Power in AC Circuits AC 2 Fundamentals Exercise 1: Power Division EXERCISE OBJECTIVE When you have completed this exercise, you will be able to determine ac power division among the components of an RLC circuit by using standard power formulas. You will verify your results with an oscilloscope. DISCUSSION In this circuit, the total real power (P T ) dissipated as heat is in the resistance (R1 and R2). The reactive components draw apparent power (S) from the generator (V GEN ) during one half-cycle and supply reactive power (Q) during the other half. The apparent power (S) supplied by the generator is computed from the generator voltage (V GEN ) and the circuit current (I). S = V GEN x I 124 FACET by Lab-Volt
AC 2 Fundamentals Power in AC Circuits Due to the phase shift introduced by the reactive components, the apparent power (S) is greater than the real power (P). In this series circuit, the circuit current (I) is common to all components. Can you determine the power value of each component by multiplying the rms current (I) by each rms component voltage drop? a. yes b. no The total real power (P T ) dissipated as heat in this circuit is a. 3 W. b. 1.5 W. c. 4.5 W. T ) delivered by V GEN, total the reactive power from each reactive component (+ for inductive and for capacitive). Q T = +Q L1 + ( Q C1 ) In the circuit shown above, the total reactive power (Q T ) introduced by C1 and L1 is a. 1 var. b. 2 var. c. 3 var. To determine the apparent power (S), which is the power supplied by the generator, multiply the rms circuit current by the rms generator voltage (V GEN ). In the circuit shown above, the apparent power (S) is a. 4.5 W. b. 1 var. c. 9.5 VA. FACET by Lab-Volt 125
Power in AC Circuits AC 2 Fundamentals 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. The rms value of the 6 ma pk-pk circuit current is 2.1. 6 mapk-pk Irms = 0.707 2 I rms = 2.1 ma rms 126 FACET by Lab-Volt
AC 2 Fundamentals Power in AC Circuits In the following steps, you will use the ADD-INVERT method to measure the voltage drop across R2. You will then convert your peak-to-peak measurement to an rms value and calculate the real power dissipated in the circuit. With the circuit current (I) 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. Then convert your measurement to an rms value. V R2(rms) = V rms (Recall Value 1) FACET by Lab-Volt 127
Power in AC Circuits AC 2 Fundamentals Using the rms value of circuit current (I) and your calculated rms voltage drop across R2 (V R2(rms) ), compute the real power (P) in R2. I rms = 2.1 ma rms V R2(rms) = V rms (Step 5, Recall Value 1) P R2 = I rms x V R2(rms) P R2 = mw (Recall Value 2) In the following steps, you will measure the individual voltage drops across L1 and C1. You will then convert your peak-to-peak measurements to rms values and calculate the reactive power (Q) in L1 and C1. 128 FACET by Lab-Volt
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. Then convert your measurement to an rms value. V L1(rms) = V rms (Recall Value 3) Using the rms value of circuit current and your calculated rms voltage drop across L1, compute the reactive power (Q) in L1. I rms = 2.1 ma rms V L1(rms) = V rms (Step 7, Recall Value 3) Q L1 = I rms x V L1(rms) Q L1 = mvar (Recall Value 4) FACET by Lab-Volt 129
Power in AC Circuits AC 2 Fundamentals 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. Then convert your measurement to an rms value. V C1(rms) = V rms (Recall Value 5) Using the rms value of circuit current and your calculated rms voltage drop across C1, compute the reactive power (Q) in C1. I rms = 2.1 ma rms V C1(rms) = V rms (Step 9, Recall Value 5) Q C1 = I rms x V C1(rms) Q C1 = mvar (Recall Value 6) Using your values of reactive power, determine the total reactive power (Q T ) delivered by the generator. Q L1 = mvar (Step 8, Recall Value 4) Q C1 = mvar (Step 10, Recall Value 6) Q T = Q L1 + ( Q C1 ) Q T = mvar (Recall Value 7) 130 FACET by Lab-Volt
AC 2 Fundamentals Power in AC Circuits Using your values of circuit current and generator voltage, calculate the apparent power (S) delivered by the generator. V GEN = 15 V pk-pk I rms = 2.1 ma rms S = I rms x V GEN(rms) S = mva (Recall Value 8) Compare your value of real power (P) to your value of apparent power (S). Is all the power delivered by the generator dissipated in R2 as heat? a. yes b. no P = mw (Step 6, Recall Value 2) S = mva (Step 12, Recall Value 8) CONCLUSION Three types of power are associated with ac circuits: real power (P), reactive power (Q), and apparent power (S). Real power (P) is the average power dissipated as heat in a resistance. It is measured in watts (W). Reactive power (Q) has no power dissipated as heat. It is measured in volt-amperes reactive (var). Apparent power (S) of the generator is the total rms circuit current multiplied by the rms generator voltage. It is measured in volt-amperes (VA). FACET by Lab-Volt 131
Power in AC Circuits AC 2 Fundamentals 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 With the oscilloscope probes as shown, use the ADD-INVERT method to measure the peak-to-peak voltage drop across R2. Then convert your measurement to an rms value. 132 FACET by Lab-Volt
AC 2 Fundamentals Power in AC Circuits V R2(rms) = V rms (Recall Value 1) Use your value of V R2(rms) (Recall Value 1) and the new value of resistance (3.2 k ) to determine the real power (P) dissipated in R2. a. 45.8 mw b. 5.7 mw c. 18.3 mw d. 5.8 mva 2. Real power (P) is dissipated as heat in a. resistance. b. reactance. c. capacitance. d. inductance. 3. The unit of measure for apparent power (S) is the a. watt. b. volt-ampere reactive. c. volt. d. volt-ampere. 4. To calculate the average power in ac circuits, use a. rms values of current and voltage. b. peak-to-peak values of current and voltage. c. peak values of current and voltage. d. average values of current and voltage. 5. A circuit consisting of three series resistors dissipating 6 W each has a total circuit dissipation of a. 6 W. b. 2 W. c. 18 W. d. 36 W. NOTE: Make sure all CMs are cleared (turned off) before proceeding to the next section. FACET by Lab-Volt 133