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아주대학교 기초전기실험 A+ 결과보고서 Ch. 15 (AC) 영문

REPORTIEEE Code of Ethics(출처: http://www.ieee.org)We, the members of the IEEE, in recognition of the importance of our technologies in affecting the quality of life throughout the world, and in accepting a personal obligation to our profession, its members and the communities we serve, do hereby commit ourselves to the highest ethical and professional conduct and agree:1. to accept responsibility in making decisions consistent with the safety, health and welfare of the public, and to disclose promptly factors that might endanger the public or the environment;2. to avoid real or perceived conflicts of interest whenever possible, and to disclose them to affected parties when they do exist;3. to be honest and realistic in stating claims or estimates based on available data;4. to reject bribery in all its forms;5. to improve the understanding of technology, its appropriate application, and potential consequences;6. to maintain and improve our technical competence and to undertake technologof others;8. to treat fairly all persons regardless of such factors as race, religion, gender, disability, age, or national origin;9. to avoid injuring others, their property, reputation, or employment by false or malicious action;10. to assist colleagues and co-workers in their professional development and to support them in following this code of ethics.위 IEEE 윤리헌장 정신에 입각하여 report를 작성하였음을 서약합니다.학 부: 전자공학과제출일:과목명: 기초전기실험교수명:분 반:학 번:성 명:CHAPTER ac 15. Passive filtersPart 1. High-pass R-C filter.R _{} `=`0.99k OMEGAThe voltageV _{o} of the resistor is obtained with respect to the voltage sourceV _{i} as shown inV _{0} =V _{i} {R} over {sqrt {X _{C}^{2} +R ^{2}}}.Thus, the ratio of the voltage between the voltage source and the resistor is equal toV _{0} /V _{i} = {R} over {sqrt {X _{C}^{2} +R ^{2}}} ``.f _{c} is a frequency atR=X _{C} . Thusf _{c} isf _{c} = {1} over {2 pi RC} = {1} over {2 pi TIMES 0.99 TIMES 10 ^{3} TIMES 0.1 TIMES 10 ^{-6}} =1607.63`Hz.Table 15.1CalculatedGraphGraph(hatA _{i} becomes 0.707 at about 1.7 khz.If you look at the graph forA _{i} below, at 2 kHz,A _{i} is 0.76.If you look at the graph oftheta below, you can see that the frequency oftheta =45 DEG is about 1.65 kHz.GRAPH 15.1Frequency (kHz)V _{0(p-p)} ` [V]A _{i} = {V _{0(p-p)}} over {V _{i(p-p)}}0.10.2730.0665850.20.5190.1265850.40.990.2414630.61.420.3463410.81.770.43170712.170.5292681.22.410.5878051.42.650.6463411.62.810.6853661.82.930.71463423.10.75609833.420.83414643.580.87317153.660.89268363.740.91219583.820.931707103.90.95122123.90.95122143.90.95122163.90.95122183.90.95122203.90.95122403.90.95122603.90.951221003.90.95122The value ofV _{i(p-p)} measured by an oscilloscope was 4.01V.Therefore, in order to obtainA _{i}, the measured voltage value of the resistor may be divided by 4.01.Input voltage atatat atFrequency`(kHz)theta0.186.440.282.910.669.53158.12238.79614.99109.13127.63204.59402.30601.531000.92If we obtain the value fortheta fromV _{0} /V _{i} = {R} over {sqrt {X _{C}^{2} +Reriment was conducted by configuring a high-pass filter that connects capacitors and resistors in series and outputs a high voltage only when a high-frequency voltage comes in. Due to the characteristics of the capacitor, when the dc voltage comes in, it becomes an open state after charging, and when the voltage is low, the capacitor takes a similarly high voltage and little voltage is applied to the resistance. Therefore, only when the voltage is high, a low voltage may be applied to the capacitor and a high voltage may be applied to the resistance.The calculated value forf _{c}, the value found in the graph, and the value at45 DEG intheta graph are slightly different from each other, but this is because of the voltage inside the capacitor and the limitation of the measuring instrument and the voltage generator, and it is not a big error. At 2 kHz, the A value is almost similar to the calculated value and the value found through the graph, so it can be said that the experiment was welrcuit that transmits voltage when a high voltage comes inThis circuit is expected to be used a lot in devices such as speakers. When speakers are connected in parallel to a resistance, a voltage is applied to a resistance when a frequency is low, and thus a small sound is generated when a voltage of the speaker is applied to a low voltage, and when a frequency is increased, a voltage of the speaker is increased.By experimenting with circuits using the characteristics of these capacitors, it is possible to understand the characteristics of capacitors, inductors, etc. in AC circuits and to solve related problems more easily.-“Impedance”, Naver Knowledge Encyclopedia, 22022.04.18. access,https://terms.naver.com/entry.naver?docId=4390151&cid=60217&categoryId=60217-“Inductor”, wikipedia, 2022.04.18. access, https://pt.wikipedia.org/wiki/Indutor-“capacitor”, Naver Knowledge Encyclopedia, 2022.04.02 access,https://terms.naver.com/entry.naver?docId=1150345&cid=40942&categoryId=32241-Robert L. C

