In: Science

Submitted By aalvar15

Words 1103

Pages 5

Words 1103

Pages 5

AC Circuits. Oscilloscope

Alvaro Alvarez

Partner: Shaine Powers

Group 1

Ta: Salman

4/1/14

Abstract:

In the experiment AC Circuits. Oscilloscope, the main objective was to use a signal generator (SG) to create a sine wae and using an oscilloscope we had to bserve the various components of a sine waves and there relationships; the component RMS, amplitude, period and frequency. Another objective was to use the RLC circuit to determine the resonant frequency. In order to complete these ojectives the circuit was first set up as in figure 1 and RMS, period, frequency and amplitude were calculated and measured. The experimental amplitude was 0.875Vac, the frequency was 1001Hz, and the period was one second. The difference beween the theoretical frequcy from the signal generator was compared to the experimental frequency giving a 0.1% error. For the second part of the lab the RLC circuit was set up (figure 2) and from there the resonant frequency was found for this was done three times (table 1). After this the voltage drop was measured in the range of 800-300Hz; the different frequencies withc corresponding average power that dissipated in the resistor was inputted into GA (table 2,). From there the power vs frequency graph was plotted. and finally the theoretical angular frequency 1.0846 x 104rad/sec was compared to the experimentally found frequency of 1.0820 x 104rad/sec, giving a 0.24% error.

Objective:

In experiment number nine “AC Circuits. Oscilloscope” The main objective in the experiment was to use a signal generator to produce an AC signal and determine the relationship between the rms value and the amplitude of the voltage of the wave and also the period and the frequency of the signal. A secondary objective was to use an RLC circuit and determine both the resonant frequency and the dissipated power of the main resistor.…...

...E663 Laboratory session 8: AC Operational Amplifiers Agrim Ganti University College London 5th December 2008 ABSTRACT The report is written on an investigation which comprises of testing three operational amplifier circuits with AC signals. The three types of circuit include the integrator, the AC inverting amplifier and the AC non-inverting amplifier circuit. The integrator circuit was tested with a square-wave and a sinusoidal wave input signal at 1kHz frequency. The results showed that the square-wave input signal produced a triangular wave output whereas the sinusoidal input produced a sinusoidal output signal with a positive 90 degree phase shift. Both output signals were showed to be the integral of their relative input signals. The output voltage gain of the AC inverting and the AC non-inverting amplifier circuits were tested with a frequency range of 100Hz to 10kHz. The results were plotted on a logarithmic scaled graph which showed that both amplifiers acted like high-pass filters, each amplifier achieving its maximum gain set by the specification at higher frequencies nearer to 10kHz. The only difference between the two op-amps was that the AC inverting amplifier achieved negative gain in comparison to the positive gain achieved by the AC non-inverting amplifier. For further investigation, the frequency was increased above 10kHz for the AC inverting amplifier circuit which showed a linear fall in gain, which was explained by the theory of slew rate......

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...klzxcvbnmqwertyuiopasdfghjklzxcvbnm Circuit City: Rejection PaperSimplicity Guaranteed4/11/2012 By: Da’Terryon Sturghill | What was Circuit City? Circuit City Stores, Inc. was an American vendor in brand-name buyer electronics, individual computers, entertainment software, and until early in 2000 came accompanied with large appliances. I mean Circuit City was an overconfident brand that America had grown to depend on through the years to offer many products which facilitate people to take pleasure in their lives. I have researched and found that the history of Circuit City is amazingly dating back to 1949. CircuitCity.com had been developed all-new and geared up to provide the customers with a wider range of products than previously offered in Circuit City stores or online in order to remain current with the continuously growth in technology as well as competition. Here is what I found of the projected items sold at the Circuit City establishment an extended variety of customer electronics, desktops, notebooks, net books, HDTVs, DVD and Blu-Ray players, DVRs, Home Theater Systems, GPS Players, Digital Cameras, Hi-Def Camcorders, Digital Picture Frames and so much more. Plus, provided a greatly expanded selection of computer mechanism, like computer cases, motherboards, memory and processors – all the things to promote your existing system, or build your own. Who were Circuit City competitors? Circuit City’s key competitor was Best Buy......

