Recommended
Recommended
More Related Content
Similar to Ruben EscaleraG00092506Lab 5 Series and Pa.docx
Similar to Ruben EscaleraG00092506Lab 5 Series and Pa.docx (20)
More from joellemurphey
More from joellemurphey (20)
Recently uploaded
Recently uploaded (20)
Ruben EscaleraG00092506Lab 5 Series and Pa.docx
- 1. Ruben Escalera G00092506 Lab 5 Series and Parallel Inductive Reactive Circuits Grantham University 2/18/2016 Introduction The main aim of this lab exercise is to perform single frequency AC analysis on series and parallel R-L circuits. The simulation results are to be compared with the calculation results. In an inductor the voltage across the inductor leads the current through it by an angle of 90°, however for a circuit containing a
- 2. resistor and an inductor the angle is less than 90°. Equipment/components used · Resistors · Inductors · 15V AC Voltage source at a frequency of 1000Hz Problem statement In this lab exercise the circuits shown in the figure below are to be constructed in MULTISIM : The circuits are to be simulated for single frequency analysis to determine currents and voltages in the circuit. The simulation results are to be compared with calculations and the differences noted. Calculated solution Parallel R-L circuit The various parameters of the parallel circuit were done as follows: L1 = 80mH, XL1 = (6.28 * 1 kHz * 80mH) = 6.28 *(1*103) * (80*10-3) = 502.4Ω∠ 90ᵒ L2= 150mH, XL2 = (6.28 * 1 kHz * 150mH) = 6.28 * (1*103) * (150*10-3) = 942Ω∠ 90ᵒ ZEQ= 1/ + + = 1/ + + = = (91.479 +27.92j) = 95.64∠ 16.97ᵒ Zeq = 95.64Ω∠ 16.97ᵒ IT = = .156A∠ -16.97ᵒ XL2= 942Ω∠ 90ᵒ XL1= 502.4Ω∠ 90ᵒ IR1 = = 0 .1A IR2= = .05A IL1 = = 29.85mA∠ -90ᵒ IL2 = = 15.92mA∠ -90ᵒ
- 3. Series R-L circuit For the series R-L circuit the calculation of the various parameters were done as follows: L2= 150mH, XL2 = (6.28 * 1 kHz * 150mH) = 6.28 * (1*103) * (150*10-3) = 942Ω∠ 90ᵒ L1 = 80mH, XL1 = (6.28 * 1 kHz * 80mH) = 6.28 *(1*103) * (80*10-3) = 502.4Ω∠ 90ᵒ Zeq = Zeq = = = = 450Ω + 1446.4Ω =1.514kΩ tan θz = = = tan-1 (3.214) = 72.72ᵒ IT = VT / ZT= 15V/1.514kΩ= 9.9mA∠ 72.72ᵒ V = I * R VR1 = 9.9mA * 150Ω = 1.49VVR2 = 9.9mA * 300Ω = 2.97V VL2 = 9.9mA * 942Ω = 9.32VVL1 = 9.9mA * 502.4Ω = 4.97V Zeq = 1.514kΩ∠ 72.72ᵒ IT= 9.9mA∠ -72.72ᵒ XL2 = 942Ω∠ 90ᵒ XL1= 502.4Ω∠ 90ᵒ VR1= 1.49V∠ -72.72ᵒ VR2= 2.97V∠ -72.72ᵒ VL1= 4.97V∠ 17.28ᵒ VL2= 9.32V∠ 17.28ᵒ
- 4. Experimental procedure 1. The parallel R-L circuit was constructed in MULTISIM software 2. The circuit was simulated for single frequency AC analysis to determine the currents and voltages in the circuit 3. The series R-L circuit was constructed in MULTISIM software 4. The circuit was simulated for single frequency AC analysis to determine the current and the voltages in the circuit. Circuit design Parallel R-L circuit Series R-L circuit Simulation results Discussion of results The results were summarized as shown in the table below. Parallel circuit Measured results Variable Magnitude Phase VR1 15V 0 VR2
- 7. Series circuit Measured results Variable Magnitude Phase (deg) VR1 1.48655V -72.70393 VR2 2.97309V -72.70393 VL1 4.98147V 17.29607 VL2 9.34025V 17.29607 IT 9.91931mA 107.29607 Calculated results Variable Magnitude Phase(deg) VR1 1.49V -72.72 VR2 2.97V -72.72 VL1 4.97V 17.28 VL2
- 8. 9.23V 17.28 IT 9.9mA -72.72 The measured results and the calculated results were very close to each other and thus the lab exercise was a success. The small deviation was due to the resistor and the inductor tolerances. Conclusion The lab was a good practice in the study of R-L circuits. The voltage across an inductor was found to lead the voltage by an angle of 90°. The leading angle in R-L circuit is however less than 90° and it depends on the frequency of the source signal and the values of the resistor and inductor. Assignment 8
