Linear Circuits 2 - How to find the impedance and power factor of an unknown AC load?

Question

We are on our last Linear Circuits 2 lab demo (Link to experiment: https://www.ece.ucf.edu/labs/EEL3123/projects/project-5-ac-load-measurement/) and in it, we will be handed a "black-box" in which an unknown/hidden configuration of reactive and resistive elements called the "AC load" resides and we have to solve for two things using the available lab equipment: the impedance value of the load and the power factor of the load (at a specified frequency). I have made a rough strategy for solving for both of these, but I was hoping someone could offer some insight, critique my strategies, and inform me if I'm on the right track or not? Thank you, it's much appreciated! Please refer to the link I provided if anything I explained is not clear.

Note: We (the students) are prohibited from directly measuring resistance, capacitance, and inductance using the lab equipment. Only voltage and current may be directly measured.

1) Strategy for finding impedance value of AC load using lab equipment:

1a) Connect the "black-box" on the breadboard and input the specified AC voltage signal using the function generator.

1b) Use the digital multimeter, set it on AC voltage reading, and measure V(rms) across AC load.

1c) Use the digital multimeter, set it on AC current reading, and measure I(rms) through AC load.

1d) Use this equation to find Z(magnitude): Z(magnitude) = V(rms)/I(rms).

Lab Objective 1 attained: impedance value, Z(magnitude), of AC load

2) Strategy for finding power factor of AC load using lab equipment:

1a) Now place a resistor (R) of 1 kOhm on the breadboard in series with the circuit, between the AC load and the ground.

1b) Use the oscilloscope, hook up channel 1 to measure the voltage across the load and resistor R.

1c) Use the oscilloscope, hook up channel 2 to measure the voltage across just resistor R.

1d) Make sure both channel 1 and 2 are scaled to the same volts/div. Our goal is to make the graph of the voltage across R look almost like a straight line (have very very small amplitude) as compared to the graph of the voltage across the load and R.

1e) If channel 2's waveform doesn't have a very small amplitude compared to channel 1's waveform, replace the resistor R with an even smaller resistor value. Do this until channel 2's waveform almost looks like a straight line as compared to channel 1's waveform (on the same volt/div scale).

1f) Now, enlarge channel 2's waveform (by decreasing ONLY channel 2's volts/div scale on the oscilloscope) until you can see it clearly, like how you see channel 1's waveform.

1g) Channel 2's waveform that you see is actually the signal representation of the current through our AC load (we have just indirectly graphed current on an oscilloscope!). Using the cursors on the oscilloscope, compute the amount of time (seconds) channel 2's waveform lags/leads channel 1's waveform.

1h) Get the power factor angle by using following equation: (amount of lag/lead in seconds) * (360 degrees)/(Period of channels' waveforms in seconds) = power factor angle (in degrees)

1i) Now that we have the power factor angle, we can compute power factor (pf): pf = cos(power factor angle)

Lab Objective 2 attained: power factor, pf, of AC load

I also made a more pictographic series of steps for my strategies, if one wishes to look at them here:

https://app.screencast.com/0qi4f1y3xwwKZ

Answer 1

Hi Kinza,

You are on the correct path. However, once you scale the current of the Rs to represent the current in the circuit, you need to change the probes in channel 1 to accurately measure the voltage across the ZL, not the ZL + Rs.

Good luck