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...Circuit City Circuit City The Circuit City story began in 1949 when entrepreneur Sam Wurtzel opened a small store in Richmond, VA, after he learned that the south’s first television station was going on the air. Over the next 20 years, Sam grew his company into a 100 store chain. His son, Alan became CEO in 1972 and created the ground-breaking superstore format that evolved into Circuit City. When Alan retired from the board in 2000, Circuit City was a fortune 500 company with more than 600 stores and 60,000 employees. Circuit City was a shining example of exceptional management practices in a 2001 best- selling book “Good to Great.” However, cracks were beginning to show in the company’s foundation. Circuit City’s rise and fall revealed leadership lessons and emphasizes the critical strategic role that training and development play in a company’s continued success. For its first 50 years, Circuit City (originally called Wards TV) was a pioneer in the rapidly changing consumer electronics industry. The following are what made the company great. Things that made the company great THE FIRST BIG BOX SUPERSTORE In 1975, Alan Wurtzel transformed the retail landscape in America by creating the first big-box superstore with a focus on Savings, Selection, Service, and Satisfaction. His “4 S’s” were captured in policies and processes that were easily understood by the customer and easily executed by his employees. Circuit City offered a low price guarantee, a 30 day......

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...(take r= infinity) to a point 0.5 m from a charge Q = 20µC? Example 17-6: Calculate the electric potential (a) At point A in the fig. due to the two charges shown and (b) at point B Potential Due to Electric Dipole; Dipole Moment Two equal point charge of opposite sign, separated by a distance l, are called electric dipole. The potential due to an electric dipole is just the sum of the potentials due to each charge, and can be calculated exactly. Approximation for potential far from dipole: Or, defining the dipole moment p = Ql, Capacitance A capacitor consists of two conductors that are close but not touching. A capacitor has the ability to store electric charge. (a) Parallel-plate capacitor connected to battery. (b) is a circuit diagram. When a capacitor is connected to a battery, the charge on its plates is proportional to the voltage: The quantity C is called the capacitance. Unit of capacitance: the farad (F), 1 F = 1 C/V Capacitance and Dielectric The capacitance does not depend on the voltage; it is a function of the geometry and materials of the capacitor. For a parallel-plate capacitor: Dielectric: A dielectric is an insulator, and is characterized by a dielectric constant K. Capacitance of a parallel-plate capacitor filled with dielectric: Dielectric strength is the maximum field a dielectric can experience without breaking down. The molecules in a dielectric tend to become oriented in a way that reduces the external field. Capacitance and......

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...The integrated circuit was invented in 1958 and 1959 by Jack Kilby and Robert Noyce. Before the integrated circuit was created, computers used single transistors and before transistors vacuum tubes were used. However, vacuum tubes were inefficient as they gave off so much heat and used enormous amounts of energy which is why the invention of transistors, which used semiconductors, was a breakthrough in the technology. Transistors are used on integrated circuits and scientists have been able to increase the amount that can fit on a circuit in record numbers. Gordon Moore, a cofounder of Intel, noticed a trend of the capacity of each new chip that was created compared to its predecessor. He found that the size nearly doubled every two years and this still holds true for today. Below is a graph showing the transistor count vs. the year the transistor was introduced and we can see that every year the count increases in a linear fashion. This is actually amazing because not only is the count of transistors increasing the chips themselves are decreasing in size. This is the result of new materials being used such as silicon and graphene. The amount of transistors that can fit on a single chip is somewhere between 2.5 billion to 3 billion, however if Moore’s law continues to hold true, which history shows that it has, this number is expected to increase in years to come. The progression of technology is really incredible as computers have gone from the size of a large...

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...when the noninverting input is 1 mV. If the open-loop gain of the op amp is 2 × 106, what is the inverting input? Chapter 5, Solution 4. v0 = Avd = A(v2 - v1) v −4 = −2μV v2 - v1 = 0 = A 2x10 6 v2 - v1 = -2 µV = –0.002 mV 1 mV - v1 = -0.002 mV v1 = 1.002 mV PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5, Problem 5. For the op amp circuit of Fig. 5.44, the op amp has an open-loop gain of 100,000, an input resistance of 10 kΩ, and an output resistance of 100 Ω. Find the voltage gain vo/vi using the nonideal model of the op amp. Figure 5.44 for Prob. 5.5 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5, Solution 5. I vd + vi + R0 Rin +......

