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184879251137.J. O'Brien; 11/22/02 Page 1 of 3 D. Day 10/18/06 07/10 HEFW J. O'Brien; Montgomery College-Rockville-PH231Page 12/1/20 Page 3 of 3 J. O'Brien; 11/22/02 D. Day 10/18/06 07/10 HEFW PH262 AC Circuits: Resonance Lab # ____ Name ______________________________________________ Date _________ Lab Partner(s) Name ___________________________________________________________ I. BACKGROUND INFORMATION When all three types of components are present in the same series circuit, it becomes possible for the inductive and capacitive reactances to at least partially cancel each other. The above graph for such an RLC circuit shows the behavior of the resistor voltage amplitude, and of the phase between current (VR/R) and emf, as the driving frequency is varied over some appropriate range. Not at ResonanceAt Resonance The left hand impedance diagram shows a circuit whose inductive reactance is about three times as large as its capacitive reactance. Since inductive reactance (XL) and capacitive reactance (XC) are always 180° out of phase, they are subtracted to obtain the value of the net reactance, X, which is then combined with R in a Pythagorean sum to give Inspection shows that reducing the frequency from this point will reduce the inductive reactance while increasing the capacitive ... hence some frequency must exist where the two reactances will be equal in size, leaving the impedance equal to the resistance as shown in the right hand diagram. Since this is the lowest possible value for Z in this particular circuit, it also will have the largest possible I, and this maximum current will be exactly in phase with the driving emf. This condition, called "resonance", occurs at the frequency such that which is seen to be identical to the natural frequency for undamped oscillations. The power delivered to the resistor by the generator is maximum when the circuit is at resonance. III. EXPERIMENTAL PROCEDURE Login to everycircuit.com 1.Connect a series combination of L = 22mH, C = .001µF, and R = 470 ohms across the sine generator in such a way that the input voltage and the resistor voltage can be simultaneously displayed on the oscilloscope. CH1 connected across function generator CH2 connected across resistor Voltage across resistor Input voltage 2.Go to the tool at the bottom left corner of the simulation. Select the sinusoidal waveform for the input signal. Choose the frequency from the menu. Rotate the flywheel at the bottom right corner of the simulation to locate resonance frequency. Look at VR on the o’scope and watch for a maximum as you vary for the input signal. At resonance frequency the amplitude of VR is almost equal to the input signal. These two signals overlap and can be separated. Hover the arrow on the signal screen. Use the zoom in or zoom out in your mouse to separate these signals as depicted below. A time bar scale is given on the x-axis. Adjust the sinusoidal wave in such a way that the crest of a wave is at the beginning of the time bar scale. Measure the length of the time bar scale carefully with the metric ruler. Find 1mm is equal to how many milliseconds. Measure the distance between adjacent hump. To read the period from the scope and compare with the expected value for this LC pair. Repeat for several other LC combinations until you are confident that you understand how to locate the resonance. Input voltage Time period 4.6cm=50us 2.7cm= 29.32us 3.Select the frequency from the tool menu. Rotate the flywheel to change the frequency and collect enough data to plot accurate graphs of VR vs . To plot a graph of vs , measure the length of the time bar scale which is given on x-axis. Adjust the sinusoidal wave in such a way that it starts from one side of the time bar scale. Measure the length of the time bar scale carefully with the metric ruler. Find 1mm is equal to how many milliseconds. Measure the time period and time interval . THINK! before you measure. You are graphing a very non-linear curve and you need to carefully choose the points to measure. Note that at resonance () the phase is 0 and VR is maximum and nearly equal to (the signal generator). Be certain to note the frequency values and VR voltages at which phases ±45° are obtained. You will be asked to show in the question at the end of the lab write-up that, at ±45°, the power delivered to the resistor is 1/2 of that delivered at resonance, and hence these frequencies are called the half-power points. Select the frequency from tool menu Rotate the flywheel to change the frequency. length of time bar scale 4.5cm=50us 1cm =11.11us 4.Finally, exchange your resistor for one