Lab Report Day 23 - Apparent Power and Power Factor

This lab assignment emphasizes the use of apparent power and power factor to quantity the AC power delivered to a load and power dissipated by the process of transmitting this power. Our goal in is to provide some necessary average power to load with minimum power lost during the power delivery process. 

To do that, we will used a 10 ohm resistor, and a series combination of 1mH inductor and resistor (10 ohm, 47 ohm, or 100 ohm) resistor to be the load. We use a sinusoidal wave with (1cos(2pi*5000t) in the circuit and calculate the power dissipated by the resistor. 

This is the set up of the lab. 

We measure the resistors to be 11.9 ohm (10), 48.6 ohm (47) and 99.8 ohm (100). Number in "()" are theoretical values. 

Those are results and calculations. The resistance is series with a load that changes each time. We measure the voltage of the load and the voltage across the resistor in series with the load.

The oscilloscope screen with 11.9 ohm resistor

The oscilloscope screen with 48.6 ohm resistor

The oscilloscope screen with 99.8 ohm resistor

This is our measured value. The phase difference are very big. 

Then we include a capacitor in parallel with the load. This can make the voltages measured more in phase.

The oscilloscope screen with 11.9 ohm resistor with a capacitor

The oscilloscope screen with 48.6 ohm resistor with a capacitor

The oscilloscope screen with 99.8 ohm resistor with a capacitor

We can see that the phase angles are very small, less than 3 degrees for each.


Summary:

Today we are told that the electric company charge us with peak voltage instead of average power, and later we are told how to avoid it. All we need to make it more efficiency and draw less current from the power source is to add a capacitor parallel to the load. Therefore, that will. And we do a lab to prove this.