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...Name: _Amogh Prabhakar____________ Date: Oct. 10, 2014___ CP: _____ Virtual Circuit Lab Learning Goals: Students will be able to understand Ohm’s Law. Students will be able to see the relationship between voltage, current, and resistance. Go to ⎝Battery-Resistor Circuit. Click on the green “Run Now” button. The simulation should look like the picture to the right. 1. Change the resistance and voltage. Observe what happens to the current. Note the relationship you observed between each of the following: (direct, inverse, none) a. resistance and current = _indirect_______________________ b. voltage and current = direct__________________________ Go to ⎝Ohm's Law. Click on the green “Run Now” button. The simulation should look like the picture to the right. 2. What is the current through a resistor with the following resistances? Let voltage = 6 V a. R = 100 ohms I = _60__ mA(current) b. R = 300 ohms I = 20____mA(current) 3. Now, determine the current through the wire with the following volts. Let resistance = 500 ohms a. Volts = 3 V I = 6___mA(current) b. Volts = 6V I = 12____mA(current) 4. Think about the formula (V=IR), does this make sense according to this formula? Explain! (Be sure to include the relationship between resistance and current, and the relationship between voltage and current in your answer) Yes,......

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...ET 145 AC Electronics | Class Work 1 | (Print) Name ______________________ | Pre Class Work Please solve for the following voltages Voltage dropped across R1 VR1 Voltage dropped across R2 VR2 Voltage at point A VA Voltage at point B VB Voltage at point C VC | | Please solve for the total resistance of three resistors that are in parallel if the first one is 6MΩ, the second one is 0.05 GΩ, and the third one is 900 kΩ. Class Work 1 DefinitionsSine Waves - The sinusoidal waveform (sine wave) is the fundamental Alternating Current (AC) and alternating voltage waveformA wave is a disturbance | | Period (T) - The duration of one complete cycle of a wave or oscillation (measured in seconds, s) The time it takes for a wave to repeat itself (and cycle through again) Sine waves are characterized by amplitude & period; The amplitude is the max value of a voltage or current, The period is the time interval for one complete cycle The period of a sine wave can be measured between any two corresponding points on the waveform By contrast, amplitude of a sine wave is only measured from the center to the maximum point Frequency ( f ) - The number of cycles a (sine) wave completes in one second (measured in hertz, Hz) Frequency ( f ) and Period (T) are reciprocals: f= 1T T= 1f | Alternating Current (AC) | Direct Current (DC) | Amount of energy that can be carried: | Can travel long distances & provide more power | Cannot travel...

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...Guided Study: Resistive circuit Working Examples 1. For the circuit shown in Fig.Q1 calculate the values for (a) the current through each resistor, (b) the p.d. across each resistor and (c) the power dissipated by the 20Ω resistor. 20V Fig.Q1 Solution: 1 1 1 R AB 25 15 R AB 9.375 1 1 1 1 RBC 10 10 20 RBC 4 R AC R AB RBC 13.375 I 20 20 1.495 A R AC 13.375 (a) By using Current Divider Rule I1 I2 I3 I4 I5 R AB 9.375 I 1.495 A 0.56 A 25 25 R AB 9.375 I 1.495 A 0.935 A 15 15 RBC 4 I 1.495 A 0.598 A 10 10 RBC 4 I 1.495 A 0.598 A 10 10 RBC 4 I 1.495 A 0.299 A 20 20 (b) The voltage across reistors 25Ω and 15 Ω is VAB V AB I R AB 1.495 9.375 14.019V The voltage across reistors 10Ω, 10Ω and 20 Ω is VBC VBC I RBC 1.495 4 5.981V (c) The power dissipated by the 20Ω resistor is 2 VBC P20 1.789W 10 or I 2 20 1.788W or I VBC 1.788W 2. Determine the p.d. between terminals E and F of the circuit in Fig.Q2. 40V E F Fig.Q2 Solution: RBF 10 10 25 1 RBCF I 1 1 15 25 RBCF 9.375 R ACF 10 RBCF 19.375 40 2.065 A R ACF VBCF I RBCF 19.355V Or by using Voltage Divider Rule R VBCF BCF 40 19.355V R ACF Then VEF 10 10 VBCF 19.355 7.742V RBF 25 3. For the circuit of Fig. Q3 calculate (a) the p.d. across the 8Ω resistor, (b)...

