Lab 15 Series RL Circuits w prelab

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Apr 3, 2024

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Lab 15, Series RL Circuits PRE-LAB Name NEELMANI BHARDWAJ__________ Date ___________________ Class ___________________ 1. What is Inductive Reactance, X L ? Show formula for X L . INDUCTIVE REACTANCE IS THE OPPOSITION FACEDBY THE CURRENT IN THE CIRCUIT DUE TO THE INDUCTOR PRESENT IN THE CIRCUIT . X L = 2x3.14xFxL 2. Does the Voltage Lead or Lag the Current in an inductor? Explain your answer VOLTAGE LEADS CURRENT BY 90 DEGREES IN AN INDUCTORBECAUSE OF AN INDUCTIVE LOAD , IT IS THE INDUCED ELECTOMOTIVE FORCE THAT HELPS THE CURRENT TO FLOW. 3. In a RL circuit, how does the Inductive Reactance manifest itself when a sinusoidal signal is applied? INDUCTIVE REACTANCE BEHAVE LIKE A OPPOSITION WHEN A SINE SIGNAL IS APPLIED TO IT. 4. In which quadrant are the voltage or impedance vector diagrams drawn and why? IT WILL BE IN THE SECOND QUADRANT BECAUSE VOLTAGE LEAD THE CURRENT BY 90 DEGREES. 5. In an RL circuit, does V R Lag or lead V S and why? V R lag Vs by 90 degrees because V r and curent are always in same phase . therefore, if source voltage leads the current by 90 so it will also lead resistor voltage by 90. Fall 2015 Lab 15 P a g e | 1

6. How does the phase angle in an RL circuit vary with Frequency? Explain your answer. If the frequency increases then X L increases and the total impedence of the circuit also increases . therefore phase angle also increases . AS phases angle = tan -1 (X L/ R) 7. How many methods are shown in this Lab for phase angle measurement in an RL circuit? Can you think of yet another method? Phase angle = tan -1 (X L/ R) 8. Can these methods be applied for a series RC circuit? Explain your answer? YES this method can als0 be applied to RC circuit but there will be Xc in place of X l. 9. Explain the basic approach of each method of step 9 (use back of page if you need more space). Fall 2015 Lab 15 P a g e | 2

10. What was the unit of τ in previous Labs? SECONDS 11. How long does it take for an RL circuit to fully react to a sinusoidal signal? 12. What happens to X L if the applied sinusoidal frequency increase to say infinity? AS THE FREQUENCY INCREASES THE IMPEDENCE OF THE INDUCTOR INCREASES AND OF CAPACITOR, IT DECREASES WHEN FREQUENCY IS TOO HIGH INDUCTOR START BEHAVING LIKE A CAPACITOR. Fall 2015 Lab 15 P a g e | 3

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Lab 15 (eBook 24) Series RL Circuits Name ____________________ Date ___________________ Class ___________________ READING Text, Sections 12–1 through 12–3 OBJECTIVES After performing this experiment, you will be able to: 1. Compute the inductive reactance of an inductor from voltage measurements in a series RL circuit. 2. Draw the impedance and voltage phasor diagram for the series RL circuit. 3. Measure the phase angle in a series circuit using either of two methods. MATERIALS NEEDED One 10 kΩ resistor One 100 mH inductor REQUIRED LAB PREPARATION (PRELAB) 1.Read all sections of the lab. 2. Read the text book, sections 12-1 through 12-3 3. Find the required resistor and inductor for this lab as indicated in the Materials Needed section above 4. Review the differential probe measurement technique from Lab 8 5. Review the oscilloscope time “base operation” or what is called Horizontal Control depending on the oscilloscope manufacturer. 6. Complete the PreLab questions at the back of this document and place in teacher’s drop box before entering the Lab SUMMARY OF THEORY When a sine wave drives a linear series circuit, the phase relationships between the current and the voltage are determined by the components in the circuit. The current and voltage are always in phase across resistors. With capacitors, the current is always leading the voltage by 90°, but for inductors, the voltage always leads the current by 90° . (A simple Fall 2015 Lab 15 P a g e | 4

