Name: S.N.: Experiment 2 INDUCTANCE AND LR CIRCUITS SECTION: PARTNER: DATE: Objectives Estimate the inductance of the solenoid used for this experiment from the formula for a very long, thin, tightly wound solenoid. Understand solutions for the transient behavior of a circuit containing inductance, resistances, a voltage source, and a switch. Use a digital storage oscilloscope to measure the time dependence of current through an inductor for increasing current and analyze this data to measure the inductance of the solenoid. Use a digital storage oscilloscope to measure the time dependence of current through an inductor for decreasing current and analyze this data to obtain a second measurement of the inductance of the solenoid. A. Self-inductance of a very long, thin, tightly wound solenoid. The solenoid used as the secondary or pickup coil in the Electromagnetic Induction lab will serve as the inductor for this experiment. Your first task will be to calculate its inductance. For any circuit with a current-carrying wire without magnetic materials such as iron, the magnetic field generated by the current is proportional to the current. In a circuit with N loops, let j be the magnetic flux set up by the current in the jth loop. Since the field B is proportional to the current, the flux through each loop is also proportional to the current. Therefore, the total flux total = 1 + 2 + 3 + 4 +..., [i.e. the sum of the flux from each of the N loops] is proportional to the current. If the flux through each loop is the same and is equal to where A is the area per turn, then the total flux is: j = BA, [1] total = j [2] But the flux through any single loop can be larger than what is caused by the current in that single loop by itself, especially if the loops are close enough together that they produce flux inside each other. It might be difficult then to calculate f j for any particular arrangement, but one thing we can be sure of, whatever the total flux from all the turns, it will be proportional to the current. The proportionality between the total flux and the current is expressed as: total = Li [3] where the constant of proportionality L is called the inductance, L. The SI unit of inductance is the Henry (symbol H). This equation is the definition of inductance. If the turns all have the same flux through them, then we can calculate the inductance: L = N j i It is just N times the inductance of a single turn, which we can calculate from j = BA. It might look as though L depends (inversely) on the current, but it does not, because is itself proportional to the current. L just depends on the geometry (if there are no magnetic materials nearby). This relation has an electrical analog with which you are familiar. The charge on two conductors is proportional to the voltage difference between them and the constant of proportionality is the capacitance, C, such that [4] 123LAB2-123LAB2-LRCIRCUIT.DOC 2008-04-14 2-1
Q = CV. Both C and L are purely geometrical in that their values in the absence of dielectric and magnetic materials depend on the configuration of conductors in the case of C and the configuration of loops for L. For a thin solenoid with a length long compared to its diameter, the magnetic field inside the solenoid is axial and uniform and has the magnitude B = µ 0 ni, where n is the number of turns per meter (n should not be confused with N, the total number of turns). B is directed along the solenoid axis. This result is easily derived using Ampere s Law. Knowledge of B and the coil geometry enable the flux and thus the inductance to be estimated. [5] where 1. Our solenoids were wound on wooden mandrels in the physics department electronic shop with a known number of turns arranged in 10 layers of #26 copper wire (diameter of each wire = 0.405 mm). Measure the outer diameter (OD) for the part of the solenoid not covered by the plastic protective layer: OD= m. Calculate the inner diameter (ID) from OD and the information about the thickness of the wire. Remember there are windings on both sides of the center. ID = m. Show here how you obtained the ID: Calculate the effective area A of the windings using the average of the OD and ID: A= m 2 Show here how you obtained the area. Record the total number of turns N written on the end of the coil N= Measure the length of the windings = m. Calculate the expected inductance L of the coil in millihenries (mh) if it were long, thin, and tightly wound: L= Show here how you obtained L. Qualitatively compare the flux enclosed by the windings of a finite length coil (for example, Tipler and Mosca, p.924) with the flux enclosed by the windings of a very long, tightly wound solenoid. Since the actual inductance is due to the sum of the fluxes linked by each coil, do you expect the correction for finite length to make the inductance larger or smaller than the calculation above? (The finite resistance of the wires leads to an energy loss to maintain the flux, but it doesn t significantly affect the inductance.) Calculating actual inductances is an important problem in designing electronic devices. The correction for the finite length of a coil such as used here is of order 15% if the finite thickness of the windings is neglected. You may find a slightly larger discrepancy between your estimate for the inductance and the values you measure in these experiments. B. LR Circuits We will now investigate the transient behavior of a circuit containing an inductance, a resistance, a voltage source and a switch. Most practical inductors have copper wire windings and, therefore, have a resistance as 123LAB2-123LAB2-LRCIRCUIT.DOC 2008-04-14 2-2
