BLOG 9 - ELECTRICAL CIRCUIT 1

THIS BLOG CONTAIN ABOUT THE SUPERPOSITION

SUPERPOSITION

 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.

to apply the superposition principle, we must keep two things in mind:

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.

3 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.

EXAMPLE:

Solution

The voltage V can be found by solving three subproblems (one circuit for each source).

In each subproblem below, just one source is kept, while the others are deactivated (zeroed out). For any superposition problem, you get as many subproblems as there are sources in the circuit. Note that even if they are the same type (like both voltage sources), they are solved separately. In this case there are three sources (1mA, 2V, 3V), so we have three subproblems.

Subproblem 1

The first subproblem is shown below, where only the current source is kept in the circuit. The 2V is zeroed out yielding a 0V source, or a short. The 3V is zeroed out, again making it a short.

The 9K ohm is shorted out now (0 in parallel with 9K is 0):

Now solve for the voltage (this is just part of the total answer). First find the current flowing right through the 5K using a current divider equation:

Now find the voltage by Ohm's law:

Subproblem 2

Now we keep the 2V source and deactivate the other two. Zeroing out the 1mA current source yields a 0 A source, or an open. Zeroing out the 3V source yields a 0 V source, or a short. We then have:

Again, the 9K resistor is shorted out. Note that the current through the 4K resistor will be zero and thus we can eliminate this resistor as well. That leaves us with:

Let's redraw the circuit with the source on the left. Be sure to maintain the same connections and the same polarity on the voltage V.

Now apply a voltage divider (twice) to get the voltage V. Note that the voltage V is the opposite direction expected for a voltage divider, so we need a negative sign as well:

Subproblem 3

Finally, keep the 3V source and deactivate the rest. The 1mA current source becomes an open, the 2V source becomes a short:

We can again eliminate the 4K resistor because no current can flow through it. We then have:

Now use a voltage divider (twice), and again note the voltage is the wrong direction, so we need a negative sign:

Final Solution

To find the total V, add the answers from each subproblem:
V = 2.27 - 0.28 - 0.56 = 1.43V

LEARNING

I learn about the superposition is we need to turn off the other independent source and we need to remain one independent source. We can use the other analysis to solve the superposition like mesh analysis and nodal analysis. When we get the output voltage and current we add the two result.