Purpose: In this lab we will get reacquainted with the oscilloscope, determine the inductance of an inductor, verify the resonance frequency and find the phase angle, φ, of an circuit. Equipment: Oscilloscope, athode ay or Digital DMM meter Function Generator Inductor (5H, toroidal) 7 KΩ resistor, 00Ω resistor, or esistance Box x 0-9 Farad capacitor or apacitance Box Patch ords, BN connectors, Alligator lips Terminal Board, Wire Jumper Kits Theory When a dc (direct current) voltage source, such as a battery, is applied to an electrical circuit, the resulting current flows only in one direction, as the polarity (+ and -) of the voltage remains constant. For an ac (alternating current) circuit voltage source, however, the polarity of the voltage alternates with time and the direction of the current flow in the circuit alternates with the same frequency as that of the voltage source. One of the most commonly used ac voltages is one that varies sinusoidally with time. This may be generally described by the equation V ( t) = Vm sin πft Eq. where V m is the maximum voltage amplitude and f is the frequency of the source. The angle θ = πft = ωt is called the phase angle. The current in an ac circuit may or may not be in phase with the voltage, depending on the nature of the components in the circuit. In any case, if the applied voltage is sinusoidal, the current I will also be sinusoidal, an as a function of time may be expressed I( t) = I m sin(πft φ) Eq. where I m is the maximum amplitude of the current and φ is the phase constant. In Eq., the phase constant of the voltage has been assumed to be zero for simplicity. In this case, φ is also the phase difference between the applied voltage and the resulting current. (More generally, φ = φ I φ V and is the angle between I m and V m, as illustrated in Fig..) of 9
V m I m V m Im π/ω π/ω 3π/ω π/ω 5π/ω 3π/ω t -I m -V m φ Phase onstant (in terms of phase angle t = θ/ω) Figure ac Voltage and current. Here, the voltage leads the current by a constant phase difference φ. The phase constant φ in Eq. can be either positive or negative. When it is positive, the maximum value of the current I m is reached at a later time than the maximum value of the voltage, V m. In this case, we say that the voltage leads the current or the current lags behind the voltage by a phase difference φ. Similarly, if φ is negative, the current leads and the voltage lags. When there is a capacitive element in the circuit [an circuit, Fig (a)], the alternate charging and discharging of the capacitor opposes the current flow. This opposition is expressed as capacitive reactance, X, and X = Eq. 3 πf where is the value of capacitance (in farads, F). The unit of X is ohms. Similarly, when there is an inductive element in an ac circuit [an circuit, Figure (b)], the self-induced counter emf in the induction coil opposes the current. The inductive reactance, X, is given by X = πf Eq. 4 where is the inductance of the coil (in henrys, H). The unit of X is ohms. Many ac circuits have both capacitive and inductive reactance elements. The combined opposition to the current flow of resistive and reactive elements in a series circuit as shown in Figure (c) (a series circuit) is expressed in terms of the impedance, Z, of the circuit, which is given by Z = [ + ( X X ) ] = + πf πf The unit of Z is also in ohms. ( is the total resistance of the circuit. In general, it is assumed that the resistance of the induction coil is negligible compared with the resistor element.) Eq. 5 of 9
H H H V G V V (a) circuit G (b) circuit G (c) circuit Figure ircuit diagrams for (a), (b) and (c) series circuits. For the series circuit, it can be shown that the phase relation of the voltage and current is given by X X tan φ = Eq. 6 This relationship is often represented in a phasor diagram, in which resistance and reactances are added like vectors (Figure 3). Note that the angle φ is either positive or negative, depending on whether the inductive or capacitive reactance is greater. If X is greater than X, φ is positive and the current lags behind the applied voltage in time. The circuit is then said to be inductive. Similarly, if X leads X, φ is negative and the current leads the voltage. In this case, the circuit is said to be capacitive. [Eq. 6 can be applied to single reactance circuits as in Figure (a) and (b) by letting the appropriate reactance be zero.] X X Z φ (X X ) (a) Inductive ircuit X X φ (X X ) Z (b) apacitive ircuit X φ = 0 Z = X (c) esistive ircuit Figure 3 Phasor diagrams for (a) an inductive circuit, (b) a capacitive circuit, and (c) a resistive circuit. In a phasor diagram, resistances and reactances are added like vectors. 3 of 9