well as an inductance. This will be indicated in our circuit diagrams by drawing the inductor with a series resistance R L, equal to the DC resistance of the inductor. Faraday s Law states that the emf generated in a loop of wire is equal in magnitude to the time rate of change of magnetic flux through the loop, = d B. If there are N loops wound in the same sense, as is the case for the solenoid, the emf generated in each loop is added to give a total emf = N d B. The negative sign follows from Lenz Law, which states that the direction of the emf is such that it would generate a current opposing the change in flux that produces it. If the defining equation for L is differentiated with respect to time, the result is dn B = L di =. Thus the emf generated in an inductor by a changing current is proportional to the time rate of change of the current and is in a direction to oppose the change. Consider the circuit shown below. 2. Draw an arrow next to L that indicates the direction in which current is increasing at the instant the switch is closed. Draw a small battery next to L oriented in the direction of the emf generated across the inductor at the moment the switch is closed. Label it L(di/).[Hint: By Lenz Law it opposes the change in current.] Write Kirchhoff s loop rule for the circuit beginning at the positive terminal of the power supply and traversing clockwise around the circuit. The first order differential equation you obtained from Kirchhoff s rule is easily integrated. However, significant insight into the problem can be gained by examining the equation without solving it. After some rearrangement, the equation you should have obtained just above can be written di = i + V L L [6] R L + R The fact that the rate of change of the current, di, is proportional to the current i is an immediate tip-off to the t / solution. It will contain an exponential (in this case a negative exponential) of the form e where is a constant that must have the dimensions of time if the argument of the exponential, -t/, is to be dimensionless. This constant can be obtained from the equation for di as indicated below. 123LAB2-123LAB2-LRCIRCUIT.DOC 2008-04-14 2-3
3. What are the dimensions of the left-hand side of equation 6 for di? The dimensions should be written in the form I x T y, where x and y are appropriate exponents. I is the current and T is the time. Use the fact that the dimensions of an equation must be the same on both sides to find the dimensions of the first term on the right-hand side in equation 6. Use your answer to write a constant in terms of L, R and R L that has the dimensions of time and is, therefore, proportional to. = 4. Predict the value of the current i 0 just after the switch is closed: i 0 = 5. Predict the value of the current i after the switch has been closed for a long time and the current is no longer changing i = 6. Show that the expression below for i as a function of time is consistent with your predictions in 4 and 5. [Hint: What is the value of e 0? Of 1 e 0?]. If not, explain what is wrong with your predictions. i(t) = t e 1 where C. LR Circuit increasing current measurement 7. Use a digital multimeter to measure three resistances that will be important in this experiment. To measure resistance R L, connect the leads as you would to measure voltage (but with nothing else connected to L) and turn the switch to the position. Also, measure the value of the resistors that are nominally 75 and 10. R(10 ) )= ± R L = ± R (75 ) = ± You are now in a position to check your solution and to obtain a value for L. To do this you will measure the timedependent behavior of the current. Use the digital multimeter to set the power supply to deliver 10.00 V. 8. In order to obtain i(t), you will measure the time varying voltage v 10 (t) at point 1 relative to point 2. Write an expression below for v 10 (t) in terms of i(t) from question 6 above. Explain your answer. Use the measured values of R L and the 10 resistor. v 10 (t) = The procedure for using the oscilloscope to obtain data for v(t) is on the next two pages. Note that point 1 is to be connected to the red terminal of the connector to the oscilloscope and point 2 to the black connector. 123LAB2-123LAB2-LRCIRCUIT.DOC 2008-04-14 2-4