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...Introduction 4 Astable Multivibrators 4 Overview of the 555 Timer 5 Integrated Circuit 5 Semiconductor material 7 Current and Resistance 9 Potentiometer 10 Calculation of the Voltages 11 Transistors 11 Light Emitting Diode (LED) 14 Capacitance 14 555 Timer Operations 15 Operation in the Astable State 17 Aim, Hypothesis, and Calculations 18 Aim 18 Hypothesis 19 Materials 20 Method 20 Variables 21 Independent variable 21 Dependant variable 22 Controlled variable 22 Results 23 Table 1: Theoretical Values of varying Resistor R1 23 Table 2: Experimental values varying resistor 1 (R1) 24 Table 3: Theoretical values varying resistor 2 (R2) 25 Table 4: Experimental values varying resistor 2 (R2) 26 Data Analysis and Discussion of Trends Using Appropriate Pot 1 27 Trend 27 Matching the Frequencies of the Chosen Songs 29 Overall Results 30 Discussion 31 Conclusion 38 References 40 Appendix 43 Error Calculations 43 The extra resistor from the wires connecting the components in the circuit 43 The effect of temperature on the resistivity of the fixed resistors in the circuit 43 Calculations of best pot 44 Choice of Resistor and Pot 44 Calculation of Frequency Ranges 44 Introduction Shaping and generation of waves is done using electronic circuits known as multivibrators. These circuits produce outputs that can be characterized as either stable or unstable in......

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...Switch-Mode DC-AC Inverters 1. 2. 3. 4. 5. Introduction Basic Concepts of Switch-Mode Single-Phase Inverters Three-Phase Inverters Effects of Blanking Time on Output Voltage in PWM 6. Rectifier Mode of Operation • Main applications: – ac motor drives; – uninterruptible power supplies; – grid connection of renewable energy sources 1 1 Switch-Mode DC-AC Inverter • Block diagram of a motor drive where the power flow is unidirectional FEUP-MIEEC. Industrial Electronics 2 2 Switch-Mode DC-AC Inverter • Block diagram of a motor drive where the power flow can be bi-directional FEUP-MIEEC. Industrial Electronics 3 3 Switch-Mode DC-AC Inverter • Four quadrants of operation FEUP-MIEEC. Industrial Electronics 4 One Leg of a Switch-Mode DC-AC Inverter • The mid-point shown is fictitious • Similar to the topology dc - dc for two-quadrants operation FEUP-MIEEC. Industrial Electronics 5 Synthesis of a Sinusoidal Output by PWM Carrier wave (triangle) Modulating wave (sine) Amplitude modulation index: ma = ˆ Vm ˆ Vtri Frequency modulation index: mf = fs f1 • Typical harmonic spectrum shows voltage components around multiples of the switching frequency FEUP-MIEEC. Industrial Electronics 6 Details of a Switching Time Period • Control voltage can be assumed constant during a switching time-period FEUP-MIEEC. Industrial Electronics 7 Analysis of output voltage • In the......

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...Q. How should Pepcid be positioned- treatment or prevention- and what are the implications? Pepcid Ac would work best when positoned as a treatment drug.We can look at it from different levels: Market demand/ Market research- - 2 tests with BASES methodology indicate that treatment has greater appeal over prevention with respect to the target groups- dual users as well as traditional antacid users. (Table D). While the focus groups and concept tests favored prevention and treatment (both together). However, JJ’s extensive experience with this BASES technique makes the findings more reliable. JJIM”s core competency- Merck’s strength is is R&D’s expertise. The treatment positioning focuses on the product’s benefits.It outperforms the existing antacids in the OTC market.However the medication’ s ability to prevent is questionable. The treatment strategy will also rely on JJ’s core competency is its consumer marketing and sales. In the recent past (1990 t0 1993) JJ successfullyswitched from prescription to OTC medications , twice. For instance,JJ drove Immonium sales from 31.5 million in 1987 $ to 125 million in 5 years. (table b) FDA as a collaborator- FDA’s rjected JJM’s to sell Pepcic AC on the grounds that the clinical trails failed to show its perforamnce as prevention medication.If JJM drops this claim it has a higher chance to be approved to be the first – to – market. Also, it is more advantageous for it to be the first-entrant to the OTC switch......