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아주대학교, 아주대, A+, 기초전기실험, 기전실

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아주대학교 기초전기실험 A+ 결과보고서 Ch. 10 (AC) 영문

REPORTIEEE Code of Ethics(출처: http://www.ieee.org)We, the members of the IEEE, in recognition of the importance of our technologies in affecting the quality of life throughout the world, and in accepting a personal obligation to our profession, its members and the communities we serve, do hereby commit ourselves to the highest ethical and professional conduct and agree:1. to accept responsibility in making decisions consistent with the safety, health and welfare of the public, and to disclose promptly factors that might endanger the public or the environment;2. to avoid real or perceived conflicts of interest whenever possible, and to disclose them to affected parties when they do exist;3. to be honest and realistic in stating claims or estimates based on available data;4. to reject bribery in all its forms;5. to improve the understanding of technology, its appropriate application, and potential consequences;6. to maintain and improve our technical competence and to undertake technological tasks for others only if qualified by training or experience, or after full disclosure of pertinent limitations;7. to seek, accept, and offer honest criticism of technical work, to acknowledge and correct errors, and to credit properly the contributions of others;8. to treat fairly all persons regardless of such factors as race, religion, gender, disability, age, or national origin;9. to avoid injuring others, their property, reputation, or employment by false or malicious action;10. to assist colleagues and co-workers in their professional development and to support them in following this code of ethics.위 IEEE 윤리헌장 정신에 입각하여 report를 작성하였음을 서약합니다.학 부: 전자공학과제출일:과목명: 기초전기실험교수명:분 반:학 번:성 명:CHAPTER ac 10. Series-Parallel Sinusoidal CircuitsPart 3. R-L-C Series-Parallel NetworkE`=`4.02`V _{(p-p)}R _{1} `=`0.99k OMEGA R _{2} `=`463.45 OMEGA"Calculated""Measured"Z _{T}718.73 ANGLE -20.98 DEG OMEGA 738.04 ANGLE -20.81 DEG OMEGAI _{s(p-p)}5.57 ANGLE 20.98 DEG mA`5.42 ANGLE 20.81 DEG mAI _{1(p-p)}3.39 ANGLE -32.14 DEG mA`3.41 ANGLE -32.40 DEG mAI _{2(p-p)}`4.33 ANGLE 59.43 DEG mA`4.34 ANGLE 59.78 DEG mAZ _{R _{1}} `=`R _{1} `=`1k OMEGA ,Z _{R _{2}} `=`R _{2} `=`470 OMEGA ,Z _{L} `=`j2 pi fL`=`j628.32 OMEGA ,Z _{C} `=` {1} over {j2 pi fC} `=`-j795.77 OMEGA Z _{1} `=`Z _{R _{1}} `+`Z _{L} `=`1000`+`j628.32` OMEGA `=`1181.01 ANGLE 32.14 DEG OMEGA Z _{2} `=`Z _{R _{2}} `+`Z _{C} `=`470`-`j795.77` OMEGA `=`924.20 ANGLE -59.43 DEG OMEGA Z _{T} `= {1} over {{1} over {Z _{1}} + {1} over {Z _{2}}} `=`671.10-j257.29 OMEGA `=`718.73 ANGLE -20.98 DEG OMEGA I _{s(p-p)} =` {4V _{(p-p)}} over {Z _{T}} `=`5.57 ANGLE 20.98 DEG mAI _{1(p-p)} `=` {4V _{(p-p)}} over {Z _{1}} `=`3.39 ANGLE -32.14 DEG mAI _{2(p-p)} `=` {4V _{(p-p)}} over {Z _{2}} `=`4.33 ANGLE 59.43 DEG mAZ _{R _{1}} `=`R _{1} `=`0.99k OMEGA ,Z _{R _{2}} `=`R _{2} `=`463.45 OMEGA ,Z _{L} `=`j2 pi fL`=`j628.32 OMEGA ,Z _{C} `=` {1} over {j2 pi fC} `=`-j795.77 OMEGA Z _{1} `=`Z _{R _{1}} `+`Z _{L} `=`990`+`j628.32` OMEGA `=`1172.56 ANGLE 32.40 DEG OMEGA Z _{2} `=`Z _{R _{2}} `+`Z _{C} `=`463.45`-`j795.77` OMEGA `=`920.89 ANGLE -59.78 DEG OMEGA Z _{T} `= {1} over {{1} over {Z _{1}} + {1} over {Z _{2}}} `=`689.89-j262.20 OMEGA `=`738.04 ANGLE -20.81 DEG OMEGA I _{s(p-p)} =` {4V _{(p-p)}} over {Z _{T}} `=`5.42 ANGLE 20.81 DEG mAI _{1(p-p)} `=` {4V _{(p-p)}} over {Z _{1}} `=`3.41 ANGLE -32.40 DEG mAI _{2(p-p)} `=` {4V _{(p-p)}} over {Z _{2}} `=`4.34 ANGLE 59.78 DEG mACONSIDERATIONThe experiment was to calculate the current flowing through the entire impedance, the current flowing through the resistor 1 and the coil, and the current flowing through the resistor 2 and the capacitor in a circuit connected in series and again in parallel.Considering KCL,I _{s(p-p)} should be equal to the sum ofI _{1(p-p)} andI _{2(p-p)}, but it is different from the value of magnitude. I think this is not the same because the phases ofI _{s(p-p)},I _{1(p-p)}, andI _{2(p-p)} are different.WhenI _{s(p-p)},I _{1(p-p)}, andI _{2(p-p)} are changed to the form ofa+jb instead of the form of magnitude and phase, the values are the same asI _{s(p-p)} `=`5.07`+`j1.93mA andI _{1(p-p)} +I _{2(p-p)} `=`5.06+j7.32mA. Therefore, it can be seen that the experiment and calculation were performed wellFrom these results, it can be seen that KCL can be used like DC in an AC R-L-C circuit.-“Impedance”, Naver Knowledge Encyclopedia, 22022.04.18. access,https://terms.naver.com/entry.naver?docId=4390151&cid=60217&categoryId=60217-“Inductor”, wikipedia, 2022.04.18. access, https://pt.wikipedia.org/wiki/Indutor-“capacitor”, Naver Knowledge Encyclopedia, 2022.04.02 access,https://terms.naver.com/entry.naver?docId=1150345&cid=40942&categoryId=32241-“Thevenin’stheorem”, Naver Knowledge Encyclopedia, 2022.03.17. access,https://terms.naver.com/entry.naver?docId=760094&cid=42341&categoryId=42341-Robert L. Boylestad and Gabriel Kousourou, 『Laboratory to Manual to Accompany Introductory Circuit Analysis Twelfth edition』,Pearson, p.385~411

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아주대학교 기초전기실험 A+ 결과보고서 Ch. 9 (AC) 영문

