Week 5 day 1 (Thevenin's Lab)

Lecture

The first activity that we do in the class is using source transformation to find Thevenin Resistance and Thevenin's voltage source.

 Our group decided to use source transformation method to solve this problem. We calculated Thevenin resistance of 20 Ω and Thevenin Potential of 40 V. In our next activity We are calculating the maximum power output of a circuit by calculating Thevenin potential and Thevenin resistance. A circuit with that consist of an adjustable load resistance and a Thevenin counterparts reached it's maximum power output when the value of load resistance is equal to its Thevenin resistance.

We calculated a Thevenin resistance of 12 Ω, Thevenin voltage of 40 V and maximum power output, when the load resistance is equal to 12 V, of 33.33 W. tHe third activity of the day is to convert Thevenin potential and Thevenin resistor to it's current counterpart, Norton's current and Norton's resistor. In order to do this, we calculated the Thevenin potential and resistance and used source transformation to convert it to Norton's current and resistance.

Thevenin Lab 

We calculated The theoretical value of Thevenin potential and resistor. 

We calculated Thevenin Resistance of 7.7 kΩ and Thevenin potential -.456 V. We then used easy circuit to simulate our Thevenin equivalent resistor and Thevenin potential.

We shorted the independent voltage source and measured equivalent resistant of 7.7 K.

 We  measured 457 mV of potential difference in our simulation circuit.. since our calculated Thevenin voltage and resistor checks out, we started our Lab:

First, we measured each of our resistors by using DMM :

Below is the table with measured values of each resistor:

Next, we assemble our circuit and measured the open circuit voltage  to find the Thevenin voltage: 

We measured .460 V across the two open circuit. We then shorted the independent voltage source to measure the Thevenin resistor:

Here we measured Thevenin resistor of 7.81 KΩ. Next, we used potentiometer to create a thevenin Equivalent Circuit :

************************************gotta do it someday *****************************************

Next, we used our first circuit and connected the two open terminals to a potentiometer.

We Then measured the potential difference across this potentiometer. we then calculated the power running through the potentiometer using equation

where Vp is equal to the voltage across potentiometer, and Rp equals to resistance the of potentiometer. We then recorded multiple valuse of P and Rp and sketch the function :

Using Matlab we can model this Curve to find Its Maximum Power and the load resistance(potentiometer resistance needed to get there:

Summary :

Finding Thevenin equivalent circuit and Norton equivalent circuit is advantageous if we want to measure  Power and current that flows through anon fixed load resistance. We can find Thevenin equivalent circuit by substituting the fixed component of the circuit with and equivalent Thevenin resistor and a voltage source. To find the value of Thevenin voltage source we simply find the potential difference across the open circuit of the fixed part of the circuit. To calculate the Thevenin resistance we shorted all the independent sources and simply measured the resistance equivalent. We can calculate the power of load resistance by multiplying the load resistance by the square of current passes through the load resistor. Maximum power output by the circuit occurs when the value of load resistor is equal to Thevenin Resistor.

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.

Week 4 Day 1

Mesh analysis practice:

Pre-lab : Find the theoretical value of current and voltage drop across 10 K ohm resistor and the voltage drop across 6.8 K ohm resistor.

We measured  the actual resistance in 6.8 K, 22 K 4.7 K and 21.6 K resistor using digital multi meter:

Below are the measured value of each resistor with uncertainty of 0.01 K ohm

Picture of our circuit :

Measured value of potential drop across 6.8 K resistor :

Measured value of current through 10 K resistor

Measured value of potential drop through 10 K resistor:

Below are the compared theoretical and experimental values; V1 represent the voltage drop on 6.8 K resistor, V2 represent the voltage drop on 10 K resistor and I1 represent the current flow through 10 K resistor:

Transistor Lecture

Calculating current through a diode and minimum resistance needed, 0.7 V comes from the voltage drop by silicon transistor :

Converting transistor into a circuit and calculating the power dissipated through resistor :

Calculating the minimum resistance needed for transistor with 0.5 Watt capacity:

Summary 

Mesh analysis is best used on circuits with multiple series component and voltage sources, while nodal analysis is best used on circuits with multiple parallel component and current sources. Instead of solving mesh analysis algebraically by going through the entire loop, we can set up mesh equation by inspecting total resistor that a current passes through the loop. Transistor works by combining positively doped and negatively doped semi conductor. There are two type of resistor: NPN, which have a positively charged base and PNP, with negatively charge base.  Transistor works to amplify current.

Week 3 day 2

Nodal Analysis Problem:

We need to calculate the Potential difference V1, V2, V3, V4, V5 using nodal analysis method:

 We make a mistake by applying nodal analysis on the reference node for our fifth equation. To fix this problem we applied Super Nodes across V1, V3 and V5 to obtain our fifth equation:

Nodal Analysis Lab

Measured resistance value using digital multimeter:

Our first attempt at setting up our Breadboard:, we used up too much ground wire:

Second Attempt at handling our breadboard, we get rid of unnecessary ground wire, since waveform have an internal ground, we connected +5V to the red wire, -5 V to the white wire, and -3 V to the yellow waveform wire:

We measured the potential difference at 10 K Ohm resistor:

We measured the potential difference at 22 K Ohm resistor:

We measured the potential difference at 6.8 K ohm resistor:

Table Comparison of experimental and calculated V across each resistor:

Mesh Analysis Lecture :

Simple Mesh Analysis with no current source :

 Mesh Analysis With Current Source; Super Mesh:

Summary:

To calculate potential difference between resistor we can apply nodal analysis on each node compared to the reference node. We can apply a mesh analysis on a planar circuit with a lot of voltage souces. When Current source exists between two non-reference nodes; we have to apply super mesh .