For the example two, the equation rearranges part on how to get V is kind confusing to me. Can someone please simply explain to me? Thank you very much!

When I did this algebra I took a shortcut I will share it with you. (I should have written out the full derivation like Sal always does, sorry.) Starting with the second equation in Example 2 Conventional Solution: - Separate the first fraction into separate numerators. Vs/R1 - v/R1 - Group together the two fractions with v in the numerator. v/R1 + v/R2 - Factor out v. v (1/R1 + 1/R2). - Here's the trick: You might recognize that resistor expression from two parallel R's Rparallel = 1/(1/R1 + 1/R2) = (R1 R2)/(R1+R2) - Substitute the last term for the resistor reciprocals, flipping it over the right way: v ( R1+R2)/(R1 R2) = Is + Vs/R1 Finish up by moving the big R fraction over to the right side. This substitution I did with the resistor expression is fully worked out in https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-resistor-circuits/a/ee-parallel-resistors where it says Special Case - Two Resistors In Parallel. It's a good one to memorize.

To be honest I was searching for the detailed information about Superposition topic which includes questions like two given point charges separated by a distance and you determine the location where the electric field is zero.I was wondering whether I am able to find it or some different topic which helps me to understand it better because I dont see this topic at school and I will be really appreciated If someone helps

This topic is covered in the physics section, under electric fields. This video might be helpful for your particular example. https://www.khanacademy.org/science/physics/electric-charge-electric-force-and-voltage/electric-field/v/net-electric-field-from-multiple-charges-in-2d

Main content

Superposition

Google Classroom

With the principle of superposition you can simplify the analysis of circuits with multiple inputs. Written by Willy McAllister.

Superposition is a super useful technique to add to your toolkit of circuit analysis methods. Use superposition when you have a circuit with multiple inputs or multiple power sources.

What we're building to

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.

Then you analyze the resulting 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 is defined using functional notation, so we talk a bit here about how circuits can be represented as functions.

Starting simple... How could we represent a lone resistor using the notation of a mathematical function? There is nothing remarkable going on here, I'm just talking about Ohm's Law using function terminology. We begin by identifying three things: the inputs, the thing performing the function, and the outputs.

I decided (arbitrarily) that voltage

v_{i}

will be the input to our resistor function. We can assume input

v_{i}

is generated by some voltage-making thing we're not showing. We assign the output to be the interesting thing we want to know. For this function, the output is the current

i

in the resistor.

The input voltage is applied to the two small circles (the circles indicate the input port to our function). The function itself comes from the resistor, by means of Ohm's Law. The output of our function will be the current,

i

, measured by some not-shown current meter.

Written as a function, our resistor is

i = f (v_{i}) = \frac{1}{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

, scaled by a constant,

R

. That means the resistor is linear. The linearity property is what triggers our ability to use superposition to help solve a circuit.

(Refresh me on the meaning of linearity.)

Using superposition to help solve a circuit

(This is a "toy" example to give you a feel for using superposition.)

Let's say the input to our function is two voltages in series:

The input to our function is two batteries in series:

v_{i} = Vs1 + Vs2

The function is

f (v) = \frac{1}{R} v

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 the KVL equation around the loop:

Vs1 + Vs2 - i R = 0

and solve for

i

i = f (Vs1 + Vs2) = \frac{Vs1 + Vs2}{R}

(conventional solution)

Solution using the principle of superposition

The principle of superposition applies to a linear function,

f

f (x_{1} + x_{2}) = f (x_{1}) + f (x_{2})

It says: If you have two inputs superimposed,

(x_{1} + x_{2})

, you can apply the inputs one at a time,

(x_{1})

followed by

(x_{2})

, and then add the individual results to get the full answer.

Now let's use the principle of superposition to solve the circuit. Since we've modeled our circuit as a function, we can say:

i = f (Vs1 + Vs2)

is the same as

i = f (Vs1) + f (Vs2)

This suggests an intriguing possibility. It says we can compute the output current the conventional way, by applying the combined inputs

f (Vs1 + Vs2)

, or, we could get the same answer by computing the function with single inputs

f (Vs1)

and

f (Vs2)

, and adding the results together. Let's try this and see what happens.

