Electrical – Node analysis in which branch contains Voltage Source

Enough time has passed, so I'll write up something for this now.

Just a couple of notes that I'll be discussing shortly, but want to bring up front right now:

• Whenever you see a current source in series with a voltage source or a resistor (and nothing else attached to the node between the two), then the current source is the only thing that matters. A current source has \$\infty\$ impedance and drives the current through either a voltage source or a resistor. In the case of the voltage source, it doesn't change the voltage source but instead determines the current through that voltage source. A voltage source and determined current through it pretty much means the voltage source is meaningless. In the case of a resistor, all the current source does is turn that resistor into a voltage drop (difference) and that's almost the same thing as a voltage source with a determined current. (The only difference being that resistors are always positive Ohms, while a voltage source and determined current through it could be negative Ohms.) You can always simplify such a series pair by removing the voltage source or resistor and replacing them with a wire.
• You are allowed to designate exactly one node as "ground" or "zero volts." Nothing special happens when you do that -- it's entirely in your mind, really -- but you get the advantages of simplifying some of the equations you develop; or making it easier to think about a good starting point for mesh analysis; or allowing yourself a convenient way to remove some distracting wiring in a schematic to help make it easier to read and understand. Choose that node well and it can really help in those ways.
• Re-drawing a schematic can greatly improve its readability to a human (not so much to a computer, though.) So it is often a good idea to redraw a provided schematic. If you don't know how better to do that, the only answer to it is to get some practice trying. Eventually, you will find your skills improving and creating readable schematics happening faster and faster. If interested, have a look here for some useful rules.

Let's start by selecting a convenient ground reference (the bottom wire of your schematic) and re-draw the schematic:

^{simulate this circuit – Schematic created using CircuitLab}

I've color-coded each box to make them stand out more distinctly, but I've also included a name for each box below to help those who may not see color differences as well as I do.

First thing to note is that I've simplified the diagram, somewhat, by removing all the busing of wire around for your bottom wire in your schematic. So long as you keep in mind that these ground connections are still connected to each other, the advantage here is that there's some distracting lines removed now and it's a little bit easier to "think about."

Looking at BOX 1, the reason you can eliminate (short across with a wire) \$V_1\$ is that the current won't change no matter what the voltage value happens to be. Whatever the value of \$V_a\$ may happen to be, the only thing \$V_1\$ does is change the voltage drop across \$I_1\$ and the current source \$I_1\$ doesn't care about that. It still provides the same current, regardless of its voltage drop. So \$V_1\$ is completely irrelevant for the purposes of analyzing \$V_a\$ or \$V_b\$. You can lose \$V_1\$ and what happens is that the voltage drop across \$I_1\$ changes by just that much. So what? Just remove it.

Note that you wrote that you are:

experiencing problem while solving the circuit problem ... because there is a voltage source on the branch which is connected to the left node

So, I'm just telling you here and now that it should NOT confuse you. Just get rid of it. However, in the interest of showing you how it doesn't matter I'll keep it in place for now. You probably need more convincing, not just hand-waving about it. So let's keep it in BOX 1 for now.

(The same arguments would go for BOX 3. But again, let's leave things as they are.)

BOX 2 and BOX 4 are each voltage sources in series with a resistor. These could technically be replaced with a Norton equivalent, with current sources and a parallel resistor. At times, that approach may help simplify the analysis. This might actually be one of those cases, too -- but again I think I'll just leave things as they are.

Okay. So that's it for my hand-waving. Outside of that, all I've really done here is re-draw the schematic. So now let's analyze it.

---

Here are the nodal equations to solve (out-flowing currents on the left side, in-flowing currents on the right side):

$$\begin{align*} \frac{V_a}{R_1}+\frac{V_a}{R_2}&=\frac{V_2}{R_1}+\frac{V_b}{R_2}+I_1\\\\ \frac{V_b}{R_2}+\frac{V_b}{R_3}+I_2&=\frac{V_a}{R_2}+\frac{V_3}{R_3} \end{align*}$$

Note that in the above pair of equations (the two unknowns are \$V_a\$ and \$V_b\$), neither \$V_1\$ nor \$R_4\$ show up. As I said, they are irrelevant for the reasons I talked about. You can see that \$I_1\$ shows up in the first equation as an in-flowing current. Take careful note that this in-flowing current is completely unaffected by \$V_1\$. You can also see that \$I_2\$ shows up in the second equation as an out-flowing current. And again take careful note that this out-flowing current is completely unaffected by \$R_4\$.

The above equations can be re-arranged into a more standard form for solution:

$$\begin{align*} \frac{V_a}{1\:\Omega}+\frac{V_a}{3\:\Omega}&=\frac{-3\:\text{V}}{1\:\Omega}+\frac{V_b}{3\:\Omega}+3\:\text{A}\\\\ V_a\cdot\left(1+\frac13\right)+V_b\cdot\left(-\frac13\right)&=0\:\text{A}\\\\ &\text{and,}\\\\ \frac{V_b}{3\:\Omega}+\frac{V_b}{2\:\Omega}+3\:\text{A}&=\frac{V_a}{3\:\Omega}+\frac{+2\:\text{V}}{2\:\Omega}\\\\ V_a\cdot\left(-\frac13\right)+V_b\cdot\left(\frac13+\frac12\right)&=-2\:\text{A} \end{align*}$$

At this point, you should find it very easy to work out that \$V_a=-\frac23\:\text{V}\$ and \$V_b=-\frac83\:\text{V}\$. Hence, the voltage across \$R_2\$ has the magnitude of \$V_a-V_b=+2\:\text{V}\$ and from this it should be trivial to calculate the indicated current you want, \$I_{a\:b}\$.

---

In case it helps, here's a highly simplified equivalent schematic:

^{simulate this circuit}

Note that in the above case, I didn't even bother with showing anything at all on one side of the current sources shown there. There's no point. It doesn't matter. It's enough to know the value of the current sources, and direction.

There's more that can be done, now. Let's first take my above comment, where I don't even "bother with showing anything at all on one side of the current sources", and now make a small change expanding on that idea:

^{simulate this circuit}

As I said, it doesn't matter. So it's just fine to hook up the other end of any current source to any voltage source I want. In this case, I just arranged it in a way that should make the next step obvious.

At this point, we can convert what now appears to be obvious Norton sources into Thevenin sources. You can treat the left side consisting of \$I_1\$ and \$R_1\$ as a Norton source and convert it to a Thevenin source, instead. You can similarly also treat the right side consisting of \$I_2\$ and \$R_3\$ as yet another Norton source and convert it to a Thevenin source, too. The result is this:

^{simulate this circuit}

Now the current is trivial to compute:

$$I_{a\:b}=\frac{0\:\text{V}-\left(-4\:\text{V}\right)}{1\:\Omega+3\:\Omega+2\:\Omega}=\frac{4\:\text{V}}{6\:\Omega}=\frac{2}{3}\:\text{A}$$

So, as you can see there are a number of ways to get the same answer. And, in fact, one way doesn't even require the simultaneous solution of two equations with two unknowns. It just becomes a nearly trivial series circuit to solve with Ohm's Law.