Node voltage analysis with multiple sources

In our previous lesson we got an introduction into node voltage analysis. Now we are going to expand the usability of this method by adding voltage and current sources into our circuit and solving them as well.

Current sources

Take the circuit above. We have a voltage source and a current source in this schematic. We have selected a reference node and labeled node \(V_A\). We have not created a second \(V_B\) node between resistor \(R_2\) and the current source. The reason for this is because we are going to use KCL to solve this node. However, if we are clever we can already solve this node. As $$1A$$ is forced into this node by the current source, it must also be exiting this node. So no need to calculate this.

For node \(V_A\) we can write down the equation: $$ \dfrac{V_A – 5}{R_1} + \dfrac{V_A – 0}{R_3} -1 = 0$$

Note something very important here. We do not write down \(\dfrac{V_A – V_B}{R_2}\) for the equation at \(R_2\). Because first off, we don’t know the voltage of a current source, and secondly we already know the current. Remember that we are just using KCL to enter currents into our equation (voltage divided by resistance), since we already know the current, there is no need to calculate it. The only thing we need to pay attention to is the direction of the current. We know it is flowing into the node, but we always calculate the current flowing out of the node, that is where the negative sign is coming from.

Now we can rearrange the equation to: $$ V_A \left( \dfrac{1}{R_1} + \dfrac{1}{R_3} \right) = \dfrac{5}{R_1} + 1$$

With only one unknown we could solve it if we have the values for the resistors.

Current source recap

What you need to know when you encounter a current source in a node-voltage analysis is that you will be happy. What is also good to remember is that you can directly enter the current in the equations for your analysis. This saves you time, but remember the direction and thus the sign of the current!

Voltage sources

Now let’s see what we have to do when voltage sources are added to our circuit.

When we have two voltage sources with a common connection (like in the circuit above) the most often used location for the reference node is in between the voltage sources. Other locations can be selected, but I advise against it.

Selecting the reference node between the voltage sources gives us the above circuit. To solve it we will do our common routine of selecting reference node (done), labeling unknown nodes (done), writing down KCL equations, rearranging the equations, and finally solving the network.

For \(V_A\) we get: $$ \dfrac{V_A – 5}{R_1} + \dfrac{V_A – 0}{R_2} + \dfrac{V_A – V_B}{R_3} = 0$$

For \(V_B\) we get: $$ \dfrac{V_b – V_A}{R_3} + \dfrac{V_B – 0}{R_4} + \dfrac{V_B – -8}{R_5} = 0$$

Again be wary when entering the values into the equations. For \(V_B\) the voltage from the voltage source is negative. The positive side of the voltage source is connected to our reference node, so it is at 0 volts. This must mean that the negative terminal of the voltage source is at -8 volts.

So far this was nothing special right? We can do this.

Voltage sources without common connection

But what happens when two voltage sources are present, but we do not have a common connection anymore.

Let’s start normally. I’ve selected a reference point at one of the voltage sources. The remaining three nodes are labeled. So next up is KCL equation time.

$$\dfrac{V_A-V_C}{R_1} + \dfrac{V_A-V_B}{R_2} + \dfrac{V_A – 10}{R_3} = 0$$

$$\dfrac{V_B-V_A}{R_2} + \dfrac{V_B – 0}{R_4} + ? = 0$$

Oh, we have run into a problem, let’s ignore it and move on:

$$\dfrac{V_C – V_A}{R_1} + \dfrac{V_C – 0}{R_5} + ? = 0$$

Crap, same problem. We don’t know the current through a voltage source. We can calculate it, but then we need to know all the currents, for which we need to voltage source current. That’s it, I’m packing my bags and I run while I still can.

 

 

You are still here? And want to know how to solve this? Ok, that is exactly what you need to become an engineer. You will face a lot of challenges, but with patience and perseverance we can solve problems. Respect for making the effort!