REPORTIEEE Code of Ethics(출처: http://www.ieee.org)We, the members of the IEEE, in recognition of the importance of our technologies in affecting the quality of life throughout the world, and in accepting a personal obligation to our profession, its members and the communities we serve, do hereby commit ourselves to the highest ethical and professional conduct and agree:1. to accept responsibility in making decisions consistent with the safety, health and welfare of the public, and to disclose promptly factors that might endanger the public or the environment;2. to avoid real or perceived conflicts of interest whenever possible, and to disclose them to affected parties when they do exist;3. to be honest and realistic in stating claims or estimates based on available data;4. to reject bribery in all its forms;5. to improve the understanding of technology, its appropriate application, and potential consequences;6. to maintain and improve our technical competence and to undertake technolog`0.99k OMEGA R _{s} `=`9.34 OMEGA"Calculated""Measured"I _{s(p-p)}6.37` ANGLE -89.96 DEG A6.37` ANGLE -89.96 DEG AI _{L(p-p)}6.37 ANGLE -90 DEG A`6.40 ANGLE -90 DEG AI _{C(p-p)}0A0AI _{R(p-p)}4mA`4.06mAV _{R _{s} (p-p)} `("for"`I _{s} )`3.99 ANGLE -3.59 DEG V`4.29 ANGLE -3.85 DEG VV _{R _{s} (p-p)} `("for"`I _{L} )`3.99 ANGLE -3.59 DEG V`4.29 ANGLE -3.85 DEG VV _{R _{s} (p-p)} `("for"`I _{C} )0V0VZ _{R} `=`R`=`1k OMEGA ,Z _{L} `=`jwL`=`j2 pi fL`=`j0.628 OMEGA `=`0.628` ANGLE 90 DEG OMEGA ,Z _{C} `=` {1} over {jwC} `=`-j {1} over {2 pi fC} `=`-j1591549.431 OMEGA `=`1.59`X``10 ^{6} ANGLE -90 DEG OMEGA Z _{T} `=` {1} over {{1} over {Z _{R}} `+` {1} over {Z _{L}} `+` {1} over {Z _{C}}} `=` {1} over {{1} over {1000} `+` {1} over {j0.628} `+` {j} over {1.59X10 ^{6}}} `=`0.00039`+`j0.628`=`0.628 ANGLE 89.96 DEG OMEGA I _{S(p-p)} `=` {4V _{(p-p)}} over {Z _{T}} `=`6.37` ANGLE -89.96 DEG AI _{L(p-p)} `=` {4} over {Z _{L}} `=`6.37 ANGLE -90 DEG A,I _{C(p-p)} `=` {4} over {Z _{C}} ` APPROX `0A,I onnected in series.I _{S(p-p)} `=` {4V _{(p-p)}} over {Z _{T,`with`sensing`R}} `=`0.399` ANGLE -3.59 DEG AI _{S(p-p)} `=` {4V _{(p-p)}} over {Z _{L,S.R.