The TDS 2002B digital storage oscilloscope is used in this experiment to acquire single sequences of voltage vs. time data and to measure selected points from such a sequence for external analysis. 1) Turn on the power with the switch on top of the scope toward the left (not shown). 2) Press the DEFAULT SETUP button marked below. 3) Press the VERTICAL Channel 1 CH 1 MENU button ONCE to display 5 menu options at the right of the screen. 4) Press the second change menu option until the bandwih limit (BW Limit) is 20 MHz (to reduce the noise). 5) Press the fourth change menu option until the Ch 1 input is the 1X probe option, 6) Turn the Ch 1 VOLTS/DIV to give 500 mv per large vertical division as indicated at bottom left of the screen. 7) Turn the CH 1 POSITION so that the arrow at the left of the screen corresponding to zero volts is at the first large division from the bottom, 8) Turn the TRIGGER LEVEL until +40.0 mv is indicated by the text in the bottom right corner of the screen, corresponding to the arrow on the right of the screen. 9) Turn the HORIZONTAL SEC/DIV to give 1.00 ms per large horizontal division as indicated to the right of M at bottom of screen. 10) Turn the HORIZONTAL POSITION until the arrow at the top of the screen is at the first large vertical division from left; this should cause the middle time position M Pos to indicate 4.000 ms. 11) Connect the LR circuit with the 10 ohm resistor wired to a two terminal banana to BNC adaptor plugged into Ch 1 input. Point 1 on the circuit diagram is connected to red (signal) side; point 2 to the black side. 12) Push the SINGLE SEQ button so that READY appears in the top center of the screen. 13) Close the switch in one decisive motion and you should obtain data. Occasionally, the data will not show a smooth rise but will have one or more glitches in it. These are the result of switch bounce. Instead of making one conducting connection, the switch will make two or more connections with breaks between like a bouncing ball. 14) If you 123LAB2-123LAB2-LRCIRCUIT.DOC 2008-04-14 2-5
need to retake your data, open the switch, press the RUN/STOP button to clear your screen, and then return to step 12. The example images below are shown with black and white inverted to save ink. 15) Cursors allow you to read the values of voltage or time displayed on the screen. Press the CURSOR button to display cursor menu options. Press the first change menu option to choose between off, voltage, and time. In cursor mode the Position Controls above Ch 1 and Ch 2 control the two cursor positions. You can only move the cursors while the CURSOR menu is displayed. In the upper left example time Cursor 1 (controlled by the left position knob in the vertical section of controls when the cursor menu is active) is set at 0.2 ms (listed with far more figures than significant in the menu) and time Cursor 2 ( marked on photo by arrow) is controlled by the right position knob in the vertical section of controls when the cursor menu is active and is set to 0.6 ms. Delta gives the magnitude of the difference between the two positions: 0.40 ms. (The khz number is relevant only to a repetitive signal.) In the lower left example voltage Cursor 1 (marked on photo by arrow) is set at the same point used for the time cursor 2 above. This corresponds to 0.48 V. The second voltage cursor is set to the asymptotic value of the voltage, 1.78 V. Delta gives the magnitude of the difference 1.30 V. which gives Note that the voltage measured by the oscilloscope across the 10 ohm resistor at time t can be related to the asymptotic voltage across the 10 ohm resistor by the following expression Taking the natural logarithm of both sides, we obtain, an equation for a straight line with a negative slope of magnitude equal to the inverse of the time constant. The time measurement can be started at any time after the current started to rise. If we use the oscilloscope zero of time which started when the voltage exceeded the trigger level we obtain for the example shown above:. 9. Using the cursors in this way, obtain 6 to 8 measurements of voltage versus time in the way shown, choosing points where the curve crosses vertical or horizontal lines. Enter your data in the table on the next page. 123LAB2-123LAB2-LRCIRCUIT.DOC 2008-04-14 2-6
t (ms) 10. Make a plot below with your axes clearly labeled, showing on the vertical axis and t on the horizontal axis. Draw the best straight line through your points with a ruler and determine the time constant of the decay from the inverse of the slope. Remember that the slope has units and is not simply the vertical distance in mm divided by the horizontal distance in mm. Estimate the uncertainty by measuring the slope of a second line that also is consistent with your data. = ms 123LAB2-123LAB2-LRCIRCUIT.DOC 2008-04-14 2-7
