Exercise 2: Power Factor

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1 Power in AC Circuits AC 2 Fundamentals Exercise 2: Power Factor EXERCISE OBJECTIVE When you have completed this exercise, you will be able to determine the power factor of ac circuits by using standard electronic formulas. You will verify your results with an oscilloscope. DISCUSSION Real power (P), reactive power (Q), and apparent power (S) can be drawn on a power triangle. Real power is drawn along the horizontal axis and represents the total real power in watts. The vertical leg of the triangle represents the reactive power (difference between Q L and Q C ) in var. 134 FACET by Lab-Volt

2 AC 2 Fundamentals Power in AC Circuits The resultant, or hypotenuse, of the triangle represents the apparent power in VA. The relationship (ratio) of a circuit s real power (P) to apparent power (S) is the power factor (PF). In ac circuits containing resistance and reactance, can the real power ever be greater than the apparent power? a. yes b. no The cosine of the angle shown ( ) equals the ratio of real power (P) to apparent power (S). Real Power P PF = = Apparent Power S PF = cos This angle ( ) equals the phase angle between the voltage and current of an ac circuit. Multiply the power factor (cos ) by the apparent power (S) to calculate real power (P) of an ac circuit. P = cos x S FACET by Lab-Volt 135

3 Power in AC Circuits AC 2 Fundamentals In this circuit, the phase angle between the generator voltage and circuit current is 45. The real power (P) dissipated as heat in the resistor is a. 20 W. b W. c W. NOTE: S (in VA) = V I GEN T The power factor (PF = cos ) of a circuit simply states what part of the apparent power (S) is real power (P). P = cos x S PROCEDURE Adjust V GEN for a 15 V pk-pk, 20 khz sine wave. With the oscilloscope probe as shown, slightly adjust the frequency of the generator so that circuit current is 6 ma pk-pk (I = V R3 GEN to 15 V pk-pk. 136 FACET by Lab-Volt

4 AC 2 Fundamentals Power in AC Circuits The rms value of the 6 ma pk-pk circuit current is mapk pk Irms = x I rms = 2.1 ma rms In the following steps, you will measure the voltage drops across R2, L1, C1, and V GEN. Your peak-to-peak voltage measurements will automatically be converted to rms values. The real, reactive, and apparent powers associated with this series ac circuit will be given to you. Using the power triangle, you will determine the power factor (PF = cos ). FACET by Lab-Volt 137

5 Power in AC Circuits AC 2 Fundamentals With the circuit current adjusted to 2.1 ma rms and the oscilloscope probes as shown, use the ADD-INVERT method to measure the peak-to-peak voltage drop across R2. V R2(pk-pk) = V pk-pk (Recall Value 1) From the circuit current and your voltage drop across R2, the real power R2 dissipates as heat can be determined. P R2 = I rms x V R2(rms) P R2 = 2.1 x [ V pk-pk (Step 5, Recall Value 1 *P R2 = mw * Certain calculated answers based on measured Recall Values are not included in the Instructor Guide. The instructor may use Recall Value nominal answers from the Instructor Guide to determine if the calculated answer by the student is within the nominal value range. 138 FACET by Lab-Volt

6 AC 2 Fundamentals Power in AC Circuits With the circuit current adjusted to 2.1 ma rms and the oscilloscope probes as shown, use the ADD-INVERT method to measure the peak-to-peak voltage drop across L1. V L1 = V pk-pk (Recall Value 2) The reactive power introduced by the inductor can be determined from the circuit current and your voltage drop across L1. Q L1 = I rms x V L1(rms) Q L1 = 2.1 x [ V pk-pk (Step 7, Recall Value 2 *Q L1 = mvar With the circuit current adjusted to 2.1 ma rms and the oscilloscope probe as shown, measure the peak-to-peak voltage drop across C1. V C1(pk-pk) = V pk-pk (Recall Value 3) * Certain calculated answers based on measured Recall Values are not included in the Instructor Guide. The instructor may use Recall Value nominal answers from the Instructor Guide to determine if the calculated answer by the student is within the nominal value range. FACET by Lab-Volt 139