`series}} `=`0.399 ANGLE -3.59 DEG AI _{C(p-p)} `=` {4} over {Z _{C}} ` APPROX `0AV _{R _{s} (p-p)} `("for"`I _{s} )`=`0.399 ANGLE -3.59 DEG A`X``10 OMEGA `=``3.99 ANGLE -3.59 DEG VV _{R _{s} (p-p)} `("for"`I _{L} )`=`0.399 ANGLE -3.59 DEG A`X``10 OMEGA `=``3.99 ANGLE -3.59 DEG VV _{R _{s} (p-p)} `("for"`I _{C} )`=`0A`X`10 OMEGA `=`0VZ _{R} `=`R`=`0.99k OMEGA ,Z _{L} `=`jwL`=`j2 pi fL`=`j0.628 OMEGA `=`0.628` ANGLE 90 DEG OMEGA ,Z _{C} `=` {1} over {jwC} `=`-j {1} over {2 pi fC} `=`-j1591549.431 OMEGA `=`1.59`X``10 ^{6} ANGLE -90 DEG OMEGA Z _{T} `=` {1} over {{1} over {Z _{R}} `+` {1} over {Z _{L}} `+` {1} over {Z _{C}}} `=` {1} over {{1} over {990} `+` {1} over {j0.628} `+` {j} over {1.59X10 ^{6}}} `=`0.00040`+`j0.628`=`0.628 ANGLE 89.96 DEG OMEGA I _{S(p-p)} `=` {4.02V _{(p-p)}} over {Z _{T}} `=`6.40` ANGLE -89.96 DEG AI _{L(p-p)} `=gher than the sensing resistance, so the sensing resistance is ignored when the capacitor and the sensing resistance are connected in series.I _{S(p-p)} `=` {4.02V _{(p-p)}} over {Z _{T,`with`sensing`R}} `=`0.429` ANGLE -3.85 DEG AI _{S(p-p)} `=` {4.02V _{(p-p)}} over {Z _{L,S.R.`series}} `=`0.429 ANGLE -3.85 DEG AI _{C(p-p)} `=` {4} over {Z _{C}} ` APPROX `0AV _{R _{s} (p-p)} `("for"`I _{s} )`=`0.429 ANGLE -3.85 DEG A`X``10 OMEGA `=``4.29 ANGLE -3.85 DEG VV _{R _{s} (p-p)} `("for"`I _{L} )`=`0.429 ANGLE -3.85 DEG A`X``10 OMEGA `=``4.29 ANGLE -3.85 DEG VV _{R _{s} (p-p)} `("for"`I _{C} )`=`0A`X`10 OMEGA `=`0VEXERCISE1. DetermineX _{L},X _{C} and thenZ _{T} for the network of Fig. 9.7 at a frequency of 10kHzX _{L} =2 pi fL=2 pi TIMES 10kHz TIMES 10mH=628.32 OMEGA X _{C} = {1} over {2 pi fC} = {1} over {2 pi TIMES 10kHz TIMES 0.01 mu F} =`1.59k OMEGA Z _{T} = {1} over {sqrt {( {1} over {R} ) ^{2} +( {1} over {X _{C}} - {1} over {X _{L}} ) ^{2}}} = {1} over {sqrt {( {1} over {1000} ) ^{2}nd the impedance of the capacitor is determined by this equationX _{C} = {1} over {2 pi fC}, if the frequency is very low at non-zero, the impedance of the coil becomes very small and the impedance of the capacitor is very high, so the overall impedance will be similar to the impedance of the coil. Since the devices are connected in parallel, the total impedance may not have a greater value than the impedance of these devices.CONSIDERATIONWhen AC power is applied while coils, resistors, and capacitors are connected in parallel, the current flowing through each element is measured, and the voltage applied to this resistance is measured by adding a sensing resistor to obtain the current flowing through each element again. However, as the frequency is low, the impedance of the coil is very low and the impedance of the capacitor is very high, most of the current flows to the coil. Unlike other experiments where the impedance of the coil is very low compared to the sensing resistance of 10 81

