Superposition is a superpower to add to your toolkit of circuit analysis methods. Use it when your circuit has multiple inputs or power sources.

Before reading this article, I recommend you review the concept of Linearity. Superposition works because of linearity.

### Contents

The principle of superposition is another name for the additivity property of Linearity,

$f(x_1 + x_2) = f(x_1) + f(x_2)$

To solve a circuit using superposition, the first step is to turn off or suppress all but one input.

• To suppress a voltage source, replace it with a short circuit.

• To suppress a current source, replace it with an open circuit.

Next, you analyze two (or more) much simpler circuits. Repeat for all inputs.
The final result is the sum of individual results.

## Describing a circuit as a function

The principle of superposition uses functional notation, so we talk a bit here about how circuits can be represented as functions. You used functional notation in algebra class when you wrote something like $f(x) = 2x + 3$.

Start simply… How might we represent a lone resistor using functional notation? There’s nothing remarkable going on here. We just talk about Ohm’s Law using function terminology.

Begin by identifying three things: the inputs, outputs, and the thing performing the function.

Input: We decide (arbitrarily) that voltage $v_i$ will be the input of our resistor function. Assume input $v_i$ is generated by some voltage source we’re not showing. The input voltage is applied to the two small circles (the circles indicate the input port to our function).

Output: We assign the output to be the interesting thing we want to know. For this function, we choose the output to be the current $i$. The output of our function will be the current, $i$, measured by some not-shown current meter.

Function: The function itself comes from the resistor, described by Ohm’s Law.

Written as a function, our resistor is

$i = f(v_i) = \dfrac{1}{\text R}\,v_i$

With this notation, we are viewing the resistor as a function that takes in a voltage and outputs a current.

## A resistor is a linear function

Looking at our resistor function, we see it has the scaling property. The output, $i$, equals the input, $v_i$, scaled by a constant, $\text R$. That means the resistor is linear. The linearity property is what lets us use superposition to help solve a circuit.

Refresh me on the scaling property and linearity.

## Example 0

(This is a “toy” example to give you a feel for how to apply superposition.)

Let’s use the same resistor function, but we change the input to two voltage sources,

The input to our function is two batteries in series, so $v_i = \text{Vs1} + \text{Vs2}$.

The function is $f(v_i) = \dfrac{1}{\text R}\,v_i$.

The output of the function hasn’t changed; it is still $i = f(v)$.

We now solve this circuit two ways: first by conventional analysis, and then using the principle of superposition.

### Conventional solution

To solve by conventional means we write Kirchhoff’s voltage law (KVL) equation around the loop,

$\text{Vs1} + \text{Vs2} - i \,\text R = 0$

and solve for $i$,

$\tag{conventional solution}i = f(\text{Vs1}+\text{Vs2}) = {\dfrac{\text{Vs1} +\text{Vs2}}{\text R}}$

### Solve using the principle of superposition

The principle of superposition applies if the function $f$ is linear.

$f(x_1 + x_2) = f(x_1) + f(x_2)$

This is the additivity property of linearity. It is also called the superposition property. The left side has two inputs superimposed, $(x_1+x_2)$. The right side of the equation says if you apply the inputs one at a time, $(x_1)$ followed by $(x_2)$, and add the individual results you get the same answer.

Now let’s use the principle of superposition to solve the circuit.

Since we know our circuit can be modeled as a linear function, we can say,

$i = f(\text{Vs1} + \text{Vs2})$
is equivalent to
$i = f(\text{Vs1}) + f(\text{Vs2})$

This suggests an intriguing possibility. It says we can compute the output current the conventional way, $i = f(\text{Vs1} + \text{Vs2})$, or, we could get the same answer applying single inputs, $f(\text{Vs1})$ and $f(\text{Vs2})$, and then add the results together.

Let’s try, and see what happens.

#### Suppress the inputs

To use superposition we need to apply one input at a time. How do you apply just one input? Easy! Just turn the other ones off. When we turn off an input we say it is suppressed.

What does it mean to turn off (suppress) a voltage source? It means we set $\text V = 0$. This is the same thing as replacing the voltage source or battery by a short circuit.

What does it mean to turn off a current source? It means we set $\text I = 0$. That’s the same as replacing the current source with an open circuit.

It may seem like suppression would completely mess up the circuit, but it doesn’t.

In the next two schematics, one of the voltage inputs is turned off (suppressed ) by replacing it with a short circuit.

When you zero out (suppress) an input it becomes $0$, allowing the other input to shine through by itself,

$f(\text{Vs1} + 0) \quad \rightarrow \quad f(\text{Vs1})$

$f(\text{0 + Vs2}) \quad \rightarrow \quad f(\text{Vs2})$

What if there are more than two inputs?