7 Power in AC Circuits AC 2 Fundamentals The reactive power introduced by the capacitor can be determined from the circuit current and your voltage drop across C1. Q C1 = I rms x V C1 Q C1 = 2.1 x [ V pk-pk (Step 9, Recall Value 3 *Q C1 = mvar Determine the total reactive power (Q T ) delivered by the generator (V GEN ). Q L1 = 2.1 x [ Q C1 = 2.1 x [ V pk-pk (Step 7, Recall Value 2 V pk-pk (Step 9, Recall Value 3 Q T = Q L1 + ( Q C1 ) Q T = mvar (Recall Value 4) * Certain calculated answers based on measured Recall Values are not included in the Instructor Guide. The instructor may use Recall Value nominal answers from the Instructor Guide to determine if the calculated answer by the student is within the nominal value range. 140 FACET by Lab-Volt

8 AC 2 Fundamentals Power in AC Circuits The apparent power (S) supplied by the generator is 11.1 mva. S = I rms x V GEN(rms) S = 11.1 mva Determine the power factor (PF) from your values of real power (P) and apparent power (S). P R2 = 2.1 x [ S = 11.1 mva V pk-pk (Step 5, Recall Value 1 PF = P R2 S PF = (Recall Value 5) The angle ( ) shown in the power triangle equals the phase angle between the generator voltage and circuit current. In the next few steps, you will measure this angle and use your measurement to determine what part of the apparent power (S) is dissipated as heat in resistor R2. FACET by Lab-Volt 141

9 Power in AC Circuits AC 2 Fundamentals With the oscilloscope probes as shown, measure the phase angle ( ) between V GEN and I T. = degrees (Recall Value 6) Use the cosine (cos) function on your calculator, and determine the power factor (PF) from PF = cos PF = (Recall Value 7) 142 FACET by Lab-Volt

10 AC 2 Fundamentals Power in AC Circuits Determine what part of the apparent power delivered by the generator is real power dissipated as heat in resistor R2. cos = (Step 16, Recall Value 7) S = 11.1 mva P R2 = cos x S P R2 = mw (Recall Value 8) CONCLUSION Apparent power (S), reactive power (Q), and real power (P) can be drawn on a power triangle. The cosine of the angle between the apparent power and real power is the power factor (PF = cos ). The power factor of a circuit simply states what part of the apparent power is real power. In an ac circuit with reactive components (inductors and capacitors), the generator has to be capable of supplying more power than is actually consumed. REVIEW QUESTIONS 1. GEN for a 15 V pk-pk, 20 khz sine wave. Place the CM switch 9 in the ON position to change the value of R2 from 1 k to 3.2 k FACET by Lab-Volt 143

11 Power in AC Circuits AC 2 Fundamentals With the oscilloscope probes as shown, measure the phase angle ( ) between V GEN and I T. = degrees (Recall Value 1) 144 FACET by Lab-Volt

12 AC 2 Fundamentals Power in AC Circuits Using the cosine function on your calculator, determine the power factor (PF) from your phase angle measurement (Recall Value 1). a b c d The power factor (PF = cos ) can be determined from the ratio of real power (P) to the a. reactive power (Q). b. phase angle ( ). c. circuit current (I). d. apparent power (S). 3. When you multiply the cosine of the angle (PF = cos ) by the apparent power (S), you obtain the a. real power of the circuit (P). b. reactive power of the circuit (Q T ). c. reactive power of the inductor (Q L ). d. reactive power of the capacitor (Q C ). 4. In this circuit, the phase angle between V GEN and I is 45. The power factor (PF = cos ) is a b c. 45. d In this circuit, the apparent power supplied by the generator (S = I x V GEN ) is a. totally consumed by the resistor. b. lower than the power consumed by the resistor. c. equal to the power consumed by the resistor. d. higher than the power consumed by the resistor. NOTE: Make sure all CMs are cleared (turned off) before proceeding to the next section. FACET by Lab-Volt 145

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