memory aid for this is ELI the ICE man, where E stands for voltage, I for current, and L and C for inductance and capacitance.) Figure 15–1(a) illustrates a series RL circuit. The graphical representation of the phasors for this circuit is shown in Figure 15–1(b) and (c). As in the series RC circuit, the total impedance is obtained by adding the resistance and inductive reactance using the algebra for complex numbers . In this example, the current is 1.0 mA, and the total impedance is 5 kΩ. The current is the same in all components of a series circuit, so the current is drawn as a reference in the direction of the x -axis . If the current is multiplied by the impedance phasors, the voltage phasors are obtained as shown in Figure 15– 1(c) . Figure 15–1 In this experiment, you learn how to make measurements of the phase angle. Actual inductors may have enough resistance to affect the phase angle in the circuit. You will use a series resistor that is large compared to the inductor’s resistance to avoid this error. PROCEDURE 1. Measure the actual resistance of a 10 kΩ resistor and the inductance of a 100 mH inductor. If the inductor cannot be measured, record the listed value. Record the measured values in Table 15–1 . 2. Connect the circuit shown in Figure 15–2 . Set the generator voltage with the circuit connected to 3.0 V pp at a frequency of 25 kHz. The generator should have no dc offset. Measure the generator voltage and frequency with the oscilloscope as many meters cannot respond to the 25 kHz frequency. Use peak-to-peak readings for all voltage and current measurements in this experiment. Component Listed Value Measured Value L 1 100 mH Fall 2015 Lab 15 P a g e | 5

R 1 10 kΩ Table 15–1 Figure 15–2 3. Using a two-channel oscilloscope, measure the peak-to-peak voltage across the resistor ( V R ) and the peak-to-peak voltage across the inductor ( V L ) . (See Figure 15–3 for the setup.) Measure the voltage across the inductor using the difference technique described in Experiment 8 . Record the voltage readings in Table 15–2 . V R pp mea sure d V L Differe ntial V mea sure d I Ohm s Law Com pute step 4 X L Ohms Law Compute step5 Z T Com pute step 6 1.6V 2.52 V 0.16 mA 15.75 18.6 5k Ω Fall 2015 Lab 15 P a g e | 6

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kΩ Table 15–2 (f = 25 kHz) 4. Compute the peak-to-peak current in the circuit by applying Ohm’s law to the resistor. That is, I = V R R Enter the computed current in Table 15–2 . 5. Compute the inductive reactance, X L , by applying Ohm’s law to the inductor. The reactance is X L = V L I Enter the computed reactance in Table 15–2 . 6. Calculate the total impedance ( Z T ) by applying Ohm’s law to the entire circuit. Use the generator voltage set in step 2 ( V S ), and the current determined in step 4. Enter the computed impedance in Table 15–2 . 7. Using the values listed in Tables 15–1 and 15–2 , draw the impedance phasors on Plot 15–1(a) and the voltage phasors on Plot 15–1(b) for the circuit at a frequency of 25 kHz. Show the plots axis name, unit and scale values. Fall 2015 Lab 15 P a g e | 7

Plot 15–1 Impedance Voltage 8. Compute the phase angle between V R and V S using the trigonometric relation θ = tan − 1 ( V L V R ) Enter the computed phase angle in Table 15–3 . 9. Two methods for measuring phase angle will be explained. The first method can be used with any oscilloscope . The second can only be used with oscilloscopes that have a “fine” or variable SEC/DIV control. Measure the phase angle between V R and V S using ONE or BOTH METHODS as you see fit. The measured phase angle will be recorded in Table 15–3 . Phase Angle Measurement—Method 1 (a) Connect the oscilloscope so that channel 1 is across the generator and channel 2 is across the resistor. (See Figure 15–3 .) Obtain a stable display showing between one and two cycles while viewing channel 1 ( V S ). The scope should be triggered from channel 1. (b) Measure the period, T, of the generator. Record it in Table 15–3 . You will use this time in step (e). Fall 2015 Lab 15 P a g e | 8