Suppressing inputs

To apply superposition we need to apply the inputs one at a time. That means we have to turn off all inputs except for one. When we turn off an input we say it is suppressed.

What does it mean to turn off a voltage source? It means we set

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

I = 0

. That's the same as replacing the current source with an open circuit.

Using superposition

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

When we zero out or suppress an input, we replace one of the inputs with

0

, allowing the one remaining input to shine through by itself.

f (Vs1 + 0) \to f (Vs1)

and

f (0 + Vs2) \to f (Vs2)

Now we solve each circuit individually,

i_{1} = \frac{Vs1}{R} i_{2} = \frac{Vs2}{R}

where

i_{1}

is the current caused by source

Vs 1

, and

i_{2}

is the current caused by source

Vs 2

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

i = i_{1} + i_{2}

i = \frac{Vs1}{R} + \frac{Vs2}{R}

i = \frac{Vs1 + Vs2}{R}

(superposition solution)

Check it out! The superposition solution is the same as the conventional solution obtained above.

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

Our example function was so simple, using 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 linear circuit with two sources: one current source and one voltage source. The two sources are the inputs to the function. For this problem we happen to want to find two outputs, currents

i_{1}

and

i_{2}

i_{1} = f_{1} (Is, Vs)

and

i_{2} = f_{2} (Is, Vs)

Let’s 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, the two output currents are:

i_{1 V} = 0 i_{2 V} = \frac{Vs}{R 2}

Where

i_{1 V}

and

i_{2 V}

are the currents in

R 1

and

R 2

caused by the voltage source.

Next, we restore the current source and suppress the voltage source, to calculate the contribution of the current source acting alone.

With just the current source, the two output currents are:

i_{1 I} = Is i_{2 I} = 0

Where

i_{1 I}

and

i_{2 I}

are the currents in

R 1

and

R 2

caused by the current source.

We complete the analysis by adding the contributions from each source:

i_{1} = i_{1 V} + i_{1 I} = 0 + Is = Is

i_{2} = i_{2 V} + i_{2 I} = \frac{Vs}{R 2} + 0 = \frac{Vs}{R 2}

The full solution looks like this:

This could have been a tricky analysis because the two sources make it more difficult to write node or loop equations. We exploited superposition, which gave us two simpler circuits to deal with.

Example 2

Conventional solution

For the following linear circuit let’s calculate the output voltage

v

We will do it the conventional way first. We write Kirchhoff's Current Law at output node

v

\begin{array}{ccc} + i_{R 1} & - i_{R 2} & + Is & = 0 \\ + \frac{Vs - v}{R1} & - \frac{v}{R2} & + Is & = 0 \end{array}

We can rearrange this to get an expression for

v

and gather like terms together on the right side:

v = \frac{R2}{R 1 + R 2} Vs + \frac{R 1 R 2}{R 1 + R 2} Is

(conventional solution)

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 ___.

The circuit collapses down to two resistors in series (a voltage divider).

Voltage

v_{V s}

is the contribution from voltage source

Vs

With just the voltage source, the output voltage is:

v_{V s} = Vs \frac{R 2}{R 1 + R 2}

Now we restore the current source and suppress the voltage source.

How would you suppress the voltage source?

Replace it with a ___.

The circuit collapses down to two resistors in parallel.

Voltage

v_{I s}

is the contribution to the output from current source

Is

v_{I s} = Is \frac{R 1 \cdot R 2}{R 1 + R 2}

We complete the superposition analysis by adding the two voltage contributions. As predicted, we get the same result as the conventional solution shown above.

v = v_{V s} + v_{I s}

v = \frac{R2}{R 1 + R 2} Vs + \frac{R 1 \cdot R 2}{R 1 + R 2} Is

(superposition solution)

There is no approximation involved. The solutions are exactly the same. The key thing to notice is that the two simpler circuits took significantly less work to analyze.

