05/24 - Lab 25: Apparent Power and Power Factor

Lab 25: Apparent Power and Power Factor

In this lab, we used a basic LR circuit to gain understanding of apparent power and power factor in order to determine the power delivered to a load. Apparent power, average power, and the power factor associated with the circuit were calculated on our white boards, and then measured during the experiment to compare our expectations with the actual values. The circuit used in this experiment is pictured below:

For this experiment, R_T = 10 Ohms, L = 1 mH, and the load resistance R_L will vary throughout the experiment from 10 Ohms to 47 Ohms, and finally 100 Ohms. The input signal V(t) had an amplitude of 1V @ 5 kHz. The circuit we built in the breadboard is pictured below (with R_L = 10 Ohms).

Our first set of calculations (with R_L = 10 Ohms) is pictured on our white board below:

We predicted average power P = 7.2 mW, apparent power S = 13.4 mVA, V_rms = 0.707, 0° V, and I_rms = 19, -57.61° mA. The actual results from measuring values on our circuit are pictured below:

The actual V_rms turned out to be 591 mV, which is slightly off but still fairly close to our predicted result (only off by ~0.1V), and our measured I_rms current turned out to be 18.3 mA which was only ~0.7 mA from our predicted result. The phase shift of the current from the input voltage can be seen as being ~-60°, which correlates with our predictions.

Next, we changed out R_L from the 10 Ohm resistor and replaced it with a 47 Ohm resistor. Our white board calculations are pictured below:

Our calculations yielded an average power P = 6.8 mW, apparent power S = 7.8 mVA, V_rms = .707, 0° V, and I_rms = 11, -28.8° mA. The actual results of our circuit are pictured below:

The actual voltage obtained from measuring the circuit was V_rms = 608 mV (again, ~0.1V off from our prediction), and our I_rms from measurements turned out to be 10.9 mA, which was extremely close to our prediction of 11 mA. The phase shift of the current can be seen as being slightly negative, which makes sense with our prediction of -28.8°.

Finally, we replaced R_L with a 100 Ohm resistor. Our white board calculations are shown below:

Our calculations yielded an average power P = 4.0 mW, apparent power S = 4.2 mVA, V_rms = .707, 0° V, and I_rms = 5.9, -15.9° mA. The actual measurements from our circuit are pictured below:

The actual voltave V_rms turned out to be 645 mV (~ .05V from our prediction), and the actual current I_rms was slightly off from our prediction at 7.1 mA (predicted 5.9 mA). The phase shift of the current from the input voltage can be seen as very small, and negative, which corresponds nicely to the value we predicted of -15.9°.

In Class Examples

1. Calculate the RMS value of the signal shown in below.

2. A relay coil is connected to a 210-V, 50-Hz supply. If it has a resistance of 30 Ω and an inductance of 0.5 H, calculate the apparent power and the power factor.

3. The voltage across a load is v(t) = 60 cos(ωt − 10◦) V and the current through the element in the direction of the voltage drop is i(t) =1.5 cos(ωt + 50◦). Find complex and apparent powers.

4. When connected to a 120-V (rms), 60-Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.