Supernode Analysis for Electric Circuit Easy Solving

Supernode analysis or super nodal analysis is still considered as nodal or node analysis but with a special case where a voltage source exists in an electric circuit. If we observe carefully or read the nodal analysis thoroughly, we avoid analyzing an electric circuit with a voltage source using nodal analysis.

Do you know why?

The simple nodal analysis can’t solve an electric circuit with a voltage source. We need to modify the circuit a bit before applying nodal analysis. The modifying circuit and solve it with the nodal analysis is called supernodal or supernode analysis.

After learning this analysis method, we don’t need to avoid using nodal analysis for a circuit which has a voltage source anymore.

Read this topic thoroughly to understand fully about supernode analysis, supernode examples and problem solving, and of course some supernode examples to help us remember how to use it.

Since it is called supernode analysis, some of you may think that this method is harder than nodal analysis. Even if we have to modify the circuit, this supernode analysis is as easy as nodal analysis.

Let’s get started right away with the theory. We will also learn about supernode analysis problems with solutions.

Supernode Analysis Theory

Just as the name implies, a supernode is still considered as nodal analysis since its main focus is still the nodes. Is nodal analysis easier to use than supernode analysis? They will be used for different cases so we can’t say which one is easier. It is just a nodal analysis with a supernode.

Nodal analysis looks simpler but it is limited only to the current source in the circuit, without a single voltage source. This is why we can’t depend on this to solve a circuit with a voltage source. This is where supernode analysis kicks in.

Let’s observe the circuit in Figure.(1) and spot the difference between the circuit below with the previous circuit we have in nodal analysis. Have you spotted the difference?

The circuit below only has voltage sources without a single current source. What can we get from the circuit? What is a supernode in circuit?

supernode analysis 1
Figure 1. Supernode electric circuit

The circuit above has two cases:

CASE 1 -Observe the 10V source on the most left branch connected to non reference node v1 and reference node (ground node).

This case is really simple, we don’t need to do any modification or advanced analysis. Since there is only a voltage source connected between these two nodes, the nodal voltage v1 has the same value as the voltage source.

(1)   \begin{align*}v_{1}=10V\end{align*}

A single voltage source in a branch really makes it simple by this knowledge of voltage.

The second case will need extra effort to solve because there is a voltage source between two nonreference nodes v2 and v3. These two nodes will form what we called generalized nodes or supernodes. We can still use KCL and KVL to calculate node voltages.

Keep in mind that:

A supernode is formed when a voltage source is connected between two nonreference nodes and any elements connected in parallel with it.

Like we learned about nodal analysis, we only need to use KCL to find the current flowing in each branch or element.

But for a supernode, it is impossible to calculate how much current is flowing through a voltage source. What can we do then? We can use any method to calculate this but make sure to satisfy the KCL on the supernode just like any other node.

First we write the KCL equation for the Figure.(1),

(2)   \begin{align*}i_{1}+i_{4}&=i_{2}+i_{3}\\\frac{v_{1}-v_{2}}{2}+\frac{v_{1}-v_{3}}{4}&=\frac{v_{2}-0}{8}+\frac{v_{3}-0}{6}\end{align*}

Then we use the KVL for the supernode in Figure.(1). But before that, we have to redraw the circuit into Figure.(2) below.

supernode analysis 2
Figure 2. Loop in the supernode circuit

The loop in the clockwise direction gives:

(3)   \begin{align*}-v_{2}+5+v_{3}&=0\\v_{2}-v_{3}&=5\end{align*}

After we obtain Equations.(1) and (2), calculating the node voltages will be easier now. After learning up to this point, we need to remember the properties of supernode analysis below:

  1. The voltage source inside the supernode provides the missing equation to solve all the node voltages.
  2. A supernode has no voltage of its own.
  3. A supernode requires the application of both KCL and KVL.

Procedure for Supernode Analysis

The procedure for supernode analysis won’t be different with the procedure of nodal analysis we have learned before such as:

  1. Identify all the nodes in the circuit including the supernode.
  2. Set a node as a reference node. It usually acts as a ground so just add a ground symbol to it.
  3. Assign node voltage to other nodes (v1, v2, v3,etc).
  4. Remove the voltage source from the circuit first.
  5. Write the KCL supernode equations (currents entering a supernode are equal to the currents leaving the supernode).
  6. Use the KVL equation for the loop where a voltage source exists if you need to find the relationship of two nodes where a voltage source exists. (To complete the missing equation).
  7. Write all the KCL equations you can find (make the sum of the leaving current from a branch equal to zero).
  8. Use substitution, elimination, Cramer’s rule, etc.

To make it easier, just jump to the examples below. Have fun!

Supernode Analysis Examples

Now we will try to understand better from the supernode problems with answers below.

