22-Mar-2016: Day 9

Exercise 1:

After redrawing a complex circuit into its Thevenin equivalent, then we may easily calculate the power of the load resister. As seen below, by simply converting current into its voltage-resistance equivalent, a value may be calculated.

The question is then, what would the graph of power as a function of the variable load resistor look like? Below is a sketch which turns out to be the product of two more well known functions.

The sketch of the graph is not exactly what is depicted on the right of the image. What was meant to happen was a graph that increases rapidly, slows as down until reaching a maximum power value, then gradually descends until it almost reaches zero... looking more like the following image.

What this tells us is that there exists a value for the load resistor that will provide a maximum value of power. 

It  turns out that when the load resister is equal to that of the Thevenin equivalent resistance, maximum power is achieved.

A simplified equation for power may then be written as below in terms of only the Thevenin voltage and resistance.

Exercise 2:

Given the circuit below we are to:

1. Obtain the Thevenin equivalent at terminals a and b.
2. Calculate the current through the load resistor at 4, 8, 12, and 16 Ohms.
3. Find the load resistance value for maximum power.
4. Determine the maximum power value.

Using some source transformations the circuit may be rewritten. Current throughout is equivalent since we are looking at a circuit that is in series, therefore, it is simply the Thevenin voltage over the total resistance. Previously it was mentioned that maximum power occurs when the load resistor is equivalent to that of the Thevenin resistor. Power is now a simply expressed in terms of Thevenin voltage and resistance.

Exercise 3:

Below, on the upper-right is a diagram with three motors connected in parallel with a single power source and some equivalent resistor.

Along with the diagram, we are given the resistance of the lower motor which is brushless. For the upper two motors, CIM, a spec chart is provided and we are to calculate the resistance of each motor.

Below is a better image of the calculations. Once the CIM motor's internal resistance is determined, the three motors may be treated as resistors and added accordingly. A maximum power may now be calculated by stating that the Thevenin resistance is equal to the already calculated load resistance and solving using a 24 Volt source.

Exercise 4:

Max Power Lab

We have learned through Thevenin's theorem that maximum power is delivered to the load when the load resistance is equal to that of the Thevenin resistance of the circuit. The purpose behind wanting to utilize maximum power is because it results in minimum loss of power as heat dissipation.

Above we constructed a simple circuit where we have a controlled voltage source, feeding into the circuit, and a variable resistor (i.e. potentiometer) that we will control at 1 kilo-Ohm  increments.

Below, is a table of voltage out, the voltage across the variable resistor, so we may calculate power at the load.

Our good friend MATLAB has come to the rescue again, providing us with the following plots.

Blue: Ideal plot
Orange: Measured plot
Light Blue: 4th Polynomial Fit
Red Dot: Max Power

Table of max values.

m1 = Voltage Out

m2 = Thevenin Resistance (constant)

m3 = Load Resistance

m4 = Power

As we can see, the load and Thevenin resistance are not exactly equal, but rather "close enough". Looking back at the graph and our data, it is possible that smaller increments of load resistance could have improved our values.