This document contains 13 multiple choice questions related to electric circuits. The questions involve calculating voltages, currents, resistances and powers in circuits containing resistors, capacitors and inductors. Complex circuit diagrams and mathematical equations are provided. The correct answers to each question are given as choices A, B, C or D without explanations.
This document provides chapter summaries and example problems from a solutions manual for a textbook on electromagnetism. It includes 20 chapters that cover topics like vector calculus, electrostatics, magnetostatics, and Maxwell's equations. The document also provides notes on mathematical expressions and references for additional resources related to the textbook.
This document contains questions for an examination on Low Power VLSI Design. It begins with instructions noting that candidates should answer any 5 questions out of 7 and state any assumptions made. The questions cover various topics related to low power VLSI design including needs for low power chips, sources of power dissipation in digital circuits, techniques to minimize power dissipation, impact of transistor sizing and technology scaling on power, low voltage circuit techniques, clock distribution schemes, and logic simulation.
This document provides solved problems from GATE (Graduate Aptitude Test in Engineering) exams from 1996-2013 in the subject of Engineering Mathematics. It contains one-mark and two-mark questions related to topics like differential equations, matrices, probability, complex analysis etc. along with their solutions. The document is available for purchase in digital format at www.nodia.co.in in separate sections or 'units'.
The document appears to be an exam question paper that covers various topics related to advanced mathematics, digital VLSI design, embedded systems, ASIC design, VLSI process technology, and related subjects. It contains 10 questions with varying point values and instructs students to answer any 5 full questions. The questions cover technical topics such as matrix operations, MOS transistor modeling, logic design, processor architecture, ASIC design flows, silicon crystal growth, and more.
This document contains a sample multiple choice question (MCQ) exam for electronics and communication engineering. It includes 15 sample questions related to basic electrical concepts such as current, voltage, resistance, capacitance and circuits. It also provides the solutions and explanations for each question. The document is from a book titled "GATE EC by RK Kanodia" which is intended to help students prepare for the graduate aptitude test in engineering exam for electronics and communication.
This document contains a summary of a student's third semester examination in field theory. Some key points:
1) The exam had two parts - Part A covered electrostatics and Part B covered magnetostatics.
2) In Part A, the student was asked to define electric field intensity, derive Maxwell's first equation, find potential due to line and point charges, and solve Laplace's equation for different boundary value problems.
3) In Part B, the student was asked to derive expressions for magnetic field and force between current elements, define displacement current density, and derive Maxwell's equations for time-varying fields.
4) The final section covered electromagnetic wave propagation - including deriving the wave
This document contains the answers to a 30 question multiple choice network theory test taken on June 10, 2017. The answers are provided in a numbered list from 1 to 30 without the corresponding questions. An answer key for the test is also included which provides the answers. The document aims to provide the answers to a network theory test for a student.
This document contains the solutions to exercises from Chapter 16. The first exercise solves for the threshold voltage (VTN) of a NMOS transistor. The second exercise calculates the output voltage and current of a common source amplifier. The third exercise analyzes the output characteristics of an inverter. The remaining exercises involve additional circuit analysis problems related to topics in Chapter 16 such as CMOS inverters, logic gates and power calculations.
This document provides chapter summaries and example problems from a solutions manual for a textbook on electromagnetism. It includes 20 chapters that cover topics like vector calculus, electrostatics, magnetostatics, and Maxwell's equations. The document also provides notes on mathematical expressions and references for additional resources related to the textbook.
This document contains questions for an examination on Low Power VLSI Design. It begins with instructions noting that candidates should answer any 5 questions out of 7 and state any assumptions made. The questions cover various topics related to low power VLSI design including needs for low power chips, sources of power dissipation in digital circuits, techniques to minimize power dissipation, impact of transistor sizing and technology scaling on power, low voltage circuit techniques, clock distribution schemes, and logic simulation.
This document provides solved problems from GATE (Graduate Aptitude Test in Engineering) exams from 1996-2013 in the subject of Engineering Mathematics. It contains one-mark and two-mark questions related to topics like differential equations, matrices, probability, complex analysis etc. along with their solutions. The document is available for purchase in digital format at www.nodia.co.in in separate sections or 'units'.
The document appears to be an exam question paper that covers various topics related to advanced mathematics, digital VLSI design, embedded systems, ASIC design, VLSI process technology, and related subjects. It contains 10 questions with varying point values and instructs students to answer any 5 full questions. The questions cover technical topics such as matrix operations, MOS transistor modeling, logic design, processor architecture, ASIC design flows, silicon crystal growth, and more.
