Superposition Theorem Formula for Electric Circuit

Superposition theorem will help us greatly when analyzing a circuit with multiple sources. If a circuit has two or more independent sources, one way to determine the value of a specific variable (voltage or current) is to use nodal or mesh analysis like has been discussed before.

Another way is to determine the contribution of each independent source to the variable and then add them up. The latter approach is known as superposition.

Make sure to read what is electric circuit first. These circuit analysis theorems are classified as:

Superposition Theorem Theory

The idea of superposition rests on the linearity property. The superposition principle states that the voltage across (or currents through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.

The principle of superposition helps us to analyze a linear circuit with more than one independent source by calculating the contribution of each independent source separately.

However, to apply the superposition principle, we must keep two things in mind:

1.We consider one independent source at a time while all other independent sources are turned off. This implies that we replace every voltage source by 0 V (or a short circuit), and every current source by 0 A (or an open circuit). This way we obtain a simpler and more manageable circuit.

2.Dependent sources are left intact because they are controlled by circuit variables.

With these in mind, we apply the superposition principle in three steps:

Steps to Apply Superposition Principle:

1.Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using the techniques in previous explanation.

2.Repeat step 1 for each of the other independent sources.

3.Find the total contribution by adding algebraically all the contributions due to the independent sources.

Analyzing a circuit using superposition has one major disadvantage: It may very likely involve more work.

If the circuit has three independent sources, we may have to analyze three simpler circuits each providing the contribution due to the respective individual source.

However, superposition does help reduce a complex circuit to simpler circuits through replacement of voltage sources by short circuits and of current sources by open circuits.

Superposition Circuit Analysis

This theorem eliminates all the independent sources but leaving one active at a time (one active independent source for one superposition circuit). We calculate the voltage and/or current drop in the desired element for each superposition circuit. The final step is calculating all the values across the desired element.

Important Question:
How many independent sources can be analyzed at a time using the Superposition Theorem?
We can only use one active independent source at a time.


Superposition is one of the strong methods to analyze a circuit consisting of multiple independent sources. Even if we get multiple equations to analyze, this theorem is very easy to use and doesn’t need high understanding to master this.

Superposition Theorem Formula

This theorem can only be used for a linear circuit. Linear circuit is a circuit where its equation will fulfill y = kx, where
k = constant
x = variable
For every linear circuit with several voltage source or current source can be analyzed using:

Algebraic sums the voltage or current generated by every independent source acting alone, when the other independent sources are replaced by their internal impedances.

If we want to explain better, then:
If there is n independent sources in a circuit then we will have n equation based on an active independent source at one time. At the end all the equations for every circuit condition will be summed together. Even if there are dependent sources, the superposition theorem only counts the n independent sources.

Linear circuits are formed from independent sources, dependent sources, and passive elements (resistor, inductor, capacitor).

Superposition Theorem Circuits:
1. Calculate the current i using superposition theorem.

superposition theorem formula

Solution:
First we activate the voltage source while we deactivate the current source (we replace it with its internal impedance, an open circuit)

superposition theorem formula

Then,

    \begin{align*}i_{1}=\frac{20}{10+10}\times 1=1A\end{align*}

Next we activate the current source while we deactivate the voltage source (we replace it with its internal impedance, a short circuit)

superposition theorem formula

    \begin{align*}&i_{2}=-\frac{20}{10+10}\times 1=-0.5A \\&\mbox{then:}\\&i=i_{1}+i_{2}=1-0.5=0.5A\end{align*}

 

2. Calculate the current i using superposition theorem.

superposition theorem formula

When we activate the source VS = 17 V then the voltage source 6V is replaced with a short circuit, and the current source 2A is replaced with an open circuit.

superposition theorem formula

    \begin{align*}&3\Omega //0\Omega\rightarrow R_{p1}=0\Omega\\&2\Omega //2\Omega\rightarrow R_{p1}=1\Omega\\&V_{Rp2}=\frac{1}{1+3}\times 17=\frac{17}{4}V\\&\mbox{then}\\&i_{1}=\frac{-V_{Rp2}}{2}=-\frac{17}{8}A\end{align*}

When the voltage source VS = 6V is activated, the voltage source 17V is replaced with a short circuit while the current source 2A is replaced with an open circuit.

superposition theorem formula

    \begin{align*}&3\Omega//2\Omega\rightarrow R_{p1}=\frac{3\times2}{3+2}=\frac{6}{5}\Omega\\&R_{s}=R_{p1}+2\Omega=\frac{6}{5}+2=\frac{16}{5}\Omega\\&R_{s}//3\Omega\rightarrow R_{p2}=\frac{\frac{16}{5}\times3}{\frac{16}{5}+3}=\frac{48}{31}\Omega\\&i_{2}=\frac{6}{R_{p2}}=\frac{6}{31/8}=\frac{31}{8}A\end{align*}

When the current source IS = 2A is activated then the voltage source 17V is replaced with a short circuit while the voltage source 6V is replaced with a short circuit.

superposition theorem formula

    \begin{align*}&3\Omega//2\Omega\rightarrow R_{p1}=\frac{3\times2}{3+2}=\frac{6}{5}\Omega\\&3\Omega//0\Omega\rightarrow R_{p2}=0\Omega\\&i_{3}=\frac{2}{2+6/5}\times2=\frac{5}{4}A\\&\mbox{then} \quad i=i_{1}+i_{2}+i_{3}\\&i=\frac{-17}{8}+\frac{31}{8}+\frac{5}{4}=3A\end{align*}

3. Find current i with superposition theorem.

superposition theorem formula

Solution:

In this circuit exists a dependent source, thus we will still follow the superposition theorem. For n independent sources, we will get n equation. For the case above, since there are 2 independent sources then there will be 2 conditions that need to be analyzed producing 2 equations.

