To me, I feel it is not my favorite way to solve a circuit.
The picture above is linearity problem. From the picture, we knew that if we doubled the Vs, the current i0 is doubled because of the linearity property.
After the linearity problem, we did our lab.
Superposition II
To use superposition, we need to follow these steps.
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.
(From Day 7 Notes Superposition and Source Transformation)
From the picture, we coverd the 5V voltage source first, and we got the voltage across 6.8k ohms resistor is 0.708V. Next, we covered the 3V voltage source, and we got the voltage across 6.8k ohms resistor is 1.99V. We added these two numbers together, and we got the final answer of the voltage across 6.8k ohms resistor = 2.698V.
Resistor Error:
9.8k(10k), 4.6k(4.7k), 0.975k(1k), 6.5k(6.8k), 21.7k(22k)
The picture above is the basic set up for this lab.
The picture above was when we took off the 5V voltage source, we got the voltage across 6.8k ohms resistor was 0.693V.
The picture above was when we took off the 3V voltage source, we got the voltage across 6.8k ohms resistor was 1.97V.
The picture above was when both 5V and 3V were active, we got the voltage across 6.8k ohms resistor was 2.66V.
The picture above is the summary of this lab. The % difference for each value are all less than 3%, which is very low, and this proves that superposition works.
Summary
We leaned how to use linearity and superposition today. To me, I feel superposition involved more work and it takes more time to solve one question. Maybe there are some of the questions that are easier to solve by using superposition, example: less power supply questions. But I will still use another method except superposition. haha
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