that is a factor of ~2 or 3 larger. Locate the resonance. To obtain half power points for this circuit. Choose the input voltage (. Find the root mean square value. . Go to the tool at bottom left corner. Select the frequency from the tool menu. First locate the resonance frequency. Then rotate the flywheel counterclockwise to adjust the frequency until reaches to . At this point you will have . Now rotate the flywheel clockwise until reaches to . At this point you will have . You should note that the Df between the half power points is significantly larger than for the lower resistance circuit. A circuit with a narrow resonance peak (a small Df) is called a high-Q circuit, where "Q" is the common abbreviation for the expression "Quality Factor" (in spite of the obvious danger of confusion with "q" for charge). Specifically, Q is given by and is thus seen to be related to the resonant frequency and to τ (1/e-life) for damped oscillations. Find the Q’s for the ~500 Ω R and for the R that is ~2 or 3 times larger. Compare the values. A narrow resonance peak is associated with a long time-constant, ie. with low damping. The Q value is of particular importance in most circuit design applications, eg. a radio tuner that must separate signals from two stations broadcasting at carrier frequencies that are close together. IV. CONCLUSIONS/QUESTIONS 1.Notice that at resonance the signal voltage, e, is pointed along VR (phase = 0). So PR (power dissipated by R) is VR2 / R = e2 / R, and this represents the maximum power dissipation. Then Pmax = e2 / R. Now suppose that = 45º = π/4. Show that PR = e2 / 2R = ½ e2 / R = ½ Pmax. Thus ±45º phase represents the half-power points on the resonance curve, VR vs. ω. 2.In section 5 above you are given the formula for Q for an RLC circuit. where and For a mass-spring system we don’t have R, L, and C, but we do have ω0 and τ where and τ = τe = the 1/e lifetime for the amplitude of the oscillation, either electrical or mechanical. Look back at your results for ω0 and τe for the mass-spring system in Lab #2 and calculate Q for the mass-spring system. You should have found that ω0 ≈ 2π rad/sec and τe ≈ 200 s. How does this value of Q compare to the Q for the RLC circuit? Q f f R L C = = = = 0 0 0 1 2 D D w w w t 2 1 0 t w = = C L R Q LC 1 0 = w R L 2 = t m K o / = w 231.ACCircuits March’99; J. O’Brien Page 6 of 6 10/06, D. Day; 07/10 HEFW PH262 AC Circuits: Phase Relationships Lab # ____ Name ______________________________________________ Date _________ Lab Partner(s) Name ___________________________________________________________ I. BACKGROUND INFORMATION The rotating phasor representation for series AC circuits should be familiar from textbook and lecture notes. A brief outline of the essential points is provided here. If a series RLC circuit is connected across a source of emf which is a sinusoidal function of time, then q and all its derivatives will also be sinusoids. Since all elements in a series circuit share the same current, it is customary to write i = I Cos(t) as the reference function against which the others will be compared. The voltage across each component is then expressed in terms of the current as: where lower-case symbols are used for time-varying quantities, and upper-case symbols for the amplitudes. The ratio of amplitudes VR/I is seen to be the DC resistance, R. The ratios of the V to I amplitudes for the other two components, which must also have units of ohms, are given the symbol X and are called "reactance", leading to the name "reactive components" to describe any L or C in a circuit. Examination of the rightmost terms in each of these equations shows the relative phase of the three voltages. vR has the same phase as i [both Cos], vC [Sin] is 90° behind i, and vL [-Sin] is 90° ahead of i. These relationships are frequently expressed by phasor diagrams, which show the voltage amplitudes summed with the appropriate phases to equal the amplitude of the driving emf. VR VL E L C E VR VC Figure 1 Above are the voltage phasor diagrams for an RL and an RC circuit. As is conventional, the phasor rotation is assumed to be counterclockwise. Thus, it is seen that the external emf will be ahead of the current in the inductive circuit, and behind the current in the capacitive circuit. Inspection of these diagrams shows that the trigonometric functions of the phase angle can be expressed as ratios of the voltages, for example It is therefore possible to determine the relative phases of such signals just by making appropriate voltage measurements, which can be done either with an oscilloscope or with an AC voltmeter or a DMM set for AC measurements. Remember that AC