1. Observe the circuit in Figure.(3) below and find the node voltages.

supernode analysis 3
Figure 3

Solution:

Just as we read before, a supernode is formed when a voltage source is connected between two nonreference nodes and any elements connected in parallel with it. In our case, the supernode consists of 2V source, nodes 1 and 2, and the 10Ω resistor. We remove those to redraw the circuit below:

supernode analysis 4
Figure 4(a)

We use the KCL for the supernode in Figure.(4a) and we get

    \begin{align*}2=i_{1}+i_{2}+7\end{align*}

Writing the KCL equation for i1 and i2 using node voltage variables gives,

(1.1)   \begin{align*}2&=\frac{v_{1}-0}{2}+\frac{v_{2}-0}{4}+7\\8&=2v_{1}+v_{2}+28\\v_{2}&=-20-v_{1}\end{align*}

We are still missing the relationship between v1 and v2, we will use KVL to the circuit in Figure.(4b),

supernode analysis 5
Figure 4(b)

From the clockwise loop, we get

(1.2)   \begin{align*}-v_{1}-2+v_{2}&=0\\v_{2}&=v_{1}+2\end{align*}

From Equations.(1.1) and (1.2) we get

    \begin{align*}v_{2}=v_{1}+2&=-20-2v_{1}\\\\3v_{1}=-22 \quad&\rightarrow\quad v_{1}=-7.333V\end{align*}

we get

    \begin{align*}v_{2}=v_{1}+2=-5.333V\end{align*}

Note that the existence of a 10Ω resistor won’t do anything because it is only connected across the supernode.


2. Find the node voltages in the circuit of Figure.(5). This circuit will take more time because we will do nodal analysis with 2 supernodes.

supernode analysis 6
Figure 5

Solution:

There are two supernodes in the circuit, they are nodes 1 and 2, also nodes 3 and 4. Observe the circuit in Figure.(6a) below and use KCL to two supernodes.

supernode analysis 7
Figure 6(a)

For supernode 1 and 2,

    \begin{align*}i_{3}+10=i_{1}+i_{2}\end{align*}

Using the terms of the node voltages for supernode 1-2 equation gives,

(2.1)   \begin{align*}\frac{v_{3}-v_{2}}{6}+10=\frac{v_{1}-v_{4}}{3}+\frac{v_{1}}{2}\\5v_{1}+v_{2}-v_{3}-2v_{4}=60\end{align*}

For supernode 3-4 we have the KCL equation along with its node voltages,

(2.2)   \begin{align*}i_{1}&=i_{3}+i_{4}+i_{5}\\\frac{v_{1}-v_{4}}{3}&=\frac{v_{3}-v_{2}}{6}+\frac{v_{4}}{1}+\frac{v_{3}}{4}\\4v_{1}+2v_{2}&-5v_{3}-16v_{4}=0\end{align*}

Next we use KVL to the branches that have voltage sources just as shown in Figure.(6b).

supernode analysis 8
Figure 6(b)

For loop 1,

(2.3)   \begin{align*}-v_{1}+20+v_{2}&=0\\v_{1}-v_{2}&=20\end{align*}

For loop 2,

    \begin{align*}-v_{3}+3v_{x}+v_{4}=0\end{align*}

Don’t forget that

    \begin{align*}v_{x}=v_{1}-v_{4}\end{align*}

then

(2.4)   \begin{align*}3v_{1}-v_{3}-2v_{4}=0\end{align*}

For loop 3,

    \begin{align*}v_{x}-3v_{x}+6i_{3}-20=0\end{align*}

We know that

    \begin{align*}6i_{3}&=v_{3}-v_{2}\\v_{x}=v_{1}-v_{4}\end{align*}

Hence,

(2.5)   \begin{align*}-2v_{1}-v_{2}+v_{3}+2v_{4}=20\end{align*}

In order to finish all the equations we have, we need the result of four node voltages (v1, v2, v3, and v4). All the Equations.(2.1) to (2.5) we got earlier, we need only four to solve the remaining equations. The fifth equation is just an extra, it can be used to double-check our calculation.

Substituting Equation.(2.3) into both (2.1) and (2.2) respectively produces

(2.6)   \begin{align*}6v_{1}-v_{3}-2v_{4}=80\end{align*}

And

(2.7)   \begin{align*}6v_{1}-5v_{3}-16v_{4}=40\end{align*}

Equations.(2.4), (2.6), and (2.7) can be solved using matrix form as shown below

    \begin{align*}\begin{bmatrix}3 & -1 & -2 \\6 & -1 & -2 \\6 & -5 & -16\end{bmatrix}\begin{bmatrix}v_{1}\\v_{3}\\v_{4}\end{bmatrix}=\begin{bmatrix}0\\80\\40\end{bmatrix}\end{align*}

Then we use the Cramer’s rule and get

    \begin{align*}\Delta&=\begin{vmatrix}3 & -1 & -2 \\6 & -1 & -2 \\6 & -5 & -16\end{vmatrix}=-18\\\\\Delta_{1}&=\begin{vmatrix}0 & -1 & -2 \\80 & -1 & -2 \\40 & -5 & -16\end{vmatrix}=-480\\\\\Delta_{3}&=\begin{vmatrix}3 & 0 & -2 \\6 & 80 & -2 \\6 & 40 & -16\end{vmatrix}=-3120\\\\\Delta_{4}&=\begin{vmatrix}3 & -1 & 0 \\6 & -1 & 80 \\6 & -5 & 40\end{vmatrix}=840\end{align*}

Hence, the node voltages are

    \begin{align*}v_{1}=\frac{\Delta_{1}}{\Delta}=\frac{-480}{-18}=26.67V\\v_{3}=\frac{\Delta_{3}}{\Delta}=\frac{-3120}{-18}=173.33V\\v_{4}=\frac{\Delta_{4}}{\Delta}=\frac{840}{-18}=-46.67V\end{align*}

and

    \begin{align*}v_{2}=v_{1}-20=6.667V\end{align*}

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