Week 4 day 2

Part 1 of Lecture:
Linearity and homogeneity on linear circuit:

kv = kiR

In our first exercise, we used mesh analysis to find the current when our  dependent voltage source is equal to 12 volts and 24 volts:

The current in each mess at 24 V is exactly doubled in value compared to 12 V.

In our next exercise,  we used this linearity property to calculate voltage at any given resistor by using proportion and backward solving method:

Using Superposition method to solve current flowing through 5 ohm resistor:

 First, we set voltage equal zero ( short circuit) and calculate the current flows through 5 ohm resistor. Next, we set current source equal to zero (open circuit).

Pre Lab: 

we used  superposition method  to solve for potential difference across the 6.8 K ohm resistor:

We calculated the potential drop when 3 V is turned on and 5 V source is turned off to be 0.708 V.

Next, we calculated the potential drop when 5 V is turned on and 3 V is turned off (shorted) to be 1.99 V. By using superposition method we expected a voltage drop of 3.7 V when both voltage source is on.

We measured each resistor value using digital multimeter and apply some pretty filter to the picture so we can upload it to instagram later:

Below are the complete values of all measured resistors:

Building the circuit on the breadboard:

Next we set our Voltmeter to measure the potential drop across 6.8 kΩ :

The voltage across 6.8 kΩ when 5V source is on and 3 V source is off, It's exactly the same with our calculated values:

The voltage across 6.8 kΩ when 3 V source is on and 5 V source is off, It's exactly the same with our calculated values:

the Measured values across the resistor is really far from our predicted values, we ( More like Tony ) deduced that there might  internal resistance or the voltage is not completely off across the 5 V. so we decided to short the circuit completely by removing the 5 V power supply out from our circuit:

We obtained a similar value with what we calculated.
Finally, we turned on both voltage source and measured the resistance across 6.8 KΩ:

Below are the table comparing our experimental voltage drop and calculated voltage drop:

Lecture Part 2:

Using Source Transformation (series to parallel and parallel to series ) to solve Linear circuit :

Summary:

In our class today we learned to use linearity property to solve a current or voltage across a certain element in a linear circuit. We also created and learned how to use every circuit app to solve a problem. We learned Superposition method to solve a current by shorting ( v=0) and open (i=0) a circuit and adding all the values together. In our Lab we prove that superposition works nicely to solve circuits that will be complicated to solve if we use either nodal analysis or mesh analysis.