Wednesday, June 10, 2015
Tuesday, June 9, 2015
26-May-2015: Apparent Power and Power Factor
PURPOSE
The purpose of this lab was to implement what we learned about apparent power and power factors in a real-life situation.
PRE-LAB
Before setting up our experiment, we did some initial calculations. We first found the Thevenin impedance, which allowed us to find the RMS current (Irms) and voltage (Vrms) across the load. In turn, we were able to calculate the average (Pavg) and apparent power (S) from these values. Finally, we took the ratio of these two variables to find the power factor (Pf). However, after looking over the values, particularly Zth, we realized that we did the calculations incorrectly and were unable to compare them to the experimental values.
PROCEDURES
After completing our initial calculations, we set up the circuit as shown in Figure 2. We used channel 1 of our oscilloscope to measure the input voltage and channel 2 to measure the voltage across the load. Then, we set up a math channel to find the current through the load. Then, we had the oscilloscope display the RMS values of these measurements.
This was the resulting graph when RT = 10 Ω. As it can be seen from the image, Vrms and Irms were 599.6 mV and 68.12 mA, respectively. From these values, we were able to find the apparent power delivered to the load (S = VrmsIrms), which was equal to 40.84 mW. When compared to the theoretical value of 11.86 mW, they were nowhere near each other. As mentioned before, we could only assume that we made an error in our pre-lab calculations. In addition, we found the average power delivered to the load with the equation PL = VrmsIrmscos(θv - θi). However, in order to do so, we had to first find the phase angle between Vrms and Irms with the equation θv - θi = Δt/T * 360°. By utilizing the graph, we were able to find the phase angle to be 17.06°. This gave us an average power of 39.04 mW, which gave us a power factor of 0.956.
When RT = 47 Ω, Vrms was 129.2 mV and Irms was 68.11 mA. This gave us an apparent power of 8.800 mW. Since there wasn't a phase shift between the voltage and the current, the average power delivered to the load was the same as the apparent power. Therefore, the power factor was equal to 1.
Next, we replaced RT with a 100 Ω resistor. The resulting graph is shown in Figure 5. From the graph, we were able to find that Vrms and Irms were 201.4 mV and 68.27 mA, respectively. From these values, we found S to be 13.75 mW. Then, we found the phase angle to be 54.18°. This gave us a Pavg of 8.047 mW and a Pf of 0.585.
The purpose of this lab was to implement what we learned about apparent power and power factors in a real-life situation.
PRE-LAB
 |
| Figure 1 |
Before setting up our experiment, we did some initial calculations. We first found the Thevenin impedance, which allowed us to find the RMS current (Irms) and voltage (Vrms) across the load. In turn, we were able to calculate the average (Pavg) and apparent power (S) from these values. Finally, we took the ratio of these two variables to find the power factor (Pf). However, after looking over the values, particularly Zth, we realized that we did the calculations incorrectly and were unable to compare them to the experimental values.
PROCEDURES
 |
| Figure 2 |
After completing our initial calculations, we set up the circuit as shown in Figure 2. We used channel 1 of our oscilloscope to measure the input voltage and channel 2 to measure the voltage across the load. Then, we set up a math channel to find the current through the load. Then, we had the oscilloscope display the RMS values of these measurements.
 |
| Figure 3: RT = 10 Ω |
This was the resulting graph when RT = 10 Ω. As it can be seen from the image, Vrms and Irms were 599.6 mV and 68.12 mA, respectively. From these values, we were able to find the apparent power delivered to the load (S = VrmsIrms), which was equal to 40.84 mW. When compared to the theoretical value of 11.86 mW, they were nowhere near each other. As mentioned before, we could only assume that we made an error in our pre-lab calculations. In addition, we found the average power delivered to the load with the equation PL = VrmsIrmscos(θv - θi). However, in order to do so, we had to first find the phase angle between Vrms and Irms with the equation θv - θi = Δt/T * 360°. By utilizing the graph, we were able to find the phase angle to be 17.06°. This gave us an average power of 39.04 mW, which gave us a power factor of 0.956.
 |
| Figure 4: RT = 47 Ω |
When RT = 47 Ω, Vrms was 129.2 mV and Irms was 68.11 mA. This gave us an apparent power of 8.800 mW. Since there wasn't a phase shift between the voltage and the current, the average power delivered to the load was the same as the apparent power. Therefore, the power factor was equal to 1.
 |
| Figure 5: RT = 100 Ω |
Next, we replaced RT with a 100 Ω resistor. The resulting graph is shown in Figure 5. From the graph, we were able to find that Vrms and Irms were 201.4 mV and 68.27 mA, respectively. From these values, we found S to be 13.75 mW. Then, we found the phase angle to be 54.18°. This gave us a Pavg of 8.047 mW and a Pf of 0.585.
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