Linearity and superposition are useful tools

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 simpler circuits that happen to be sitting on top of each other. It seems like magic, but this property means that overlapping inputs and 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, we can use superposition to simplify the analysis. This is especially useful for circuits with multiple input sources.

To analyze a linear circuit with multiple inputs, you suppress all but one input or 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.

Suppressing sources

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

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

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Sort by:

Jack Wu
Posted 7 years ago. Direct link to Jack Wu's post “For the example two, the ...”
For the example two, the equation rearranges part on how to get V is kind confusing to me. Can someone please simply explain to me? Thank you very much!
Button navigates to signup pageComment on Jack Wu's post “For the example two, the ...”
(4 votes)
Answer
- Willy McAllister
  Posted 7 years ago. Direct link to Willy McAllister's post “When I did this algebra I...”
  When I did this algebra I took a shortcut I will share it with you. (I should have written out the full derivation like Sal always does, sorry.)
  Starting with the second equation in Example 2 Conventional Solution:
  - Separate the first fraction into separate numerators. Vs/R1 - v/R1
  - Group together the two fractions with v in the numerator. v/R1 + v/R2
  - Factor out v. v (1/R1 + 1/R2).
  - Here's the trick: You might recognize that resistor expression from two parallel R's
  Rparallel = 1/(1/R1 + 1/R2) = (R1 R2)/(R1+R2)
  - Substitute the last term for the resistor reciprocals, flipping it over the right way:
  
  v ( R1+R2)/(R1 R2) = Is + Vs/R1
  
  Finish up by moving the big R fraction over to the right side.
  
  This substitution I did with the resistor expression is fully worked out in
  https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-resistor-circuits/a/ee-parallel-resistors where it says Special Case - Two Resistors In Parallel. It's a good one to memorize.
  Comment on Willy McAllister's post “When I did this algebra I...”
  (8 votes)
asadkhan11223369
Posted 7 years ago. Direct link to asadkhan11223369's post “plz upload video tutorial...”
plz upload video tutorial its very helpfull to understand.. :)
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Glenn D'Abrera
  Posted 7 years ago. Direct link to Glenn D'Abrera's post “Great link, thanks for th...”
  Great link, thanks for that. Had'nt come across that before myself .
  Button navigates to signup page
  (1 vote)
GrayJedi
Posted 8 years ago. Direct link to GrayJedi's post “To be honest I was search...”
To be honest I was searching for the detailed information about Superposition topic which includes questions like two given point charges separated by a distance and you determine the location where the electric field is zero.I was wondering whether I am able to find it or some different topic which helps me to understand it better because I dont see this topic at school and I will be really appreciated If someone helps
Button navigates to signup pageComment on GrayJedi's post “To be honest I was search...”
(1 vote)
Answer
- Jingbin Huang
  Posted 8 years ago. Direct link to Jingbin Huang's post “This topic is covered in ...”
  This topic is covered in the physics section, under electric fields. This video might be helpful for your particular example. https://www.khanacademy.org/science/physics/electric-charge-electric-force-and-voltage/electric-field/v/net-electric-field-from-multiple-charges-in-2d
  Comment on Jingbin Huang's post “This topic is covered in ...”
  (7 votes)
mohamad
Posted 5 years ago. Direct link to mohamad's post “it is written that" R2 is...”
it is written that" R2 is shorted out. There is no voltage across R2, and therefore there is no current through it."how do we know that there is voltage across R1?
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(3 votes)
Answer
Yashvin Vedanaparti
Posted 8 years ago. Direct link to Yashvin Vedanaparti's post “How do you deal with depe...”
How do you deal with dependent sources when doing superposition?
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(3 votes)
Answer
Aetriox
Posted 5 years ago. Direct link to Aetriox's post “Hello, I have been going ...”
Hello, I have been going through the modules for Circuit Analysis but do not seem to see anything on Source Transformations, Thevenin and Norton Theorems and Operational Amplifiers. I was wondering if someone could guide me to a link on Khan Academy or perhaps external links. Thank you and fantastic work.
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(2 votes)
Answer
- Willy McAllister
  Posted 5 years ago. Direct link to Willy McAllister's post “This section of spinningn...”
  This section of spinningnumbers covers Thévenin, Norton, source transformation, and an improved version of the superposition article.
  
