NETWORK THEOREMS (Superposition Theorem)

Superposition Theorem

The superposition theorem states that; “the current through, or voltage across, an element in a linear bilateral network is equal to the algebraic sum of the currents or voltages produced independently by each source”.

The strategy used in the Superposition Theorem is considering only one voltage source in a network at a time, eliminating all other sources, using series/parallel analysis to determine voltage drops (and/or currents) within the modified network for each power source separately. The voltage drops and/or currents are determined considering all the voltage sources separately, the values are then “superimposed” on top of each other (added algebraically) to find the actual voltage drops/currents with all sources active.

The number of networks to be analysed using this theorem is equal to the number of independent voltage sources considered.

We consider each of the sources independently by removing and replacing the sources without affecting the final result. To remove a voltage source when applying this theorem, we “short circuit” and to remove a current source we “open circuit”.

   (a) Voltage Source                  (b) Current Source

Figure 1: Removing a Practical voltage and current source.

Let’s illustrate the use of this theorem in examples.

Example 1:

Determine I₁ for the network of figure 1 bellow.

Figure 1

Removing the voltage source E as in figure 1.1;

Figure 1.1

Since,  R_sc||R₁, the share current I from the source between each other

  (See:Current divider rule).

This indicates the obvious, that all the current from the source flow through the path of zero resistance.

Now, removing the current source I as in figure 1.2;

Figure 1.2

The circuit in figure 1.2 depicts the 6Ω resistance in series with the voltage source E, no current flows through the open circuit as it imposes an infinite resistance.

Then superimposing the figure 1.1 on figure 1.2, we have;

I = I’ + I’’ = (0 + 5)A  =  5A.

Example 2:

Determine the current flowing in the resistors R₁, R₂ and R₃, in the circuit in figure 2 bellow.

Figure 2

Solution:

Let the current flowing in R₁, R₂, R₃ be represented by I₁, I₂, I₃ respectively. Assuming the current flow directions for the current as shown in figure;

Removing E₂ as in figure 2.1 (Network 1);

Figure 2.1

Current I flowing out of the circuit is given by;

While R_T = R₁ + R₂||R₃

Calculating R_T;

Then, 

  (See figure 2.1(b)).

I= I₁ is shared between R₂ and R₃, and the current through each of the resistors can be calculated thus,

Current I₂ flowing through R₂ :

 (See:Current divider rule).

Removing E₁ as in figure 2.2 (Network 2);

Figure 2.2

 Here, R_T  =  R₃ + R₂||R₁

Hence,  

I₃ is divided between R₂ and R₁ (see figure 2.2a).

With the result we can fill the table 1.1 bellow as;

           

Current I (A)	Network 1	Network 2	Total Current IT (A)
I1	6.0	1.0	I1_T= 6-1 = 5
I2	2.0	2.0	I2_T= 2+2=4
I3	4.0	3.0	I3_T= 4-3 =1

In the table above, I_{1_T}, I_{2_T}, I_{3_T} is the current flow through R₁, R₂ and R₃ respectively with sources E₁ and E₂ in operation. We chose the direction of current in “network 1” as positive, we have I_{1_Total}  = 6-1, because the direction of current I₁ as labelled in network two is opposite with respect to that of “network 1”, ditto for I_{2_Total} = 4-3. Obviously I_{2_Total} = 2+2 because the direction of current I₂ in the two networks are the same.

Example 3:

Using the principle of superposition, find the current I₁, I₂, I₃, I₄,  through the 6kΩ, 12kΩ, 14kΩ and 35kΩ  resistors respectively in figure 3.

Figure 3

Solution:

Considering the effect of 6mA current source without the 9V source, we have can redraw the circuit diagram as in figure 3.1, lets say network 1;

Figure 3.1: Network 1

The direction of current in the circuit has been assumed, our result will show if the assumptions are correct or needs to be reversed.

Similarly,


See current division rule @ Current Divider)

We can see that R₁||R₂ are in series with R₃||R₄ and thus the total current I is also been shared between R₃ and R₄.

Now, let’s consider the 9V supply without the current source, as in figure 3.2

Figure 3.2

Here, 9V is across R₁ in series with R₂ are connected across 9V (sharing 9V supply), and similarly R₃ in series with R₄ are also connected across 9V (sharing 9V supply).

To determine the current through each of the resistors, we will calculate the voltage drop across each of them, using voltage divider rule (Voltage Divider).

Current (A)	Network 1	Network 2	IT	Resistors
I1	I’1=4	I’’1=0.5	4.5A	R1
I2	I’2=2	I’’2=0.5	2.5A	R2
I3	I’3=4.286	I’’3=0.184	4.47A	R3
I4	I’4=1.714	I’’3=0.184	1.898A	R4

See Also:
Current divider rule
Voltage divider rule
Thevenin's Theorem