**The Superposition Theorem makes it easy for us to analyse circuits with multiple power sources. Before using it, first, we have to understand two important terms: Linearity and Superposition.**

## Linearity & Superposition

### Linearity

Let’s have an **amplifier**. Usually, an amplifier has a volume knob on it but for the sake of simplicity our amplifier does not. **It amplifies the input music by A.**

**The input is essentially just a voltage signal**– the music that is fed in the amplifier – is just quickly changing voltage levels.

**Think of A as a number:**The signal from your phone or mp3 player is weak. It has to be amplified multiple times (example 40 times) for your speakers to play it loud enough.

If our amplifier is linear, it does not matter what level of signal we feed into it (what is the volume level on your phone). It will always amplify the music** A** times. The speakers will always play A times louder than your phone’s output is.

In other words: the gain is always **A** and is __not dependent__ on the input voltage.

### Superposition

2 or more different signals can be **superposed**. What does it mean? It simply means they can be **added up**. So we add them up, feed them in the amplifier and the amplifier will amplify the sum **A** times.

If the amp is linear, what we get at the output is the same what we would get if we amplified the signals separately and added them up after they’ve been amplified.

**Linearity: ****A component is linear if the output is proportional to the input. **In case of an amplifier, the output will always be **A** times louder regardless of the input signal’s level.

**Superposition: **A linear system (amplifier) has a certain response ( amount of amplification). **The Superposition Theorem states that in case of a linear system, the system’s response to multiple sources separately is the same as the response for the same sources inputed at the same time.**

♦

**When to use the Superposition Theorem?**

- We’ve got a circuit below where we have
**two Power Sources**(voltage sources in this case)

- We want to find the
**Current I1**

- The Superposition Theorem is a 3 step process:
**Short E2 and calculate I1 using E1 only****Short E1 and calculate I1 using E2 only****Add the two I1 together**

**When analysing a circuit with multiple power sources, we can use the Superposition Theorem. **

**We calculate the current in question considering only one power source at a time and add up the results. We do this by replacing all other voltage sources with a short and current sources with an open circuit.**

♦

**Solved Example 1**

**Find the current in the 4 Ohm branch.**

Let’s solve this problem first using **Kirchhoff’s Law** and next using the **Superposition Theorem**:

**Solution 1: Using Kirchhoff’s Law:**

- We have
**2 voltage sources: E1**and**E2**

**r1**and**r2**are the internal resistances of**E1**and**E2**

- First, let’s mark the nodes:
**ABCDEF**

- Next, let’s define the current flow:
- It is up to us in which direction we label the current, but here we do it this way:
- Let’s say the main current
**I**comes from**E1**and at node**B**it splits to**I1**and**I2**. **E2**is also a source, that is true. It would fire out the current in the opposite direction, but let’s say**E1**is stronger.- Choosing our own interpretation of the current flow works because we are going to take 2 different loops and as long as the currents have the same direction in the two loops we get the correct result.

The idea is to choose two loops and write down two voltage equations.

- Let’s work with the following loops:
**ABCDEF**and**ABEF**

- Using
**Kirchhoff’s Voltage Law**(2nd law) we write down the voltage equation for both loops by simply going around the loop.

#### Voltage equation in the **ABCDEF** loop:

- On the
__left__we have the__EMFs__

- On the
__righ__t we add up the__voltage drops__across the resistors

- The direction of the voltage drops matter.

- If we start from
**E1**and go around the loop, we see the voltage drop across**E2**is in the other direction as**E1**. Pointing against the**I2**current (red arrow).

- Since we start from
**E1***(the**reference point)*it will be positive by default.**E2**‘s voltage drop is in the opposite direction, thus,**E2**will be negative and we**subtract E2 from E1**.

- On the
__right hand side__of the equation, we simply**add up the voltage drop across the resistors**around the**ABCDEF**loop.

- We only have 2 resistors around this loop:
**r1**and**r2**.

- The voltage drop for resistors are always in the opposite direction of the current

- Voltage drop for an individual EMF goes from the negative terminal towards the positive terminal.

- From Ohm’s Law (
**V=IR**), the voltage drop for**r1**is:**V(R1)=1 x (I1+I2)**and for**r2**:**V(r2)=2 x I2**

- Note, if we were on the right side of the equation we would subtract the
**r1**and**r2**voltage drops from**E1**as the voltage drop is opposing the current that originates from**E1**. Although, we are on the right side of the equation, thus, both**r1**and**r2**will have a positive sign in front of them.