This document contains a sample multiple choice question (MCQ) exam for electronics and communication engineering. It includes 15 sample questions related to basic electrical concepts such as current, voltage, resistance, capacitance and circuits. It also provides the solutions and explanations for each question. The document is from a book titled "GATE EC by RK Kanodia" which is intended to help students prepare for the graduate aptitude test in engineering exam for electronics and communication.
This document contains a summary of a student's third semester examination in field theory. Some key points:
1) The exam had two parts - Part A covered electrostatics and Part B covered magnetostatics.
2) In Part A, the student was asked to define electric field intensity, derive Maxwell's first equation, find potential due to line and point charges, and solve Laplace's equation for different boundary value problems.
3) In Part B, the student was asked to derive expressions for magnetic field and force between current elements, define displacement current density, and derive Maxwell's equations for time-varying fields.
4) The final section covered electromagnetic wave propagation - including deriving the wave
This document contains the answers to a 30 question multiple choice network theory test taken on June 10, 2017. The answers are provided in a numbered list from 1 to 30 without the corresponding questions. An answer key for the test is also included which provides the answers. The document aims to provide the answers to a network theory test for a student.
This document contains the solutions to exercises from Chapter 16. The first exercise solves for the threshold voltage (VTN) of a NMOS transistor. The second exercise calculates the output voltage and current of a common source amplifier. The third exercise analyzes the output characteristics of an inverter. The remaining exercises involve additional circuit analysis problems related to topics in Chapter 16 such as CMOS inverters, logic gates and power calculations.
This document contains solutions to problems in Chapter 13 regarding computer simulations and analyses of bipolar junction transistor circuits. The key points summarized are:
1) Problem 13.1 provides a computer simulation of a circuit and calculates the overall gain as -1.59x106.
2) Problem 13.6 calculates various currents in a BJT reference circuit, including the reference current of 0.5 mA and collector current of 21.1 μA.
3) Problem 13.8 calculates currents in another BJT reference circuit, including the reference current of 0.22 mA, collector current of 14.2 μA, and other node currents.
This document contains information about an engineering mathematics exam for a fourth semester bachelor's degree program. It provides details about the exam such as the duration, maximum marks, and instructions to answer questions from each part of the exam. The document then lists the questions in two parts - Part A and Part B. Part A contains questions on topics like Taylor series, Runge-Kutta method, Adams-Bashforth method, systems of differential equations, and Bessel functions. Part B contains questions on Laplace's equation in cylindrical coordinates, Legendre polynomials, probability, distributions, hypothesis testing, and curve fitting.
This document contains solutions to exercises from Chapter 13. The solutions involve calculating various currents and voltages in BJT amplifier circuits. Equations are provided and values are calculated for currents, voltages, gains, and other circuit parameters. Detailed steps are shown for each calculation.
The document is from a solutions manual for a power electronics textbook. It provides step-by-step solutions to example problems from Chapter 2 on power semiconductor diodes and circuits and Chapter 3 on diode rectifiers. The problems cover topics like calculating diode voltage and current characteristics, analyzing diode circuits, designing rectifier circuits, and calculating output voltages and currents for various rectifier configurations.
This document contains information about a computer aided engineering drawing examination, including instructions, questions, and diagrams. Question 1 involves drawing projections of points and lines. Question 2 involves drawing projections of hexagonal and frustum pyramids. Question 3 involves drawing isometric projections of a pentagonal pyramid or reducing a frustum of a square pyramid to development of its lateral surfaces. The examination tests skills in technical drawing, geometry, and spatial visualization.
This document contains solved examples of circuit analysis problems involving nodes, meshes, and supernodes/supermeshes. Key results include:
- Node analysis of a two-node circuit gives voltages of v1 = -6 V and v2 = -42 V.
- Mesh analysis of a three-mesh circuit with a dependent source gives currents of i1 = 4.632 A, i2 = 631.6 mA, and i3 = 1.4736 A.
- Circuit analysis using the concept of a supernode yields voltages of v = -400 mV and v1 = 5.6 V for a two-node circuit.
This document contains the solutions to various exercises related to circuit analysis and electronics. Exercise problems cover topics such as diode circuits, rectifiers, filters, and transistor circuits. The solutions involve calculating voltages, currents, power, capacitance values, and other circuit parameters through applying circuit laws and relationships. PSpice simulations are also used to verify some of the circuit responses.
The document contains solutions to example problems from Chapter 17 of an exercise book. Problem EX17.1 involves calculating currents in a circuit with a voltage source and two resistors. Part a calculates the currents with different input voltages. Part b calculates the node voltages. EX17.2 calculates power dissipation for a logic gate circuit with different input voltages.