  https://spinningnumbers.org/t/topic-dc-analysis-special.html
  
  Here is coverage of opamps. https://spinningnumbers.org/t/topic-amplifiers.html (Same as the KA section of the same name.)
  Button navigates to signup page
  (1 vote)
kgillo
Posted 3 years ago. Direct link to kgillo's post “Is it necessary to turn o...”
Is it necessary to turn off all but one source? If I thought it might be a quicker path to the answer, would the superposition theorem be valid if I turned off all but two or three sources to minimize the number of times I need to reanalyze a similar circuit?
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(1 vote)
Answer
- Willy McAllister
  Posted 3 years ago. Direct link to Willy McAllister's post “I suppose you could gener...”
  I suppose you could generate a sub-circuit with two or more independent sources and solve it. There's nothing stopping you from doing that. You should be confident, though, that you can solve that circuit quicker than two simpler circuits.
  
  The Superposition Method is based on single-source sub-circuits. I don't think there would be a way to describe a reliable method for knowing when an n>1-source sub-circuit would for-sure lead you to a solution.
  
  For a spectacular application of Superposition, see this proof of Thevenin's Theorem... https://spinningnumbers.org/a/thevenin-proof.html
  Button navigates to signup page
  (1 vote)
Sahil Shah
Posted 3 years ago. Direct link to Sahil Shah's post “How do you solve example ...”
How do you solve example 1 using a conventional solution?
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(1 vote)
Answer
- Willy McAllister
  Posted 3 years ago. Direct link to Willy McAllister's post “The simple circuit in Exa...”
  The simple circuit in Example 1 (I-source, V-source, and 2 resistors) is actually super easy to solve with Mesh Current. (So maybe it wasn't the best example for a superposition problem.)
  
  Assign the bottom node to ground, and define two mesh currents, i1 and i2.
  
  i1 falls out easily because it flows through the current source. i1 = IS.
  
  i2 is almost as simple. The voltage across R2 is VS. So i2 = VS/R2.
  
  Done.
  
  The one somewhat interesting variable is the voltage at the top of the current source. What's that?
  Button navigates to signup page
  (1 vote)
Joshua Kimeto
Posted 3 years ago. Direct link to Joshua Kimeto's post “examples of non linear el...”
examples of non linear elements include?
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(1 vote)
Answer
- Willy McAllister
  Posted 3 years ago. Direct link to Willy McAllister's post “Examples of on-linear ele...”
  Examples of on-linear elements: diodes, transistors, switches, heater elements (incandescent bulbs).
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  (1 vote)
Thomas Burs
Posted 7 years ago. Direct link to Thomas Burs's post “Hi - thanks for those gre...”
Hi - thanks for those great tutorials.
Small mistake in the 'Using Superposition' section:
"Now we solve each circuit individually,
i1=R⋅Vs1i2=R⋅Vs"
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(1 vote)
Answer
- Willy McAllister
  Posted 7 years ago. Direct link to Willy McAllister's post “Thanks for catching the e...”
  Thanks for catching the error. It has been fixed.
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  (1 vote)

Electrical engineering

Course: Electrical engineering > Unit 2

What we're building to

Describing a circuit as a function

A resistor is a linear function

Using superposition to help solve a circuit

Conventional solution

Solution using the principle of superposition

Suppressing inputs

Using superposition

Example 1

Example 2

Conventional solution

Solution using superposition

Linearity and superposition are useful tools

Summary

Suppressing sources

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