When the current source IS = 8A active then current source 4A is replaced with an open circuit.

superposition theorem formula

    \begin{align*}&i_{1}=\frac{3}{3+2}\times(3i_{1}-8)\\&i_{1}=\frac{3}{5}\times(3i_{1}-8)\\&5i_{1}=9i_{1}-24\rightarrow i_{1}=\frac{24}{4}=6A\\\end{align*}

When the current source IS = 4A active then current source 8A is replaced with an open circuit.

superposition theorem formula

    \begin{align*}i_{2}&=\frac{3}{3+2}\times(3i_{2}+4)\\i_{2}&=\frac{3}{5}\times(3i_{2}+4)\\5i_{2}&=9i_{2}+12\rightarrow i_{2}=\frac{-12}{4}=-3A\\&\mbox{then}\\i&=i_{1}+i_{2}=6-3=3A\end{align*}

Superposition Theorem Circuit Examples

For better understanding let us review examples below:
1. Use the superposition theorem to find v in the circuit of Figure.(1)

superposition theorem formula
Figure 1

Solution :
Since there are two sources, let

    \begin{align*}v=v_{1}+v_{2}\end{align*}

where v1 and v2 are the contributions due to the 6V voltage source and the 3 A current source, respectively. To obtain v1, we set the current source to zero, as shown in Figure.(2a).

superposition theorem formula
Figure 2

Applying KVL to the loop in Figure.(2a) gives

    \begin{align*}12i_{1}-6=0 \quad\rightarrow\quad i_{1}=0.5A\end{align*}

Thus,

    \begin{align*}v_{1}=4i_{1}=2V\end{align*}

We may also use voltage division to get v1 by writing

    \begin{align*}v_{1}=\frac{4}{4+8}(6)=2V\end{align*}

To get v2, we set the voltage source to zero, as in Figure.(2b). Using current division,

    \begin{align*}i_{3}=\frac{8}{4+8}(3)=2A\end{align*}

Hence,

    \begin{align*}v_{2}=4i_{3}=8V\end{align*}

And we find

    \begin{align*}v=v_{1}+v_{1}=2+8=10V\end{align*}

2. Find io in the circuit of Figure.(3) using superposition.

superposition theorem formula
Figure 3

Solution :
The circuit in Figure.(3) involves a dependent source, which intact. We let

(1)   \begin{align*}i_{o}=i^{'}_{o}+i^{"}_{o}\end{align*}

where i’o and i”o are due to the 4 A current source and 20 V voltage source respectively. To obtain i’o, we turn off 20 V source so that we have the circuit Figure.(4a). We apply mesh analysis in order to obtain i’o. For loop 1,

(2)   \begin{align*}i_{1}=4A\end{align*}

For loop 2,

(3)   \begin{align*}-3i_{1}+6i_{2}-1i_{3}-5i^{'}_{o}=0\end{align*}

superposition theorem formula
Figure 4

For loop 3,

(4)   \begin{align*}-5i_{1}-1i_{2}+10i_{3}+5i^{'}_{o}=0\end{align*}

But at node 0,

(5)   \begin{align*}i_{3}=i_{1}-i^{'}_{o}=4-i^{'}_{o}\end{align*}

Substituting (2) and (5) into (3) and (4) gives two simultaneous equations

(6)   \begin{align*}3i_{2}-2i^{'}_{o}=8\end{align*}


(7)   \begin{align*}i_{2}+5i^{'}_{o}=20\end{align*}

which can be solved to get

(8)   \begin{align*}i^{'}_{o}=\frac{52}{17}A\end{align*}

To obtain i”o, we turn off the 4 A current source so that the circuit becomes that shown in Figure.(3b). For loop 4, KVL gives

(9)   \begin{align*}6i_{4}-i_{5}-5i^{"}_{o}=0\end{align*}

and for loop 5,

(10)   \begin{align*}-i_{4}+10i_{5}-20+5i^{"}_{o}=0\end{align*}

But i5 = –i”o. Substituting this in (9) and (10) gives

(11)   \begin{align*}6i_{4}-4i^{"}_{o}=0\end{align*}



(12)   \begin{align*}i_{4}+5i^{"}_{o}=-20\end{align*}

which we solve to get

(13)   \begin{align*}i^{"}_{o}=-\frac{60}{17}A\end{align*}

Now substituting (8) and (13) to (1) gives

    \begin{align*}i_{o}=-\frac{8}{17}=-0.4706A\end{align*}

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