When a circuit has two or more independent sources, we can determine the contribution of each source applying nodal analysis or mesh analysis. Another way is to find out the contribution of the independent sources (voltage or current) is to apply superposition theorem.
Superposition theorem is based on circuit linearity property.
The principle of superposition theorem is “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.
We can analyze the contribution of each source separately in a linear circuit with superposition. We should keep in mind two things applying superposition
- We analyze one independent source at a time and all other sources considering turned off. We replace voltage source as short circuit or 0V and current source as a short circuit or 0A. In this way we will take reading of each independent source.
- As dependent sources are controlled by other source so we will left intact dependent source.
Three steps applying superposition principle
Step 1: Turn on only one independent source and turn off all other independent sources. Find the output current or voltage for that source using KVL or KCL or other basic method.
Step 2: Do step 1 for all the independent source separately.
Step 3: Add all the contributions of independent sources algebraically and find total contribution.
I said before that the law of superposition theorem is only applicable for linear circuit. So apply this method in linear circuit.
Superposition method reduces complex equations. Although it has a disadvantage also is that we have to analyze the contributions one by one for each independent source, so here we have to do more work.
Now we solve a problem using superposition theorem and it will give clear concept about superposition.
Superposition theorem problem
Find v in the circuit in figure 1 using superposition theorem.
Let, v1 and v2 are the contributions of two sources 12V voltage source and 6A current source. According to linearity we can write,
v = v1 + v2
As per rule we will find only one independent contribution at a time and we set the all sources turned off. To obtain v1 we set current source open circuit or 0A as figure 2 circuit.
Applying KVL we get in the loop,
-12 + 16i1 + 8i1 = 0
or i1 = 0.5A
So, v1 = i18 = 4V
We can find out the value of v1 using voltage divider rule,
v1 = {8/(16+8)} 12 = 4V
To obtain v2, we set voltage source short circuit as figure 3 or 0V.
Now using current divider rule we can find i3,
i3 = {16/(8+16)} 6 = 4A
Thus, v2 = 8i3 = 32V
Finally, v = v1 + v2 = 4 + 32 = 36V
In this way we find voltage or current for a specific resistance for linear circuit.
