09-1 Name Date Partners ab 9 A FITES AND ESONANE OBJETIES OEIEW To understand the design of capacitive and inductive filters To understand resonance in circuits driven by A signals In a previous lab, you explored the relationship between impedance (the A equivalent of resistance) and frequency for a resistor, capacitor, and inductor. These relationships are very important to people designing electronic equipment. You can predict many of the basic characteristics of simple A circuits based on what you have learned in previous labs. ecall that we said that it can be shown that any periodic signal can be represented as a sum of weighted sines and cosines (known as a Fourier series). It can also be shown that the response of a circuit containing resistors, capacitors, and inductors (an circuit) to such a signal is simply the sum of the responses of the circuit to each sine and cosine term with the same weights. ecall further that if there is a current of the form I ( t) = I sin ωt (1) ( ) flowing through a circuit containing resistors, capacitors and/or inductors, then the voltage across the circuit will be of the form ( ) sin( ω ϕ ) t = I Z t +. (2) Z is called the impedance (and has units of resistance, Ohms) and φ is called the phase shift (and has units of angle, radians). The peak voltage will be given by = I Z. (3)
09-2 ab 9 - A Filters & esonance Figure 1 shows the relationship between and I for an example phase shift of +20. We say that leads I in the sense that the voltage rises through zero a time t before the current. When the voltage rises through zero after the current, we say that it lags the current. Figure 1 The relationship between ϕ and t is given by t ϕ = 2π or ϕ = 360 f t T where T is the period and f is the frequency. For a resistor, Z (4) = and there is no phase shift ( ϕ = 0 ). For a capacitor, Z = X = 1 ω and ϕ = 90 while for an inductor, Z = X = ω and ϕ = + 90. In other words: = I sin( ωt) (5) = I X cos( ωt) (6) and = I X cos( ωt) (7) X is called the capacitive reactance and X is called the inductive reactance. et us now consider a series combination of a resistor, a capacitor and an inductor shown in Figure 2. To find the impedance and phase shift for this combination we follow the procedure we established before. Figure 2
ab 9 - A Filters & esonance 09-3 From Kirchhoff s loop rule we get: = + + (8) Adding in Kirchhoff s junction rule and Equations (2), (5), (6), and (7) yields sin ( ωt + ϕ ) = I sin ( ωt ) + ( X X ) cos ( ωt ) Once again using a trigonometric identity 1 and equating the sin t cos ω t, we get coefficients of ( ω ) and ( ) and cos( φ ) = I sin ( φ ) = ( ) I X X Hence the phase shift is given by X X tan ( ϕ ) = (9) and the impedance of this series combination of a resistor, an inductor, and a capacitor is given by: ( ) 2 Z = I = + X X (10) 2 The magnitudes of the voltages across the components are then and = I = (11) Z X = I X = (12), Z X = I X = (13), Z Explicitly considering the frequency dependence, we see that = ( ω 1 ω ) 2 + 2 (14) and,, = = ω ( ω 1 ω ) + 2 1 ω ( ω 1 ω ) 2 + 2 2 (15) (16) 1 sin( α + β ) = sin( α )cos( β ) + cos( α )sin( β )
09-4 ab 9 - A Filters & esonance This system has a lot in common with the forced mechanical oscillator that we studied in the first semester. ecall that the equation of motion was F = ma + bv + kx = mx ɺɺ + bxɺ + kx (17) Similarly, Equation (8) can be written as 1 = qɺɺ + qɺ + q (18) We see that charge separation plays the role of displacement, current the role of velocity, inductance the role of mass (inertia), capacitance (its inverse, actually) the role of the spring constant, and resistance the role of friction. The driving voltage plays the role of the external force. As we saw in the mechanical case, this electrical system displays the property of resonance. It is clear that when the capacitive and inductive reactances are equal, the impedance is at its minimum value,. Hence, the current is at a imum and there is no phase shift between the current and the driving voltage. Denoting the resonant frequency as ω and the common reactance of the capacitor and inductor at resonance as X, we see that, at resonance so and ( ω ) ( ω ) X X = X ω = 1 (19) X = (20) At resonance the magnitude of the voltage across the capacitor is the same as that across the inductor (they are still 180 out of phase with each other and ±90 out of phase with the voltage across the resistor) and is given by X, ( ω ) =, ( ω ) = (21)