- 9. Capacitive Circuits 1 Capacitive Circuits 2 Ruben Escalera G00092506 Lab 3 Capacitive Circuits
- 10. Grantham University 2/16/2016 Introduction The purpose of this lab exercise is to perform single frequency AC analysis on parallel R-C circuit. The circuit is to be constructed in MULTISIM software and then simulated. The simulation results are to be compared with the calculated results. In a capacitor the current lead the voltage across it by an angle of 90°. Equipment/components used · AC voltage source of varying frequencies · Resistor(300Ω) · Capacitor(5uF) Problem statement The circuit shown in the figure below is to be constructed in MULTISIM: Single frequency AC analysis is to be done on the circuit for frequencies (100Hz, 500Hz, 1000Hz, 1500Hz, 10000Hz and 20000Hz). From the single frequency AC analysis the following is to be determined: IT-total current flowing through the circuit IR-current flowing through the resistor IC-current through the capacitor
- 11. Calculated solution: Capacitive reactance XC = The equivalent impedance is evaluated as follows: Zeq = The total current flowing in the circuit is evaluated as follows: IT = The current flowing through the resistor is: IR1 = The current flowing through the capacitor is: IC1 = Experimental procedure 1. The circuit schematic was constructed in MULTISIM. 2. The circuit was simulated for single frequency AC analysis to determine the currents (IT, IC and IR). 3. The previous step was repeated for different frequencies
- 12. (100Hz, 500Hz, 1000Hz, 1500Hz, 10000Hz and 2000Hz). Circuit design The circuit was constructed in MULTISIM as shown in the figure below: Simulation results: 100Hz 500Hz 1000Hz 1500Hz 10000Hz 20000Hz Discussion of results The simulation results are shown in the table below: Frequency
- 13. IT IR1 IC1 100 Hz 500 Hz 1000 Hz 1500 Hz 10000 Hz 20000 Hz The measured values and the calculated values for various frequencies were close to each other. Answers to questions: a. Describe the relationship between the frequency and IT. An increase in frequency decreases the capacitive reactance of the capacitor. The overall is an increase in the current flowing through the circuit. b. What effect does frequency have on Zeq?
- 14. An increase in frequency results in results in a decrease in the capacitive reactance of the circuit. The overall effect is a decrease in the equivalent impedance Zeq. c. How could the circuit be modified to bring the phase angle closer to 0○? Increasing the frequency of the source signal and connecting a resistor in series with the capacitor. d. How could the circuit be modified to bring the phase angle closer to -90○? Decreasing the frequency of the source signal and connecting a capacitor in series with the resistor. Conclusion The lab exercise was a good insight in the study of the capacitive reactance of a capacitor. The current in a capacitor leads the voltage by an angle of 90°. The voltage lags behind the current by an angle 90°. The overall lag angle in the circuit is less than 90°. The measured results and the calculated results deviated from each other slightly. This could be because of the capacitor and the resistor tolerances. V1 120 Vrms 500 Hz 0° R1 300Ω 5% C1 5µF 5%
- 15. Series and Parallel Inductive Reactive Circuits in Complex Form 1. Watch the video: · Week 7 Video Lecture – Multisim Series Capacitive Reactive Circuit Repeat the operations from Week 3 and Week 5 Labs. Capture a screenshot of the output of the analysis to confirm your calculations in complex form. Create a table of expected and measured results. 2. Include a discussion of any differences noticed between the calculations and the simulations. Include all calculations, your table, and screenshots of the analysis in a word processing document and submit as EE115W7LabYourGID.docx, or an equivalent word processing file extension.