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아주대학교 기초전기실험 A+ 결과보고서 Ch. 2 (AC)

REPORTIEEE Code of Ethics(출처: http://www.ieee.org)We, the members of the IEEE, in recognition of the importance of our technologies in affecting the quality of life throughout the world, and in accepting a personal obligation to our profession, its members and the communities we serve, do hereby commit ourselves to the highest ethical and professional conduct and agree:1. to accept responsibility in making decisions consistent with the safety, health and welfare of the public, and to disclose promptly factors that might endanger the public or the environment;2. to avoid real or perceived conflicts of interest whenever possible, and to disclose them to affected parties when they do exist;3. to be honest and realistic in stating claims or estimates based on available data;4. to reject bribery in all its forms;5. to improve the understanding of technology, its appropriate application, and potential consequences;6. to maintain and improve our technical competence and to undertake technolog employment by false or malicious action;10. to assist colleagues and co-workers in their professional development and to support them in following this code of ethics.위 IEEE 윤리헌장 정신에 입각하여 report를 작성하였음을 서약합니다.학 부: 전자공학과제출일:과목명: 기초전기실험교수명:분 반:학 번:성 명:CHAPTER AC 2. The OscilloscopePart 3. Sinusoidal Waveforms-Magnitudev`=`0.2sin`2 pi 500t"Vertical"`"Sensitivity"`=`100"mV"`"/"`"div""Horizontal"`"Sensitivity"`=`500 mu s`"/"`"div"As the peak voltage is 0.2V, the peak-peak voltage was set to 0.4V. According to this equationv = A sin 2 pi ft, the frequency is 500 Hz. According to this equationT= {1} over {f}, period isT`=`0.002sec`=`2000 mu sec.Because of peak-peak voltage400mV and"Vertical"`"Sensitivity"`=`100"mV"`"/"`"div", the oscilloscope displays the voltage of the graph over two spaces. Because of periodT`=`0.002sec`=`2000 mu sec and"Horizontal Sensitivity"`=`500 mu "V"`"/"`"div", The oscilloscope displays the period of the graph over four spaces.Part 4. Sinusoidal Waveforms-Frequencyvu sec.Because of peak-peak voltage800mV and"Vertical"`"Sensitivity"`=`200"mV"`"/"`"div", the oscilloscope displays the voltage of the graph over two spaces. Because of periodT`=100 mu sec and"Horizontal"`"Sensitivity"`=`20 mu "V"`"/"`"div", The oscilloscope displays the period of the graph over five spaces.CHAPTER AC 3. R-L-C ComponentsPart 1. ResistanceV _{R _{s} (DMM)}V _{R _{s} (peak)}V _{R _{s} (p-p)}I _{(p-p)}R42.75mV60.46mV120.92mV1.21mA3.31k OMEGAAccording to this equationV _{(peak)} `=` sqrt {2} V _{rm "rms"}, the peak voltage applied to the resistance is60.46mV. The peak-peak voltage is 120.92mV because it is twice the peak voltage. According to this equationI _{p-p} `=` {V _{R _{s} (p-p)}} over {R _{s}}, the peak-peak value of the voltage is 1.21mA. According to this equationR`=` {4V _{(p-p)}} over {I _{p-p}}, the resistance value of the resistance R is3.31k OMEGA . The resistance value of the resistance R calculated in this way is very similar to the resistance value measuree waveform defined byv`=`-1.5`+`2.5sin`2 pi (20`X`10 ^{3} )t on the scope pattern of Fig. 2.13. Include the vertical and horizontal sensitivities."Vertical"`"Sensitivity"`=`500mV`"/"`"div""Horizontal"`"Sensitivity"`=`25 mu s`"/"`"div"The maximum value of this waveform voltage is 1 V and the minimum value is -4 V. If you look at the drawn graph, it is drawn over 10 spaces for the voltage. Therefore, the vertical sensitivity is500mV`"/"`"div". According to this equationT= {1} over {f}, period isT`=0.00005sec. If you look at the drawn graph, it is drawn over 2 spaces for the period. Therefore, the horizontal sensitivity is25 mu s"/"`"div".Chapter AC 2.3. Write the sinusoidal expression for the waveform appearing in Fig. 2.12.Since the peak-peak voltage of the resistor Rs is 4V, according to this equationI _{p-p} `=` {4V} over {R _{s}}, the current flowing through the resistor Rs is0.2A. Since the rms voltage of the voltage source is 10V, the peak voltage of the voltage source is14V accord peak voltage peak-peak voltage and peak current with it, and then calculate the resistance value of the resistance R.In Chapters Ac 2, you can learn exactly what these mean and learn how to use the oscilloscope by taking various sinusoidal voltage equations, calculating frequencies and periods, and then looking at graphs that appear on the oscilloscope.In Chapter Ac 3, when a peak-peak voltage of 4V is applied to the resistor R, the rms voltage of the resistor Rs is measured, and a peak-peak voltage of the resistor Rs is calculated using this value and the resistance value. As the resistor R and the resistor Rs are connected in series, a peak-peak current of the same value flows, so the resistance R value was recalculated with it, and it was very similar to the measured value. Therefore, it can be seen that the experiment was performed well.Through this experiment, we learned how to operate the oscilloscope and the rms, peak, peak-peak voltage, and peak-peak current at AC voltage, whi

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아주대학교 기초전기실험 A+ 결과보고서 Ch. 15 (DC)