ab 9 - A Filters & esonance 09-5 In analogy with the mechanical case, we call the ratio of the amplitude of the voltage across the capacitor (which is proportional to q, our displacement ) at resonance to the driving amplitude the resonant amplification, which we denote as Q, Q ω (22) Hence, ( ), Q = X = (23) Figure 3 (below) shows the voltage across a capacitor (normalized to the driving voltage) as a function of frequency for various values of Q. Figure 3 In this lab you will continue your investigation of the behavior of resistors, capacitors and inductors in the presence of A signals. In Investigation 1you will explore the relationship between peak current and peak voltage for a series circuit composed of a resistor, inductor, and capacitor. You will also explore the phase difference between the current and the voltage. This circuit is an example of a resonant circuit. The phenomenon of resonance is a central concept underlying the tuning of a radio or television to a particular frequency. INESTIGATION 1: THE SEIES ESONANT (TUNE) IUIT In this investigation, you will use your knowledge of the behavior of resistors, capacitors and inductors in circuits driven by various A signal frequencies to predict and then observe the behavior of a circuit with a resistor, capacitor, and inductor connected in series.
09-6 ab 9 - A Filters & esonance The series circuit you will study in this investigation exhibits a resonance behavior that is useful for many familiar applications such a tuner in a radio receiver. You will need the following materials: oltage probes Multimeter 510 Ω resistor test leads 800 mh inductor 820 nf capacitor onsider the series circuit shown in Figure 4 (below). [For clarity, we don t explicitly show the voltage probes.] Figure 4 Prediction 2-1: At very low signal frequencies (less than 10 Hz), will I and be relatively large, intermediate or small? Explain your reasoning. Prediction 2-2: At very high signal frequencies (well above 3,000 Hz), will the values of I and be relatively large, intermediate or small? Explain your reasoning.
ab 9 - A Filters & esonance 09-7 1. On the axes below, draw qualitative graphs of X vs. frequency and X vs. frequency. learly label each curve. X and X Frequ ency 2. On the axes above (after step 1) draw a curve that qualitatively represents X X vs. frequency. Be sure to label it. 3. ecall that the frequency at which Z is a minimum is called the resonant frequency, f and that the common reactance of the inductor and the capacitor is axes above, mark and label f and X. X. On the Question 2-1 At f will the value of the peak current, I, in the circuit be a imum or minimum? What about the value of the peak voltage,, across the resistor? Explain. 4. Measure the 510 Ω resistor (you have already measured the inductor and the capacitor): 5. Use your measured values to calculate the resonant frequency, the reactance of the capacitor (and the inductor)
09-8 ab 9 - A Filters & esonance at resonance, and the resonant amplification factor. Show your work. [Don t forget the units!] f X Q Activity 2-1: The esonant Frequency of a Series ircuit. 1. Open the experiment file 09A2-1 Filter. 2. onnect the circuit with resistor, capacitor, inductor and signal generator shown in Figure 4. [Use the internal generator.] 3. Adjust the generator to make a 50 Hz signal with amplitude of 2. 4. onnect voltage probe P A across the resistor, P B across the inductor, and P across the capacitor. 5. Use the Smart Tool to determine the peak voltages (,,, and, ). Enter the data in the first row of Table 2-1. 6. epeat for the other frequencies in Table 2-1. Table 2-1 f (Hz) (), (), () 50 100 200 400 800
ab 9 - A Filters & esonance 09-9 7. Measure the resonant frequency of the circuit to within a few Hz. To do this, slowly adjust the frequency of the signal generator until the peak voltage across the resistor is imal. [Use the results from Table 2-1 to help you locate the resonant frequency.] f, exp Question 2-2: Discuss the agreement between this experimental value for the resonant frequency and your calculated one. 8. Use the Smart Tool to determine the peak voltages at resonance.,, Question 2-3: From these voltages, calculate Q and discuss the agreement between this experimental value and your calculated one.
