# Simple question about nodal analysis

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

• FYI, what you're doing here is essentially constructing a supernode. This is the standard way to deal with voltage sources when doing nodal analysis. – The Photon Nov 4 '15 at 2:22

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.

• Now you said it this way... kinda I feel you stated the obvious :0 That's interesting to see that if we inverted the order of vs and R8 it would be the same expression (Va - vs)/R8, and that's what confused me. Yet, since R8 is at potential 0 (next to the ground), would have it been okay to simply "remove" it from the schematic i.e. I would guess it won't appear in the equations right, as (if I stand correct) no voltage goes through it? – Yannick Nov 4 '15 at 1:33
• No, if you removed R8 (or reduced its value to 0), you'd just have $V_a=v_s$. It would change the behavior of the circuit. – The Photon Nov 4 '15 at 2:05
• But isn't $V_{a}$ at the same potential than $v_{s}$ on the schematic? I would have said so since nothing "separates" $v_{s}$ and $V_{a}$ (no impedance between), and besides, $v_{s}$ is right next to the ground. – Yannick Nov 4 '15 at 2:35
• No, $V_a$ is the potential difference between node Va and ground. $v_s$ is the potential difference between Va and Vd (as I defined it in my answer). – The Photon Nov 4 '15 at 2:36
• A voltage source creates a potential difference between the two nodes it is connected to. If you want it to force one node to have a particular voltage relative to ground, you have to connect the source's other terminal to ground, not to any other node. – The Photon Nov 4 '15 at 2:38