REPORTIEEE Code of Ethics(출처: http://www.ieee.org)We, the members of the IEEE, in recognition of the importance of our technologies in affecting the quality of life throughout the world, and in accepting a personal obligation to our profession, its members and the communities we serve, do hereby commit ourselves to the highest ethical and professional conduct and agree:1. to accept responsibility in making decisions consistent with the safety, health and welfare of the public, and to disclose promptly factors that might endanger the public or the environment;2. to avoid real or perceived conflicts of interest whenever possible, and to disclose them to affected parties when they do exist;3. to be honest and realistic in stating claims or estimates based on available data;4. to reject bribery in all its forms;5. to improve the understanding of technology, its appropriate application, and potential consequences;6. to maintain and improve our technical competence and to undertake technologSeries-Parallel R-C dc NetworkR _{1``measured} :1.18k OMEGA ,R _{2``measured} :3.24k OMEGAV _{1}V _{2}V _{3}V _{4}Calculated0.00V0.00V10.00V10.00VMeasured0.00V0.00V10.00V10.01VThe circuit is open when both capacitors are fully charged. So, the circuit will no longer have current flowing. Thus, it can be calculated that bothV _{1} andV _{2} will take0.00V and actually the voltage value0.00V was measured. Since capacitorsC _{1} andC _{2} are connected in parallel, the same voltage will be charged, and the charge of c will be charged because the supplied power is10.00V and actually the voltage value10.00V was measured.Part 4. Determining C (Actual Value)R _{``measured} :97.78k OMEGA TheorticalMeasuredtau 9.78sec10.95secTheorticalMeasuredtau 21.51sec25.62sec100 mu F220 mu FC _{meas}111.99 mu F262.02 mu FThe time constanttau is determined by the equationtau `=`RC. Thus, when the capacitance of the capacitor is100 mu F, we can calculatetau `=`97.78k OMEGA `X`100 mu F`=`9.78sec. Since the meatime constant value and the measured time constant value, and between the electric capacity of two given capacitors and the value calculated. This error is thought to be caused by the fact that, as a current flows in the resistance, heat is lost and that, in measuring the time when a voltage of 6.32 V is applied accurately, a human shall see and press to measure it.EXERCISE1. For the circuit shown in Fig. 15.5, determine the following:(a) Calculate the time constant for the circuit in Fig. 15.5The time constanttau is determined by the equationtau `=`RC. Thus, when the capacitance of the capacitor is10 mu F and the resistance is100k OMEGA , we can calculatetau `=`100k OMEGA `X`10 mu F`=`1sec.(b) Determine the capacitor voltage after ten time constants.When 5 time constants, the capacitor is fully charged. So, when ten time constants, the capacitor will be fully charged and the the capacitor voltage will be 10V.(c) What is the value of the circuit current after five time constants?When 5 agree with my answer in part 3(c).(e) How much electrical energy is stored in the capacitorThe electrical energyU stored in the capacitor can be obtained by equationU`=` {1} over {2} CV ^{2}. SoU`=` {1} over {2} `X`10 mu F`X`(10V) ^{2} `=`5`X`10 ^{-4} J.2. For the circuit in Fig. 15.5, draw the effective circuit at t = 0s, and t = 10s.(a) Calculate the initial current in the circuit.When t = 0s, the capacitor is not charged and current flow. Therefore the initial current isI`=` {10V} over {100k OMEGA } `=`0.1mA.(b) What is the initial voltage across the resistor and the capacitor?When t = 0s, the capacitor is not charged and current flow. So, the voltage across resistance is 10V and the capacitor voltage is 0V.(c) Relate these voltages to the initial current in the circuit. Is the data consistent?Using the initial current and resistance values, the initial voltage across the resistor is0.1mA`X`100k OMEGA `=`10V. So, the data is consistent.(d) What is the initial energy stored in the ctructed as shown in FIG. 15.3, and after a sufficient time, the resistors and the voltages applied to the capacitors were measured. And by comparing these values with the calculated values, the capacitor and resistance were found in the RC circuit.In Part 4, the time constant was obtained by constructing an RC circuit as shown in FIG. 15.4 and measuring the time until the voltage applied to the capacitor reached 6.32 V. And we could see how to calculate the time constant by comparing it with the calculated value. And, the measured time constant value was divided by the resistance value to obtain the electric capacity of the capacitor and compared with the actual electric capacity of the capacitor. Some errors occurred between the calculated time constant value and the measured time constant value, and between the electric capacity of two given capacitors and the value calculated. This error is thought to be caused by the fact that, as a current flows in the resistance, heat is lost and193

공학/기술| 2023.03.09| 6페이지| 2,500원| 조회(159)

아주대학교, 아주대, A+, 기초전기실험, 기전실

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