'll find this is exactly what is accounted for in the node equation.
In your version you seem to have assumed \$I_8 = \frac{v_s-0}{R_8}\$. This doesn't make any sense because \$v_s\$ isn't applied across R8.