Since the voltage and the current are continually changing in an ac circuit, it is convenient to consider effective or time-averaged values of the voltage and current. These root-mean-square (rms) values are given by V = Vm and I = I m, where V m and I m are the maximum or peak voltage and current, respectively. The rms values of V and I are the values read on most ac voltmeters and ammeters. An Ohm s law type relationship holds between these values and the impedance, Z: V = IZ Eq. 7 and it follows that V = m I mz Thus, for a given applied voltage, the smaller the impedance of a circuit, the greater the current in the circuit ( I = V/Z ). Notice in the reactance term in Eq. 5 that there is a minus sign and that the individual reactances are reciprocally frequencydependent. As a result, for given and values, the total reactance can be zero for a particular frequency when X = X. That is, and solving for f, π f = 0 Eq. 8 πf f r = Eq. 9 π where f r is called the resonance frequency. In this condition, the circuit is said to be in resonance. The impedance is then equal to the resistance in the circuit, Z =, and the circuit is resistive [Figure (c)]. Since the impedance is a minimum at the resonance condition, the maximum current flows in the circuit from the voltage source, and maximum power P = I Z = I. For fixed values of and, resonance occurs at the particular resonance frequency f r given by Eq. 9. However, notice from Eq. 8 that for a given source frequency, resonance can also be obtained by varying and/or in the circuit. In this experiment, it is desired to measure the phase difference between the applied voltage and current in ac circuits, and to investigate the resonance condition in a series (or parallel) circuit using an oscilloscope. Notice from Eq. 6 that for the resonance condition, X = X, the voltage and current are in phase (φ = 0), since tan φ = 0 (and tan 0 = 0). Suppose that different voltage signals are applied to the horizontal and vertical inputs of an oscilloscope. If the ratio of the horizontal and vertical frequencies is an integral or half-integral, then a stationary elliptical pattern such as in Figure 4 is observed. Assume that the applied voltages have the forms x = Asin πft y = B sin(πft φ) Eq. 0 where A and B are the amplitudes (i.e. the voltages V x and V y ). [ompare with Eqs. and that give the voltage V(t) and circuit current I(t), respectively.] Note that the y 4 of 9
intercepts, +b and b, occur when x = 0, or when πft = 0 (since sin 0 = 0). Then from the second equation, b = B sin(-φ). Hence, b sin φ = ± B and similarly a sin φ = ± Eq. A You should be able to prove and understand that if A = B and φ = 90 o, the trace would be a circle. Also, if A = B and φ = 0 o, the trace is a straight line. To measure the phase angle difference of the voltage and current in a circuit, the voltage signal from the signal generator is applied to the horizontal input. For example, for the circuits in Figure, connections are made to points H and G, where G is the horizontal ground terminal. The voltage input signal has the form of Eq.. B Y x = A sin πft y = B sin (πft φ) b a A X Ellipse centered at origin Figure 4 The elliptical pattern used to measure the phase angle difference The current signal is applied to the vertical input with connections made to points V and G, where G is to ground. This is actually a voltage input from across the resistor in the circuit. However, it is proportional to and in phase with the current through the resistor (and therefore through the entire series circuit). Hence, the phase angle difference between the voltage and current can be determined from the shape of the resulting elliptical oscilloscope pattern, as described above. Experiment and Analysis: Part A:. With the meter, determine a good value of the resistance of the 7KΩ resistor (unless you are using the resistance box), the inductor, and capacitor. If these values are considerably different than the stated values, check the batteries of your meter!. Hook up circuit as shown in the sketch, remembering to connect the resistor to the negative (ground) output of the function generator. Use the nano-farad capacitor and 5 of 9