This document appears to be notes for a chemistry experiment involving the hydrolysis of tert-butyl chloride. It provides the chemical reaction, initial concentrations of reagents, volumes used, conductivity measurements over time, and asks the student to:
1. Write the ionic equation for the hydrolysis reaction
2. Calculate the amount of product formed
3. Derive an expression relating conductivity to concentration of ions
4. Calculate the concentration of product over time
5. Determine the time taken for complete reaction
6. Plot concentration against time and determine the rate law
7. Calculate values at t=50s and the initial rate of reaction
8. Use conductivity and ion concentrations to calculate molar conductivity values
This document appears to contain exam questions for the subject "Electronic Circuits". It includes questions related to BJT operating point, UJT construction and operation, MOSFET and CMOS characteristics, photoconductors and optocouplers. Some sample calculations are provided related to photodiode parameters like NEP, detectivity, quantum efficiency. The document tests knowledge of fundamental electronic devices and circuits.
This doctoral thesis models pulse propagation in optical fibers using finite-difference methods. It summarizes the numerical model, which solves the nonlinear Schrödinger equation describing pulse propagation using Crank-Nicholson and split-step Fourier methods. It tests the model's accuracy by comparing results to analytic solutions and a commercial simulation program. Effects like dispersion, loss, self-phase modulation and polarization mode dispersion are modeled. Additional models are presented for optical amplifiers and filters used in the fiber links.
Solutions manual for microelectronic circuits analysis and design 3rd edition...Gallian394
Solutions Manual for Microelectronic Circuits Analysis and Design 3rd Edition by Rashid IBSN 9781305635166
Download at: https://goo.gl/ShMdzK
People also search:
microelectronic circuits analysis and design rashid pdf
microelectronics circuit analysis and design 3rd edition pdf
microelectronic circuits analysis and design pdf
microelectronic circuits analysis and design rashid solution manual
electronic circuit analysis and design
1) The document contains mathematical equations and symbols relating to electrical circuits, energy, and particle physics.
2) Variables include resistance, current, time, capacitance, and energy levels with subscripts and primes denoting separate but related quantities.
3) Equations appear to relate these variables mathematically, such as relating current over time to resistance and voltage in a circuit.
This document contains solutions to 10 exercises involving bipolar junction transistors and MOSFET circuits. The key points summarized are:
1) Exercise 10.1 calculates the reference and base currents in a BJT circuit. Exercise 10.2 calculates the reference current, zero bias current, and small signal output resistance.
2) Exercise 10.3 calculates the various currents in a multi-emitter BJT. Exercise 10.4 calculates the early effect resistance and voltage difference between two BJTs.
3) Exercise 10.5 again calculates reference current and zero bias current. Exercises 10.6 through 10.11 involve various MOSFET and BJT circuit calculations.
This document contains solutions to problems from Chapter 11. Problem 11.1 solves for resistances in a common emitter amplifier circuit. Part c) finds the maximum and minimum voltages. Problem 11.2 calculates the common-mode gain of a differential amplifier as -0.01012. Problem 11.3a) solves for currents and voltages in a multi-emitter transistor. Part b) solves for currents when the collector voltage is changed.
This document provides information about digital electronics and logic gates:
- It defines analog and digital signals and gives examples of each.
- It provides truth tables for basic logic gates like AND, OR, NAND, NOR and explains how digital inputs can represent 1s and 0s.
- It discusses logic expressions and how they can be simplified using Boolean algebra identities. Diagrams show how logic expressions can be implemented using logic gates.
- It provides details about common logic IC packages like their pin configurations and what types of gates they contain.
- Threshold voltages are discussed for interpreting 1s and 0s between different logic circuits.
- A truth table is given as an example to summarize the
This document appears to be an exam paper for a basic thermodynamics course consisting of multiple choice and numerical problems. Some key points:
1) It asks students to differentiate between control mass/volume and intensive/extensive properties and classify some examples.
2) It includes problems on gas thermometry, gas expansion processes, calculating work done by gases, and the steady flow energy equation as applied to systems like turbines and nozzles.
3) Questions cover concepts like the zeroth law of thermodynamics, definitions of work, the first law of thermodynamics as an equation, and analyzing compressor processes.
This document contains 25 multiple choice questions from a GATE EE exam, along with explanations for each answer. The questions cover topics such as probability, complex variables, circuit analysis, differential equations, and electric machines. For each question, the correct answer is identified and a brief explanation of the solution steps is provided.
1. The question document contains details about an engineering mathematics examination including 5 questions from Part A and 3 questions from Part B.
