Pre-Lab Read through this entire lab. Perform all of your calculations (calculated values) prior to making the required circuit measurements. You may need to measure circuit component values to obtain your calculated values. All calculations should be written on a separate piece of paper (or in your lab notebook). They should be legible and written so that someone else can clearly understand your thought process. This is to demonstrate your understanding of the material, as well as aid in the troubleshooting process. Introduction Real Power: Recall that power is the flow of energy. You should be familiar with real or average power, which is measured in watts (or joules per second). Real power is the only type of power than can do useful work. Of the resistor, capacitor, and inductor, the resistor is the only circuit element that can dissipate real power. A resistor in a circuit could represent any load that takes power from the circuit, such as a light bulb, a heating element, or a motor. For a purely resistive load, the current through and voltage across it will always be in phase. This means the product of voltage across and current through the load, known as instantaneous power p(t), will always be positive. p(t) Average power T 1 T 2 t (s) Figure 1: Instantaneous Power for a Resistor Reactive Power: Inductors and capacitors are circuit elements that store energy temporarily, then give all of their stored energy back to the circuit. The net result of this energy transfer is always zero joules, so inductors and capacitors dissipate zero real power. However, since power is continuously flowing in and out of these circuit elements in an AC circuit, there is still movement of charge (current) that occurs. When a circuit is built, it must be able to handle all possible current loads, and so ignoring the effects of inductors and capacitors on a wire s current load could be a costly mistake. The power that flows to and from capacitors and inductors must be accounted for. This type of power is called reactive power and is measured in volt-amps reactive (AR). In an AC circuit, all voltages and currents fluctuate at the same frequency as the circuit s source. While all of the capacitors in a circuit are absorbing power from the circuit, all of the inductors are putting power back into the circuit. Likewise, while all of the capacitors are putting power 2014 Dan Kruger 1
into the circuit, the inductors are absorbing power from the circuit. For this reason, opposite signs are given to the reactive power of each of these elements. Inductors dissipate positive reactive power, and caps dissipate negative reactive power. Q 2 2 L L = ILXL = X L Eqn 1 p(t) i L (t) v L (t) t (s) Figure 2: Instantaneous Power for an Inductor Q 2 2 C C = ICXC = X C Eqn 2 i C (t) v C (t) p(t) Figure 3: Instantaneous Power for a Capacitor t (s) Apparent Power, Power Factor, and the Power Triangle: Most actual electrical loads have a combination of resistances, capacitances, and inductances. Both real and reactive loads must be considered when determining how much current flow a circuit s conductors must be able to handle. Ironically, complex numbers can be used to simplify AC power analysis. We will use the complex plane (real and imaginary axes) to represent these quantities. 2014 Dan Kruger 2
ince real power (P) is the only type of power that can do useful work, it is given the real axis. Reactive power (Q) does not do useful work, so it is given the imaginary axis. The hypotenuse connecting these two orthogonal vectors is known as apparent power () and is measured in voltamps (A) Q P P Q Figure 4: Power Triangle for a Resistive & Inductive Load Figure 5: Power Triangle for a Resistive & Capacitive Load P: real power, in watts (W) Q: reactive power, in volt-amps reactive (AR) : apparent power, in volt-amps (A) : angle between apparent and real power vectors, in degrees ( ) Apparent power is often used because it is what can indicate how much current will actually be drawn by a load. Loads are often specified in how much reactive power they dissipate. For example, if you are supplying a machine with 240 rms at 60Hz, and it is rated to dissipate 7.2kA, you could figure out how much current this requires as follows. 7200 I = = = 30A 240 Remember that even though reactive power does no useful work, its presence requires a higher current load. A large amount of reactive power requires more current, and thus more losses in power transmission lines (wires) as the current flows back and forth between the power source and the reactive elements. ince real power is what does useful work, it is ideal to have a load dissipate 100% real power, and 0% reactive power. The power factor (F P, no units) is a measure of how much real power a load s apparent power consists of. The power factor is the ratio of real power to apparent power. F p P = Eqn 3 When a load has 0 AR of reactive power, then its real and apparent powers will be equal, and a power factor of unity (one) is achieved. This is ideal from an efficiency standpoint. When a load is dissipating 100% reactive power, the power factor will be zero. As you can see from Figures 4 and 5, the power triangle is a right triangle. This lets us use many trigonometric functions to relate all the quantities shown. 2014 Dan Kruger 3
cos sin ( ) ( ) P FP = = Q = = P + Q 2 2 Eqn 4 Eqn 5 Eqn 6 Complex Power and the Complex Conjugate: The vector is a complex number power and is measured in volt-amps (A), just like its magnitude. Eqn 7 = P + jq = Another way to find the complex power dissipated by a load is to multiply the voltage across the load times the complex conjugate of the current through the load. * = I The asterisk is the symbol for complex conjugate. It operates on a complex number and flips the sign on its angle, if expressed polar form. If the number is in rectangular form, the complex conjugate flips the sign on the imaginary term. ee the example below. H = 4+ j3= 5 36.9 * H ( j ) j ( ) * 36.9 * 36.9 = 4+ 3 = 4 3= 5 = 5 o here is another way of looking at complex power. = I = I I * I I = I = I = I = so, = I Eqn 8 2014 Dan Kruger 4
Required Equipment: LCR Meter Oscilloscope Function Generator R = 1kΩ C = 47nF Procedure 1) In the circuit in Figure 6, use the nominal values to solve for the circuit current s magnitude and phase for a source frequency of 1kHz. 0 = 5 RM C 47nF R I 1kΩ R Figure 6: eries RC Circuit 2) Knowing that the current is the same everywhere in this series circuit, use Equation 8 to calculate the complex power dissipated by this circuit. Use your results with Equation 7 to obtain the circuit s real and reactive powers. These will be your theoretical results for real, reactive, and apparent power. 3) Create an Excel table like the one in Figure 7, and populate the Theoretical column with your results. f = 1kHz % or Absolute Theoretical Measured Error I (ma) 1.416 I ( ) 73.55 Figure 7 (ma) 7.08 ( ) -73.55 P (mw) 2.00 Q (mar) -6.79 4) Assemble the circuit in Figure 6 using the proper magnitude and frequency for your voltage source. 2014 Dan Kruger 5
5) Devise a way to measure the circuit current s magnitude (RM) and phase angle and record these values into your table. 6) Use your measured current s magnitude and phase along with Equations 7 and 8 to obtain your measured values for real, reactive, and apparent power. Fill these values into your Measured column. Compare magnitudes using percent error and phase angles using absolute error. 7) Repeat steps 1 through 6 for a circuit frequency of 3386.3Hz. 8) Repeat steps 1 through 6 for a circuit frequency of 20kHz. 9) As frequency increases, is the circuit getting more or less capacitive? How do you know this? Is the circuit current leading or lagging the source voltage? 2014 Dan Kruger 6