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

Linearity explanation with amplifier example

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

Superposition explanation with amplifier example

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:
    1. Short E2 and calculate I1 using E1 only
    2. Short E1 and calculate I1 using E2 only
    3. Add the two I1 together
Superposition Theory example ciruit

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.

Superposition Theory solved example circuit

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.
Solved Example solution circuit by Kirchhoff's Law

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 right 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.
Voltage equation around the ABCDEF loop
  • 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)
Voltage equation around the ABEF loop
  • 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.
Finding I1 by subtracting the two voltage equations from each other

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.

Solved Example solution circuit by using the Superposition Theorem

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.
Case 1: finding the equivalent resistance in
  • 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:

Case 1: Finding IA
  • If you scroll back to the circuit above where we shorted E2, you see that this IA current splits to IA1 and IA2.
  • We are looking for IA1, so there is a simple current divider equation to use:
Case 1: Finding I1A

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:
Case 2: Finding the equivalent resistance
  • Next, calculate the current IB by using Ohm’s law:
Case 2: Finding IB
  • Using the current divider equation (true only for 2 branches) to find I1A:
Case 2: Finding 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.
Finding I1

Solved Example 2

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

Example 2 circuit

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.
Example 2: Finding the total resistance for voltage source only
Calculation: finding the potental at X and Y as well as the current I3/

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.
Calculation: Calculating Resistors in Parallel and Series to find an Equivalent Circuit for the case when we consider the Current Source only..
  • 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).
Calculation: Finding I2, I3, I4 and I5 for the case when we considering the current source only.
Calculation: Finding the potential at Y and X for the current source only.
  • 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)
Calculation: Final values for X, Y and the current through the R2 branch.

optional reading: Success in Electronics book by Tom Duncan
NEXT TOPIC: Thévenin’s Theorem