11. Using the measured values of R L, the actual resistance of the 10 resistor, and the formula for given earlier, determine the value of inductance L measured in this way and compare with the estimate you made earlier for a long, thin, tightly wound solenoid. L(measured with increasing current) mh Show how you calculated this: L(predicted for a long, thin, tightly wound solenoid). In the next part you will make an independent measurement of the inductance and compare the two experimental values. D. LR Circuit decreasing current measurement To measure the time dependence of the decreasing current in an LR circuit one cannot simply use the circuit from part C and open the switch. di becomes very large, in fact, infinite. For example, for a sufficiently rapid change in the current, a voltage could be generated that would cause a spark to jump across the switch. (Recall that.) The circuit to the right avoids this problem. 12. For the circuit shown, assume the switch has been closed for a long time. If the power supply is set at 10.00V, calculate the current through the inductor and calculate the current through the 75 resistor R, using the actual values of resistance you measured earlier. i L = A i 75 = A Draw an arrow next to the inductor indicating the direction the current is increasing immediately after the switch is opened (i.e., an arrow in the opposite direction to the decreasing current). Also draw a battery next to the inductor that shows the direction of the emf generated by the inductor immediately after the switch is opened. Redraw below the circuit diagram for the situation with the switch open (the power supply and the switch should be missing) and indicate on the diagram the direction of the current through the inductor and through the 75 resistor immediately after the switch is opened. 123LAB2-123LAB2-LRCIRCUIT.DOC 2008-04-14 2-8
13. Write Kirchhoff s loop rule for the circuit with the switch open and solve it for di/. di/ = 14. Calculate the value of the current i 75 (0) through the 75- resistor immediately after the switch is opened and the voltage v 75 (0) across it. Which end of the 75 resistor has the higher potential immediately after the switch is opened? i 75 (0) = ± A v 75 (0) = ± V Set up the circuit shown with the supply voltage at 10.00 V. The red terminal is now connected to point 2 and the black lead to 1. Change the vertical scale to 2 V/div, move the voltage zero to the bottom of the screen, change the horizontal sec/div to 500 µs/div, and move the horizontal time zero to 2 ms. Set the trigger level to -2.0 V. 15. The switch should be closed initially. Push the SINGLE SEQ button so that READY appears in the top center of the screen. Open the switch. You should get an exponentially decaying signal. If not, check your connections and the supply voltage, push the SINGLE SEQ button to that READY appears in the top center of the screen, and open the switch.. In this case, where (The time constant is different than for the increasing current case because the resistance of the circuit is different.) This equation can be transformed to. From the oscilloscope record of the decay of the current make measurements using the cursor as before, taking readings for times the curve crosses convenient horizontal or vertical lines, and record your data in the table on the next page. 123LAB2-123LAB2-LRCIRCUIT.DOC 2008-04-14 2-9
t (ms) v 75 (t) in volts ln(v 75 (t)) 16. How does the magnitude of the initial voltage across the 75 resistor just after the switch is opened compare to the 10 V supply? Where did this voltage come from? 17. Make a plot below with your axes clearly labeled, showing on the vertical axis and t on the horizontal axis. 123LAB2-123LAB2-LRCIRCUIT.DOC 2008-04-14 2-10
Draw the best straight line through your points with a ruler and determine the time constant of the decay from the inverse of the slope. Estimate the uncertainty by measuring the slope of a second line that also is consistent with your data = ms 18. Using the measured values of R L, the actual resistance of the 75 resistor, and the formula for you found earlier, determine the value of inductance L measured in this way and compare with the estimate you made earlier for a long, thin, tightly wound solenoid. L(measured with decreasing current) Show how you calculated this: 19. Compare this with the earlier value of L(measured with increasing current). Are the two consistent within your experimental uncertainties? CLEAN UP CHECKLIST Turn off the oscilloscope and the power supply. Return equipment and parts to the places shown in the photograph. Make sure you have answered all questions. Check the first page to see that you ve written your name and student number, your lab section, and your partner s name. Staple all pages together. Call the TA to your lab station to initial your report. Turn it in before leaving the lab. 123LAB2-123LAB2-LRCIRCUIT.DOC 2008-04-14 2-11