If a circuit has two or more independent sources, one way to determine the value of a specific variable (voltage or current) is to use nodal or mesh analysis. Another way is to determine the contribution of each independent source to the variable and then add them up. The latter approach is known as the superposition theorem.
The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.
The principle of superposition helps us to analyze a linear circuit with more than one independent source by calculating the contribution of each independent source separately. However, to apply the superposition principle,
We must keep two things in mind while dealing with superposition theorem:
1. We consider one independent source at a time while all other independent sources are turned off. This implies that we replace every voltage source by 0 V (or a short circuit), and every current source by 0 A (or an open circuit). This way we obtain a simpler and more manageable circuit.
2. Dependent sources are left intact because they are controlled by circuit variables.
With these in mind, we apply the superposition principle in three steps:
Steps to Apply Superposition Principle :
1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis.
2. Repeat step 1 for each of the other independent sources.
3. Find the total contribution by adding algebraically all the contributions due to the independent sources.
Analyzing a circuit using superposition has one major disadvantage: it may very likely involve more work. If the circuit has three independent sources, we may have to analyze three simpler circuits each providing the contribution due to the respective individual source. However, superposition does help reduce a complex circuit to simpler circuits through replacement of voltage sources by short circuits and of current sources by open circuits.
Keep in mind that superposition is based on linearity. For this reason, it is not applicable to the effect on power due to each source, because the power absorbed by a resistor depends on the square of the voltage or current. If the power value is needed, the current through (or voltage across) the element must be calculated first using superposition.
EXAMPLE:Use the superposition theorem to find v in the circuit.
Since there are two sources, let
v = v1 + v2
where v1 and v2 are the contributions due to the 6-V voltage source and
the 3-A current source, respectively. To obtain v1, we set the current
source to zero, as shown in Fig.(a). Applying KVL to the loop in Fig.
Note: While applying any theorem to a network other sources effect must be zero.so we make voltage source short circuit and current source open circuit.
12i1 − 6 = 0 ⇒ i1 = 0.5 A
v1 = 4i1 = 2 V
We may also use voltage division to get v1 by writing
v1 = 4/(4 + 8)*(6) = 2 V
To get v2, we set the voltage source to zero, as in Fig.(b). Using
i3 = 8/(4 + 8)*(3) = 2 A
v2 = 4i3 = 8 V
And we find
v = v1 + v2 = 2 + 8 = 10 V.