To solve a circuit like in Figure 4, we need to be smart. Remember that we use KCL, this means that all the current flowing from node \(V_B\), must flow into node \(V_C\) or vice versa. We also know
$$I_{R_2} + I_{R_4} + I_{V_2} = 0$$ and
$$ I_{R_1} + I_{R_5} + I_{V_2} = 0$$
Since $$ I_{V_2}$$ must be equal in both nodes (it is the same current after all), we can combine the nodes together to form a supernode. The equation for the supernode is:
$$ I_{R_2} + I_{R_4} + I_{R_1} + I_{R_5} = 0$$.
Why can we leave out $$ I_{V_2}$$? Because it will either flow from \(V_B\) to \(V_C\) or the other way around. In both cases, it will leave one node, and enter the other node.
If we combine the equations, we have to add it in one node, but subtract it in the other, which after combining them cancels them out like this:

\(I_{R_2} + I_{R_4} + I_{V_2} – I_{V_2} + I_{R_1} + I_{R_5} = 0\) or
\(I_{R_2} + I_{R_4} – I_{V_2} + I_{V_2} + I_{R_1} + I_{R_5} = 0\)

What we are left with is two KCL equations. Our original KCL equation for node \(V_A\):
$$\dfrac{V_A-V_C}{R_1} + \dfrac{V_A-V_B}{R_2} + \dfrac{V_A – 10}{R_3} = 0$$
And our new equation which combined node \(V_B\) and \(V_C\):
$$\dfrac{V_B-V_A}{R_2} + \dfrac{V_B – 0}{R_4} + \dfrac{V_C – V_A}{R_1} + \dfrac{V_C – 0}{R_5} = 0$$

Now we have two equations with three unknown. We are stuck once more. But we have an ace up our sleeves. Because we know that:
$$V_B – V_C = 8V$$

To show that we solved the circuit with this method let try this with \(R_1 = 4k\Omega\), \(R_2 = 2k\Omega\), \(R_3 = 1k\Omega\), \(R_4 = 8k\Omega\), \(R_5 = 10k\Omega\)

For node \(V_A\):

$$\dfrac{V_A-V_C}{R_1} + \dfrac{V_A-V_B}{R_2} + \dfrac{V_A – 10}{R_3} = 0$$

Entering the values for the resistors:

$$\dfrac{V_A-V_C}{4k} + \dfrac{V_A-V_B}{2k} + \dfrac{V_A – 10}{1k} = 0$$

$$V_A \left(\dfrac{1}{4k} + \dfrac{1}{2k} + \dfrac{1}{1k} \right) + V_B \left( \dfrac{-1}{2k} \right) + V_C \left( \dfrac{-1}{4k} \right) = \dfrac{10}{1k}$$

For the supernode we get:

$$\dfrac{V_B-V_A}{R_2} + \dfrac{V_B – 0}{R_4} + \dfrac{V_C – V_A}{R_1} + \dfrac{V_C – 0}{R_5} = 0$$

$$\dfrac{V_B-V_A}{2k} + \dfrac{V_B – 0}{8k} + \dfrac{V_C – V_A}{4k} + \dfrac{V_C – 0}{10k} = 0$$

$$V_A \left( \dfrac{-1}{2k} + \dfrac{-1}{4k} \right) + V_B \left( \dfrac{1}{2k} + \dfrac{1}{8k} \right)+ V_C \left( \dfrac{1}{4k} + \dfrac{1}{10k} \right) = 0 $$

And the last one:

$$ V_B – V_C = 8V$$

Combining everything in a matrix gives us:

\begin{gather} \begin{bmatrix} \dfrac{7}{4k} & \dfrac{-1}{2k} & \dfrac{-1}{4k} \\ \dfrac{-3}{4k} & \dfrac{5}{8k} & \dfrac{7}{20k} \\ 0 & 1 & -1 \\ \end{bmatrix} \begin{bmatrix} V_A \\ V_B \\ V_C \end{bmatrix} = \begin{bmatrix} \dfrac{10}{1k} \\ 0 \\ 8 \\ \end{bmatrix} \end{gather}

Solving this matrix gives us: \(V_A = 8.656V, V_B = 9.530V, V_C = 1.530V\)

This was a bit harder, but once you know the concept you can apply this method of a supernode to solve also circuits with more than 1 voltage source, without a common ground.

Voltage source recap

Dealing with voltage sources can be more difficult, but it is not impossible. Let this list give you some guidance to solve circuits:

1. When you encounter a single voltage source in your circuit:
   Put the reference node on either side, and you can solve it without any problem!
2. When you encounter two or more voltage sources, but they share a common connection:
   Put the reference node at the common connection, and solve it without any problem! (do note the possible negative voltage)
3. When you encounter two or more voltage sources, but they don’t share a common connection:
   – Place the reference node at any of the voltage sources
   – For each voltage source without a connection to a reference node, create a supernode. 
   – Add an equation for the voltage difference between the two merged nodes. 
   – Solve the circuit!