09-10 ab 9 - A Filters & esonance Question 2-4: alculate your experimental value of X and discuss the agreement between this value and your calculated one. Prediction 2-5: What will we get for Q if we short out the resistor? Show your work. 9. Short out the resistor. 10. Measure Q. [You may have to lower the signal voltage to 0.5.] Show your work. Explicitly indicate what you had to measure. Q Question 2-6: Discuss the agreement between this experimental value and your predicted one.
ab 9 - A Filters & esonance 09-11 Activity 2-2: Phase in an ircuit In previous labs, you investigated the phase relationship between the current and voltage in an A circuit composed of a signal generator connected to one of the following circuit elements: a resistor, capacitor, or an inductor. You found that the current and voltage are in phase when the element connected to the signal generator is a resistor, the current leads the voltage with a capacitor, and the current lags the voltage with an inductor. You also discovered that the reactances of capacitors and inductors change in predictable ways as the frequency of the signal changes, while the resistance of a resistor is constant independent of the signal frequency. When considering relatively high or low signal frequencies in a simple circuit, the circuit element (either capacitor or inductor) with the highest reactance is said to dominate because this element determines whether the current lags or leads the voltage. At resonance, the reactances of capacitor and inductor cancel, and do not contribute to the impedance of the circuit. The resistor then is said to dominate the circuit. In this activity, you will explore the phase relationship between the applied voltage (signal generator voltage) and current in an circuit. onsider again our circuit (it is the same as Figure 4). Figure 5 Question 2-7: Which circuit element (the resistor, inductor, or capacitor) dominates the circuit at frequencies well below the resonant frequency? Explain.
09-12 ab 9 - A Filters & esonance Question 2-8: Which circuit element (the resistor, inductor, or capacitor) dominates the circuit at frequencies well above the resonant frequency? Explain. Question 2-9a: In the circuit in Figure 5, will the current through the resistor always be in phase with the voltage across the resistor, regardless of the frequency? Explain your reasoning. Question 2-9b: If your answer to Question 2-9a was no, then which will lead for frequencies below the resonant frequency (current or voltage)? Which will lead for frequencies above the resonant frequency (current or voltage)? Question 2-10a: In the circuit in Figure 5, will the current through the resistor always be in phase with applied voltage from the signal generator? Explain your reasoning. Question 2-10b: If your answer to Question 2-10a was no, then which will lead for frequencies below the resonant
ab 9 - A Filters & esonance 09-13 frequency (current or voltage)? Which will lead for frequencies above the resonant frequency (current or voltage)? 1. ontinue to use 09A2-1 Filter. 2. econnect the circuit shown in Figure 5. onnect voltage probe P A across the resistor, P B across the inductor, and P across the capacitor. 3. Start the scope and set the signal generator to a frequency 20 Hz below the resonant frequency you measured in Investigation 2, and set the amplitude of the signal to 2. Question 2-11: Which leads applied voltage, current or neither when the A signal frequency is lower than the resonant frequency? Discuss agreement with your prediction. 4. Set the signal generator to a frequency 20 Hz above the resonant frequency. Question 2-12: Which leads applied voltage, current or neither when the A signal frequency is higher than the resonant frequency? Discuss agreement with your prediction.
09-14 ab 9 - A Filters & esonance Question 2-13: At resonance, what is the phase relationship between the current and the applied voltage? 5. Use this result to find the resonant frequency. f, phase Question 2-14: Discuss how this experimental value compares with your calculated one. Question 2-15: How does this experimental value for the resonant frequency compare with the one you determined by looking at the amplitude? omment on the relative sensitivities of the two techniques.