Figure 15–3 (c) Set the oscilloscope to view both channels. ( Do not have channel 2 inverted.) Adjust the amplitudes of the signals using the VOLTS/DIV, VERT POSITION, and the vernier controls until both channels appear to have the same amplitude as seen on the scope face. (d) Spread the signal horizontally using the SEC/DIV control until both signals are just visible across the screen. The SEC/DIV control must remain in the calibrated position . Measure the time between the two signals, Δ t, by counting the number of divisions along a horizontal graticule of the oscilloscope and multiplying by the SEC/DIV setting. (See Figure 15–4 .) Record the measured Δ t in Table 15–3 . Figure 15–4 (e) The phase angle may now be computed from the equation θ = ( ∆t T ) × 360 ° Enter the measured phase angle in Table 15–3 under Phase Angle— Method 1. Computed Phase Angle θ Measured Period T Time Difference Δ t Phase Angle Method 1 θ Method 2 θ 57.50degrees 40us 6.32us 56.88 56.88 Table 15–3 Fall 2015 Lab 15 P a g e | 9

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Phase Angle Measurement—Method 2 (a)In this method the oscilloscope face will represent degrees, and the phase angle can be measured directly. The probes are connected as before. View channel 1 and obtain a stable display. Then adjust the SEC/DIV control and its vernier until you have exactly one cycle across the scope face. This is equivalent to 360° in 10 divisions, so each division is worth 36°. 1 (b) Now switch the scope to view both channels. As before, adjust the amplitudes of the signals using the VOLTS/DIV, VERT POSITION, and the vernier controls until both channels appear to have the same amplitude. Fall 2015 Lab 15 P a g e | 10

(c) Measure the number of divisions between the signals and multiply by 36° per division. Record the measured phase angle in Table 15–3 under Phase Angle—Method 2. 1 For even better resolution, you can set one-half cycle across the screen, making each division worth 18°. Care must be taken to center the waveform. CONCLUSION .IN THIS LAB I LEARNT HOW TO FIND PHSAEANGLE IN A RL CIRCUIT BY DIFFERNT TYPES OF METHOD . FOR THESE DIFFERENT ANGLETHE RESULTS ARE APPROXEMETLY THE SAME . MOREOVER , I GET TO KNOW HE PHSOR RELATIONSHIP OF INDUCTOR WITH RESISTOR , CURRENT AND VOLTAGE SOURCE OF THE CIRCUIT. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EVALUATION AND REVIEW QUESTIONS 1. (a) What will happen to the impedance in this experiment if the frequency increases? Fall 2015 Lab 15 P a g e | 11

(b) What would happen to the impedance if the inductance were larger? 2. (a) What will happen to the phase angle in this experiment if the frequency increases? (b) What would happen to the phase angle if the inductance were larger? 3. Compute the percent difference between the computed phase angle and the method 1 phase angle measurement. See Lab 1 4. The critical frequency for an RL circuit occurs at the frequency at which the resistance is equal to the inductive reactance. That is, R = X L . Since X L = 2π fL for an inductor, it can easily be shown that the circuit frequency for an RL circuit is Compute the critical frequency for this experiment. What is the phase angle between V R and V S at the critical frequency? Show your calculations. f crit = . . . . . . . . . . . . . . . . θ = . . . . . . . . . . . . . . . . . Fall 2015 Lab 15 P a g e | 12

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5. A series RL circuit contains a 100 Ω resistor and a 1.0 H inductor and is operating at a frequency of 60 Hz. There are 3.0 V are across the resistor. Draw the circuit, compute and show your calculations for the following: (a) the current in the inductor (b) the inductive reactance, X L (c) the voltage across the inductor, V L (d) the source voltage, V S (e) the phase angle between V R and V S Fall 2015 Lab 15 P a g e | 13