#### Voltage Equation in the **ABEF** loop:

- There is only one EMF:
**E1**in the loop, we write this on the left side. - There are two resistors:
**r1**and**R**. They go on the right side. - Voltage drop across
**r1**is**V(r1)=****4 x I1**

- Voltage drop across
**R**is**V(R)= 1 x (I1+I2)**

- Lastly, we subtract the two equations from each other in such way that we can get rid of
**I2**as we are calculating**I1**.

- If the subtraction does not cancel out
**I2**, simply multiply all members, both left and right side in one equation with a number that zeroes out**I2**after subtraction.

**Solution 2: Using the Superposition Theorem:**

- As explained above, let’s define the flow of current first (see below).

- According to the Superposition Theorem, first we replace
**E2**with a short circuit and calculate**I1**using**E1**.

- Next we short
**E1**and calculate**I1**using only**E2**.

- Finally, we
__add up__the two results.

**Note, when replacing an EMF with a short, we leave the internal resistance alone and still consider it in our calculations.**

Below you can see the two cases that we have to solve **I1** for. In the left circuit we replaced **E2** with a short. In the right circuit **E1** is replaced with a short.

#### Case 1: Consider E1 only (E2 is shorted)

- We now have a circuit with only one voltage source (
**E1**)

- We are looking for the current through
**R (I1A)**

- First, find the total current (
**IA**)

- To do this, let’s come up with an
**equivalent circuit**

- An equivalent circuit is purely theoretical. It means we can substitute on paper a bunch of components (example resistors) with one component (resistor) that has the same value as the bunch summed up.

- In an equivalent circuit we only have a power source and a resistor. Therefore, it is easy to find the current using Ohm’s law.

- First, let’s
**replace R and r2 with only one resistor**. This will be the parallel combination of the two (see in the picture above).

- Note in the solution, there is a simple trick to calculate the parallel combination: multiply the two values in the nominator and add them up in the denominator.

- Next, we see that the new resistor that we got is in series with
**r1**. We simply add this to the other resistor.

- The single resistor that we got is the
**equivalent resistance**of the circuit.

**Using Ohm’s law to find IA:**

- If you scroll back to the circuit above where we shorted
**E2**, you see that this**IA**current splits to**IA1**and**IA2**.

- This is a
**current divider**. We covered this in a previous topic.

- We are looking for
**IA1**, so there is a simple current divider equation to use:

#### Case 2: Consider E2 only (E1 is shorted)

- These steps are the same as in Case 1.
- Draw the
**equivalent circuit**and find the equivalent resistor value:

- Next, calculate the current
**IB**by using Ohm’s law**:**

- Using the
**current divider equation**(true only for 2 branches) to find**I1A**:

- Finally, sum up the results from Case 1 and Case 2 to get
**I1**. - This is our final result which is the same what we got using Kirchhoff’s Law.

**Solved Example 2**

**Find the current in the R2 branch as well as the potential at X and Y.**

**Solution:**

- Using the Superposition Theorem, we will divide the circuit into 2 individual circuits.
- First, we take out the current source and find the required voltages and currents.
- Next, we carry out the same calculation, only with the current source.
- Lastly, we sum the values together.

#### The circuit with only the voltage source only:

- Calculate the total resistance.
- Draw the equivalent circuit.
- Find the current
**I1**.

#### The circuit with the current source only

- We already know a current,
**I=5A**. - Find the total resistance in a way so we have a junction where current splits.

- Draw the circuit that we got.
**5A**will split between the two branches based on the resistor values in each branch.- Use the Curren Divider Law to calculate each current.
- Next, form
**I3**, calculate**I4**and**I5**(from above circuit).

- So we got an
**X**and**Y**value when calculating only with the voltage source and also got another**X**and**Y**value when calculating with the current source.

- Add the
**X**values together to get the potential at**X**in the circuit with both sources present.

- Add the
**Y**values together to get the potential at**Y**in the circuit with both sources present.

- Finally, we need to calculate to current in the
**R2**branch. We do this in a similar way: add**I2**from the voltage source circuit and**I4**from the current source circuit together.

- Note, I4 flows in different direction than I2, thus, it will be negative.

- We did not find I2 earlier, but it is essentially the the voltage at X in the voltage source circuit divided by the resistor value, which is 4Ω. (Ohm’s Law)

**optional reading:** Success in Electronics book by Tom Duncan

**NEXT TOPIC:** Thévenin’s Theorem

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