KCL - calculating current when voltage source is present
Since you are "familiar with nodal analysis," I'll approach your question from that perspective. Let's draw this up two ways:







simulate this circuit &ndash; Schematic created using CircuitLab

(We'll assume that you can specify values for the three resistors and the two voltage sources.)



In the left schematic, the values of \$V_A=V_1\$ and \$V_B=V_2\$ are known, since you have them already referenced from the ground reference, directly. So the solution for \$V_X\$ using KCL (nodal analysis here) is:

$$\frac{V_X}{R_1}+\frac{V_X}{R_2}+\frac{V_X}{R_3}=\frac{V_A}{R_1}+\frac{V_B}{R_2}+\frac{0\:\text{V}}{R_3}$$

Note the symmetry here. The voltage at \$V_X\$ drives a current outward through each of the three resistors. But this must equal the current inward arriving through each of those three resistors from the other sources. But you are probably more familiar with this:

$$\frac{V_X-V_A}{R_1}+\frac{V_X-V_B}{R_2}+\frac{V_X-0\:\text{V}}{R_3}=0\:\text{A}$$

Note that there is no difference. These two above equations are identical. And they solve out as:

$$V_X=R_3\:\frac{R_1\:V_2 + R_2\:V_1}{R_1\:R_2 + R_1\:R_3 + R_2\:R_3}$$

But you could also have solved this as a resulting central voltage, given voltage sources with series impedances. The general equation is:

$$V_X=\frac{\sum_{i=1}^N \left[V_i\cdot \prod_{j=1,j\ne i}^N R_j\right]}{\sum_{i=1}^N\left[\prod_{j=1,j\ne i}^N R_j\right]}$$

In this case, that would turn into:

$$V_X=\frac{V_1\:R_2\:R_3 + V_2\:R_1\:R_3 + 0\:\text{V}\:R_1\:R_2 }{R_1\:R_2 + R_1\:R_3 + R_2\:R_3}=R_3\:\frac{R_1\:V_2 + R_2\:V_1}{R_1\:R_2 + R_1\:R_3 + R_2\:R_3}$$

Same result.

Mostly, I just want to point out that there are lots of ways to approach the left side schematic. Keep your mind flexible. Look at things in different ways and find the equivalent approaches.



To the right side schematic now. Nodal analysis would make you write two equations now, instead of one. \$V_B=V_2\$, so we don't need to worry about node \$V_B\$. But we do now have to worry about a new unknown, \$V_A\$. I have to choose a direction for the current in \$V_1\$ and I'm going to say that it is flowing from (-) to (+), or that it is pointing inward relative to \$V_X\$. So:

$$\begin{align*}\frac{V_X}{R_2}+\frac{V_X}{R_3}&amp;=I_{V_1}+\frac{V_B}{R_2}+\frac{0\:\text{V}}{R_3}\\\\\frac{V_A}{R_1}+I_{V_1}&amp;=\frac{0\:\text{V}}{R_1}\end{align*}$$

But \$V_A=V_X-V_1\$, so:

$$\begin{align*}\frac{V_X}{R_2}+\frac{V_X}{R_3}&amp;=I_{V_1}+\frac{V_B}{R_2}+\frac{0\:\text{V}}{R_3}\\\\\frac{V_X-V_1}{R_1}+I_{V_1}&amp;=\frac{0\:\text{V}}{R_1}\end{align*}$$

Solving these simultaneously (or solving the 2nd equation for \$I_{V_1}\$ and substituting that into the 1st equation) will provide the same answer as before:

$$V_X=R_3\:\frac{R_1\:V_2 + R_2\:V_1}{R_1\:R_2 + R_1\:R_3 + R_2\:R_3}$$

The difference here is that you had to create a variable for the current (unknown, just yet) of \$V_1\$ and you had to recognize that there is a very direct relationship then between \$V_X\$ and \$V_A\$ that is created by the fact that there is a voltage source holding them apart by a fixed value.

It's an extra step. But the answer is the same.

The actual circuit is the same complexity, regardless. But being allowed to call a node "ground" and assign it the special value of \$0\:\text{V}\$ provides one way to help simplify the equations a little. And sometimes, if you move things around (without materially changing the circuit topology) you can simplify them a little more. But regardless of how you approach it, if the question being asked isn't changed and if the circuit remains equivalent, the same results should be obtained.