the 5 Henry inductor if available. Otherwise, using the lowest capacitance value on the capacitance box. H V G Figure 5 3. onnect hannel of the oscilloscope across the function generator [positive (red) lead to point H, negative (ground) lead to point G]. Set the function generator to 000 Hz and 5 Vpp. 4. onnect hannel of the oscilloscope across the resistor [positive (red) lead to point V, negative (ground) lead to point G]. Display the voltage across the resistor. Note that by changing the frequency of the function generator, you can change the amplitude of the voltage. Why? 5. Display both channels of the oscilloscope at the same time, and measure the time phase shift t between the current (actually the voltage across the resistor), and the voltage across the function generator. 6. onvert this to an angular phase shift, φ, from the relation: φ = π t T Where t is the horizontal displacement between the two peaks (hannel and hannel ), and where T is the period of the sin wave generator (/f). Make sure that you have a good value for the period (i.e. check the setting on the generator vs. the oscilloscope reading). 7. ompare the value of φ from Step 5 to the value expected from theory (see Part below). 8. epeat Steps 5-7 for a frequency of 500 hertz. 6 of 9
Part B: esonant frequency. eplace the 7 KΩ resistor with a 00 Ω resistor and in order to make the resonant peak sharper (Why will this happen?). Also be sure to use a nano-farad capacitor (Use the meter to get a good value) to increase the resonance frequency.. Using the best value for, calculate the resonant frequency. 3. With the oscilloscope connected across the resistor, vary the frequency of the function generator and determine the highest voltage across the resistor. ecord this value (don't worry about being too precise). 4. eplace the oscilloscope across the resistor with a DMM, and use it to indicate the voltage across the resistor (essentially, you a measuring the current across the resistor since I = V/ and is constant.) 5. Vary the frequency, in increments of -5% of your resonance frequency f (in Hz), from about 75% below your value for resonant frequency to about 75% above. Try to get a total of about 0 points. ecord values of V, and f. When you are finished with this section, plot V vs. f and determine the fractional half width of the resonance curve: ω/ω. ompare this value to the theoretical value ω ω = Z 3. Note that Z equals the total resistance in the circuit this includes the resistance of the capacitor and inductor (be sure to measure them!). Part : Phase Angle, The Old-Fashioned Way. Set f = 000 Hz. eturn the 7 KΩ resistor to the circuit. Use the same capacitance as in Part A.. Set hannel and hannel of the oscilloscope to ground, and be sure that both horizontal lines are along the x-axis (i.e. the vertical deflection for both channels when grounded should equal zero!). 3. eset the voltage levels to D, and set the time/div knob to "X-Y mode.". Obtain an elliptical pattern on the screen of appropriate size for measurement by adjusting the Horizontal and Vertical Gains. Adjust the Intensity and Focus controls to obtain a sharp pattern. 4. Make the measurements on the shape of the elliptical trace required to determine the phase difference angle φ. (Measure b and B or a and A for convenience, sin b/b = sin b/b). Use Figure 4 as a reference, if needed. 5. ompare φ as measured in Part A and Part, Step 4 to each other and to the theoretical value: X X φ = tan. 7 of 9
NOTE: Be sure to compare radians with radians, or degrees with degrees! 6. If time permits, repeat Part Step for the inductor and capacitor (instead of measuring the voltage across the resistor). Sketch a graph of v (t), v (t) and ε(t). esults: Write at least one paragraph describing the following: what you expected to learn about the lab (i.e. what was the reason for conducting the experiment?) your results, and what you learned from them Think of at least one other experiment might you perform to verify these results Think of at least one new question or problem that could be answered with the physics you have learned in this laboratory, or be extrapolated from the ideas in this laboratory. The theory section of this lab was shamelessly borrowed from: Jerry D. Wilson. Physics aboratory Experiments, nd Edition. exington MA: D.. Heath and ompany, 986. Experiment 40. 8 of 9
lean-up: Before you can leave the classroom, you must clean up your equipment, and have your instructor sign below. How you divide clean-up duties between lab members is up to you. lean-up involves: ompletely dismantling the experimental setup emoving tape from anything you put tape on Drying-off any wet equipment Putting away equipment in proper boxes (if applicable) eturning equipment to proper cabinets, or to the cart at the front of the room Throwing away pieces of string, paper, and other detritus (i.e. your water bottles) Shutting down the computer Anything else that needs to be done to return the room to its pristine, pre lab form. I certify that the equipment used by has been cleaned up. (student s name),. (instructor s name) (date) 9 of 9