Superposition Theorem

Superposition theorem is one of those strokes of genius that takes a complex subject and simplifies it in a way that makes perfect sense. A theorem like Millman’s certainly works well, but it is not quite obvious why it works so well. Superposition, on the other hand, is obvious.

The strategy used in the Superposition Theorem is to eliminate all but one source of power within a network at a time, using series/parallel analysis to determine voltage drops (and/or currents) within the modified network for each power source separately. Then, once voltage drops and/or currents have been determined for each power source working separately, the values are all “superimposed” on top of each other (added algebraically) to find the actual voltage drops/currents with all sources active. Let’s look at our example circuit again and apply Superposition Theorem to it:

Since we have two sources of power in this circuit, we will have to calculate two sets of values for voltage drops and/or currents, one for the circuit with only the 28 volt battery in effect. . .

. . . and one for the circuit with only the 7 volt battery in effect:

When re-drawing the circuit for series/parallel analysis with one source, all other voltage sources are replaced by wires (shorts), and all current sources with open circuits (breaks). Since we only have voltage sources (batteries) in our example circuit, we will replace every inactive source during analysis with a wire.

Analyzing the circuit with only the 28 volt battery, we obtain the following values for voltage and current:

Analyzing the circuit with only the 7 volt battery, we obtain another set of values for voltage and current:

When superimposing these values of voltage and current, we have to be very careful to consider polarity (voltage drop) and direction (electron flow), as the values have to be added algebraically.

Applying these superimposed voltage figures to the circuit, the end result looks something like this:

Currents add up algebraically as well, and can either be superimposed as done with the resistor voltage drops, or simply calculated from the final voltage drops and respective resistances (I=E/R). Either way, the answers will be the same. Here I will show the superposition method applied to current:

Once again applying these superimposed figures to our circuit:

Source Transformation

• Recall that we discussed how to combine networks of resistors to simplify circuit analysis – Series combinations 

– Parallel combinations – 

Delta ↔ Wye Transformations 

• We can also apply this idea to certain combinations of sources and resistors

• Performing source transformations.

 • Replacing series or parallel resistors by an equivalent resistor. 

• Replacing series voltage sources by an equivalent voltage source.

 • Replacing parallel current sources by an equivalent source source

We will first go over voltage source transformation, the transformation of a circuit with a voltage source to the equivalent circuit with a current source.

In order to get a visual example of this, let's take the circuit below which has a voltage source as its power source:

Using source transformation, we can change or transform this above circuit with a voltage power source and a resistor, R, in series, into the equivalent circuit with a current source with a resistor, R, in parallel, as shown below:

We transform a voltage source into a current source by using ohm's law. A voltage source can be changed into a current source by using ohm's formula,I=V/R.

Example

Let's do an actual example to demonstrate the mathematics of ohm's law, using the circuit shown below:

Here, we have a circuit with a voltage source of 10V with a resistor in series of 2Ω.

To calculate what the equivalent current source would be, we calculate it using the formula: I=V/R, which is I= 10V/2Ω= 5A. So the equivalent circuit would be:

The new power source is now a 5A current source. The resistor value, however, as with all source transformations stays the same. The only thing that changes is it is now in parallel for a current source transformation.

Try out our calculator below. With this calculator, you can try out as many examples as you want. The calculator does source transformations and presents the new circuits with the new values.

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Current Source Transformation

We will now go over current source transformation, the transformation of a circuit with a current source to the equivalent circuit with a voltage source.

In order to get a visual example of this, let's take the circuit below which has a current source as its power source:

Using source transformation, we can change or transform this above circuit with a current power source and a resistor, R, in parallel, into the equivalent circuit with a voltage source with a resistor, R, in series, as shown below:

We transform a current source into a voltage source by using ohm's law. A voltage source can be changed into a current source by using ohm's formula,V= IR.

Example

Let's do an actual example to demonstrate the mathematics of ohm's law, using the circuit shown below:

Here, we have a circuit with a current source of 2A with a resistor in parallel of 3Ω.

To calculate what the equivalent current source would be, we calculate it using the formula: V= IR, which is V= 2A*3Ω = 6V. So the equivalent circuit would be:

The new power source is now a 6-volt voltage source. The resistor value, however, again, as with all source transformations stays the same. The only thing that changes is it is now in series for a voltage source transformation.

Again, you can try as many examples as you would like if our calculators below, which do source transformations. 

With Dennis Matildo