If the function (circuit) has more than two inputs, you isolate a single input by suppressing all the others. If the function has three inputs, you suppress two at time, and you do that three times. If the function is,

$out = f(in_1, in_2, in_3)$

then you isolate $in_1$ by suppressing $in_2$ and $in_3$ and so on,

$out_1 = f(in_1, 0, 0)$

$out_2 = f(0,in_2,0)$</p> $out_3 = f(0,0,in_3)$</p>

$out = out_1 + out_2 + out_3$

#### Solve the two circuits

Now we solve each circuit individually,

$i_1 = \text R\cdot\text{Vs1} \qquad$ and $\qquad i_2 = \text R\cdot\text{Vs2}$

where $i_1$ is the current caused by source $\text{Vs}1$, and $i_2$ is the current caused by source $\text{Vs}2$.

(You can come up with your own system for naming all the different $i$ and $v$ variables that make up the two new circuits.)

#### Superimpose (add) the two solutions

The total current comes from superimposing (adding) the currents from each circuit.

$i = i_1 + i_2$

$i = \dfrac{\text{Vs1}}{\text R} + \dfrac{\text{Vs2}}{\text R}$

$\tag{superposition solution}{i = \dfrac{\text{Vs1}+\text{Vs2}}{\text R}}$

Check it out! Compare this superposition solution to the conventional solution we got above. They’re the same!

What we did here is called the linear superposition of two circuits.

Here’s a key thing to appreciate. Notice how taking out one voltage source did not modify or mess up the contribution from the other voltage source. It’s like those two circuits we created with suppression are living inside/with each other, without affecting the other one. Very cool to think about.

In this toy example our function was so simple that superposition really didn’t save much (if any) effort. In the following examples the circuits are more complicated, and the difference in effort becomes more apparent.

## Example 1

Consider the following circuit composed of two linear components (resistors) and two sources: a current source and a voltage source. The two sources are the inputs. We define two outputs to be the currents $i_1$ and $i_2$. In functional notation we write this as,

$i_1 = f_1(\text{Is},\text{Vs)}\quad$ and $\quad i_2 = f_2(\text{Is},\text{Vs)}$

### Solve using superposition

Let’s go through the steps to analyze this circuit using superposition.

First, we suppress the current source and analyze the circuit with just the voltage source acting alone. To suppress the current source, we replace it with an open circuit.

With just the voltage source, we solve for the two output currents,

$i_{1V} = 0 \qquad i_{2V} = \dfrac{\text{Vs}}{\text R2}$

Where $i_{1V}$ and $i_{2V}$ are the currents in $\text R1$ and $\text R2$ caused by the voltage source.

Next, we restore the current source and suppress the voltage source by replacing it with a short circuit.

With just the current source, we solve for the two output currents,

$i_{1I} = \text{Is} \qquad i_{2I} = 0$

Where $i_{1I}$ and $i_{2I}$ are the currents in $\text R1$ and $\text R2$ caused by the current source.

zero current in R2?

The ends of $\text R2$ are connected together. There is no voltage across $\text R2\,$ and therefore there is no current through it.

We finish by adding the contributions from each source,

$i_1 = i_{1V} + i_{1I} = 0 + \text{Is}= \text{Is}$

$i_2 = i_{2V} + i_{2I} = \dfrac{\text{Vs}}{\text R2} + 0 = \dfrac{\text{Vs}}{\text R2}$

The full solution looks like this,

This could have been a tricky analysis because of the two sources. Superposition gave us two simpler circuits to deal with. Notice how whole chunks of the original circuit fall away when we suppress the sources.

## Example 2

For the following linear circuit let’s calculate the output voltage $v$.

### Conventional solution

We will do it the conventional way first. We write Kirchhoff’s current law (KCL) at output node $v$,

$\quad +i_{\text R1} \qquad - i_{\text R2} \qquad +\text{Is} \quad = 0$

$+\dfrac{\text{Vs}-v}{\text{R1}} \quad - \dfrac{v}{\text{R2}} \qquad + \text{Is} \quad = 0$

We can rearrange this to get an expression for $v$ by gathering like terms together on the right side,

$\tag{conventional solution}v = \dfrac{\text{R2}}{\text R1 + \text R2}\,\text{Vs} + \dfrac{\text R1\,\text R2}{\text R1+\text R2}\,\text{Is}$

I took a shortcut I will share it with you.