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..."Bad" and "Burned Out" Integrated Circuits Some Conventions - - - - - - - CHAPTER 2 SOME TTL INTEGRATED CIRCUITS . . . . . . . . . 38 Type Numbers and Descriptions CHAPTER 3 LOGIC. . . . . . . . . . . . . . . . . . 122 - The Two-Input Gate as a Simple Switch State Definitions: One-Input Logic - Two-Input Logic Other What Is a Zero? A Trick Called DeMorgan9s TheoTwo-Input Logic Functions Advanced Logic rem - Open-Collector Logic - Tri-State Logic Design: Data-Selector Logic - Advanced Logic Design: The ReadOnly Memory - Some Examples and Logic Design Rules The ASCII Computer Code - - - - - CHAPTER 4 GATE TIMER AND CIRCUITS . . . . . . . . . . . . 158 - Two Cross-Coupled Inverters - Improved Triggering The SetEdge Triggering Using RS Flip-Flops - The Reset Flip-Flop A High-Impedance Interface - Other Interface Schmitt Trigger A Wide-Range Voltage-Controlled Circuits - Signal Sources Oscillator - Another Crystal Oscillator - The 555 and MC1555 -Two-Tone Alarm Tempo Generator or Electronic Metronome Digital Capacitance Measurement - Brightness Control for a Electronically Variable Time Constant: A Music Digital Display Monostable Multivibrators and Pulse Attack-Decay Generator The Half-Monostable Multivibrator The 555 as a Generators Monostable Multivibrator Frequency Meter or Tachometer Digital Thermometer - Negative-Recovery Circuits - l T L Monostable......

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...SUBMITTING YOUR WORK. 1. Given the resistive network below with the given values. R1 = R8 = 1000 Ohms, R2 = R3 = 400 Ohms, R4 = R6 = 100 Ohms and R5 = 300 Ohms. Find Rab and Rac. (20 points) [pic] 2. For the given circuit below, find the following: [pic] 3. For the given figure below, find the following: [pic] 4. The resistance of a certain device is equal to 1000 Ohms at a temperature of 15OC. Its temperature coefficient is 0.001/OC and displays a positive response towards a change in temperature. Find the following under the assumption that a linear behavior is observed: a. The slope of the line. (5 points) b. The temperature and resistance intercepts. (10 points) c. The resistance at the temperature T = 80OC. (5 points) 5. The charge flowing in a wire is given by the figure below. Determine the following a) current flowing through the wire at each time interval and the average current through the wire. (15 points) [pic] ----------------------- 1. The current i and v. (10 points) 2. The current through the 8-Ohm resistor. (5 points) 3. The power dissipated by each resistive element. (10 points) 4. The power supplied by the 9-A source. (5 points) 5. How many nodes, branches and loops are in the circuit? (10 points) 1. The total current supplied by the battery. (5 points) 2. The current through each bulb. (10 points) 3. The resistance of each bulb. (10 points)...

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...[pic] Figure 4 7. The circuit in Figure 5 consists of a battery with EMF ε = 3.0 V, a resistor with resistance R = 10 Ω, an ammeter, and a voltmeter. The voltmeter and the ammeter (labeled V and A, respectively) can be considered ideal; that is, their resistances are infinity and zero, respectively. The current in the resistor is I, and the voltage across it is V. The internal resistance of the battery rint = 0.2 Ω. a. What will the ammeter read?= 0.294A b. What is the terminal voltage?= 3V c. What will the voltmeter read? = 2.94V d. What is the power lost in the battery? = 0.0173W e. What is the power PR dissipated in the resistor? = 0.8644W f. What is the total power generated by the battery? = 0.882W [pic] Figure 5 8. For the circuit shown in Figure 6, find q. Equivalent resistance of the circuit.= [(1/820 +1/680)^-1 + 470] = 841.73ohm r. Current through the 470 Ω resistor. = 0.01426 A (using current divider rule) s. Current through the 820 Ω resistor. = 0.006463A t. Current through the 680 Ω resistor. = 0.007795A [pic] Figure 6 9. For the circuit shown in Figure 7, R1 = R2 = R3 = R4 = 6 Ω, and the battery voltage V = 6 V. u. Find equivalent resistance of the circuit when the switch is open as shown. = 9ohm v. Find current through R1 when the switch is open. = 0.66A w. Find equivalent resistance of the circuit when the switch is closed. =......

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