Starting from the KCL equation,

$\tag{1}+\dfrac{\text{Vs}-v}{\text{R1}} - \dfrac{v}{\text{R2}} + \text{Is}= 0$

Separate the first fraction into separate numerators,

$\tag{2}+\dfrac{\text{Vs}}{\text{R1}} - \dfrac{v}{\text{R1}} - \dfrac{v}{\text{R2}} + \text{Is}= 0$

Group together the two fractions with $v$ in the numerator,

$\tag{3}+\dfrac{\text{Vs}}{\text{R1}} - \left ( \dfrac{v}{\text{R1}} + \dfrac{v}{\text{R2}} \right ) + \text{Is}= 0$

Factor out $v$ from the group,

$\tag{4}+\dfrac{\text{Vs}}{\text{R1}} - v\left ( \dfrac{1}{\text{R1}} + \dfrac{1}{\text{R2}} \right ) + \text{Is}= 0$

Now the trick: You should recognize the resistor expression in parentheses as two resistors in parallel. You can write this two ways, as the sum of reciprocals, or as the product over the sum,

$\tag{5}\text R_{\text{parallel}} = \dfrac{1} {\left (\dfrac{1}{\text{R1}} +\dfrac{1}{\text{R2}} \right )} = {\dfrac{\text{R1}\cdot\text{R2}} {\text{R1} + \text{R2}}}$

Substitute the last term for the resistor reciprocals, flipping it over the right way,

$\tag{6}+\dfrac{\text{Vs}}{\text{R1}} - v\left ( {\dfrac{\text{R1} + \text{R2}} {\text{R1} \cdot \text{R2}}} \right ) + \text{Is}= 0$

Move some stuff over to the right side and keep the signs right,

$\tag{7}v\left ( {\dfrac{\text{R1} + \text{R2}} {\text{R1} \cdot \text{R2}}} \right ) = \dfrac{\text{Vs}}{\text{R1}} + \text{Is}$

Move and flip the parallel $\text R$ expression over to the right side to isolate $v$,

$\tag{8}v = \left ( {\dfrac{\text{R1} \cdot \text{R2}} {\text{R1} + \text{R2}}} \right ) \left ( \dfrac{\text{Vs}}{\text{R1}} + \text{Is} \right )$

Multiply through to get the form we want for later comparison to the superposition result,

$\tag{9}v = \left ( {\dfrac{\text{R1} \cdot \text{R2}} {\text{R1} + \text{R2}}} \right ) \dfrac{\text{Vs}}{\text{R1}} + \left ( {\dfrac{\text{R1} \cdot \text{R2}} {\text{R1} + \text{R2}}} \right ) \text{Is}$

$\tag{10}v = \dfrac{\text{R2}}{\text R1 + \text R2}\,\text{Vs} + \dfrac{\text R1\cdot\text R2}{\text R1+\text R2}\,\text{Is}$

The expression for two parallel resistors is fully worked out in this article on parallel resistors. It’s a good one to memorize.

### Solution using superposition

Now we will solve the same problem using the principle of superposition. As before, we suppress the input sources and solve new simpler circuits.

How would you suppress the current source?
Replace the current source with a ______. (short circuit or open circuit?)

Replace the current source with an open circuit.

The circuit collapses down to two resistors in series (also known as a voltage divider.

Voltage ${v_{Vs}}$ is the contribution from voltage source $\text{Vs}$. With just the voltage source, the output voltage is,

$v_{Vs}= \text{Vs} \,\dfrac{\text R2}{\text R1+ \text R2}$

Now restore the current source and suppress the voltage source.

How would you suppress the voltage source?
Replace the voltage source with a ______. (short circuit or open circuit?)

Replace the voltage source with a short circuit.

The circuit collapses down to two resistors in parallel.

Voltage ${v_{Is}}$ is the contribution to the output from current source $\text{Is}$.

$v_{Is} = \text{Is}\,\dfrac{\text R1\cdot\text R2}{\text R1 + \text R2}$

We complete the superposition analysis by adding the two voltage contributions.

$v = v_{Vs} + v_{Is}$

$\tag{superposition solution}v = \dfrac{\text{R2}}{\text R1 + \text R2}\,\text{Vs} + \dfrac{\text R1\cdot\text R2}{\text R1+\text R2}\,\text{Is}$

As predicted, we get the same result as the conventional solution shown above. With superposition, there is no approximation involved. The solutions are exactly the same. The key thing to notice is that the two simpler circuits took less work to analyze.

## Closing thoughts

If you have a circuit made from linear elements, you get to use the principle of superposition. This means the original complicated circuit is really made of simpler circuits that happen to be sitting on top of each other. It seems like magic, but this means superimposed circuits don’t affect each other or intertwine at all. Every simple circuit is blissfully unaware of the others until you do the final addition.

This is a marvelous property of linear circuits, and it is one of the reasons we love linearity so much. Circuits that are not linear (non-linear circuits) don’t have this property, and superposition cannot be applied. (But don’t worry, we love non-linear circuits, too, just in a different way.)

## Summary

If a circuit is made of linear elements you can use superposition to simplify the analysis. This is especially useful for circuits with multiple sources. Whenever you have a circuit with two energy sources, like two batteries, or one input and some internal energy, the possibility of using superposition should pop into your head.

To analyze a linear circuit with multiple inputs, suppress all but one input or energy source and analyze the resulting simpler circuit. Repeat for all inputs and sources. Then add the results to find the total response for the full circuit.

To suppress a voltage source, replace it with a short circuit,

To suppress a current source, replace it with an open circuit,