2. The questions cover topics such as Fourier series, numerical methods, differential equations, and Laplace transforms.
3. Students are required to answer 5 full questions by selecting at least 2 questions from each part.
This document contains the instructions and questions for a mathematics examination. It includes 12 questions split into two sections. The questions cover topics like differential equations, derivatives, integrals, and vector calculus. Students are instructed to answer 1 question from each section, show working, use diagrams where needed, and assume suitable data if questions don't provide enough information. Calculators, tables and other aids are permitted.
The document describes a proposed social website project that would allow people to connect with others and share information. It discusses the objectives of allowing existing members to view new members and increase social connections. It outlines some potential issues with security, portability and compatibility. The benefits are seen as improved accessibility to information and keeping in touch with low cost. It provides an overview of the key modules to be included and proposes future enhancements around privacy features and administrative functions to prepare for a successful launch.
This document contains solutions to problems in Chapter 13 regarding computer simulations and analyses of bipolar junction transistor circuits. The key points summarized are:
1) Problem 13.1 provides a computer simulation of a circuit and calculates the overall gain as -1.59x106.
2) Problem 13.6 calculates various currents in a BJT reference circuit, including the reference current of 0.5 mA and collector current of 21.1 μA.
3) Problem 13.8 calculates currents in another BJT reference circuit, including the reference current of 0.22 mA, collector current of 14.2 μA, and other node currents.
This document contains information about an engineering mathematics exam for a fourth semester bachelor's degree program. It provides details about the exam such as the duration, maximum marks, and instructions to answer questions from each part of the exam. The document then lists the questions in two parts - Part A and Part B. Part A contains questions on topics like Taylor series, Runge-Kutta method, Adams-Bashforth method, systems of differential equations, and Bessel functions. Part B contains questions on Laplace's equation in cylindrical coordinates, Legendre polynomials, probability, distributions, hypothesis testing, and curve fitting.
This document contains solutions to exercises from Chapter 13. The solutions involve calculating various currents and voltages in BJT amplifier circuits. Equations are provided and values are calculated for currents, voltages, gains, and other circuit parameters. Detailed steps are shown for each calculation.
The document is from a solutions manual for a power electronics textbook. It provides step-by-step solutions to example problems from Chapter 2 on power semiconductor diodes and circuits and Chapter 3 on diode rectifiers. The problems cover topics like calculating diode voltage and current characteristics, analyzing diode circuits, designing rectifier circuits, and calculating output voltages and currents for various rectifier configurations.
This document contains information about a computer aided engineering drawing examination, including instructions, questions, and diagrams. Question 1 involves drawing projections of points and lines. Question 2 involves drawing projections of hexagonal and frustum pyramids. Question 3 involves drawing isometric projections of a pentagonal pyramid or reducing a frustum of a square pyramid to development of its lateral surfaces. The examination tests skills in technical drawing, geometry, and spatial visualization.
This document contains solved examples of circuit analysis problems involving nodes, meshes, and supernodes/supermeshes. Key results include:
- Node analysis of a two-node circuit gives voltages of v1 = -6 V and v2 = -42 V.
- Mesh analysis of a three-mesh circuit with a dependent source gives currents of i1 = 4.632 A, i2 = 631.6 mA, and i3 = 1.4736 A.
- Circuit analysis using the concept of a supernode yields voltages of v = -400 mV and v1 = 5.6 V for a two-node circuit.
This document contains the solutions to various exercises related to circuit analysis and electronics. Exercise problems cover topics such as diode circuits, rectifiers, filters, and transistor circuits. The solutions involve calculating voltages, currents, power, capacitance values, and other circuit parameters through applying circuit laws and relationships. PSpice simulations are also used to verify some of the circuit responses.
The document contains solutions to example problems from Chapter 17 of an exercise book. Problem EX17.1 involves calculating currents in a circuit with a voltage source and two resistors. Part a calculates the currents with different input voltages. Part b calculates the node voltages. EX17.2 calculates power dissipation for a logic gate circuit with different input voltages.
This document appears to be notes for a chemistry experiment involving the hydrolysis of tert-butyl chloride. It provides the chemical reaction, initial concentrations of reagents, volumes used, conductivity measurements over time, and asks the student to:
1. Write the ionic equation for the hydrolysis reaction
2. Calculate the amount of product formed
3. Derive an expression relating conductivity to concentration of ions
4. Calculate the concentration of product over time
5. Determine the time taken for complete reaction
6. Plot concentration against time and determine the rate law
7. Calculate values at t=50s and the initial rate of reaction
8. Use conductivity and ion concentrations to calculate molar conductivity values
This document appears to contain exam questions for the subject "Electronic Circuits". It includes questions related to BJT operating point, UJT construction and operation, MOSFET and CMOS characteristics, photoconductors and optocouplers. Some sample calculations are provided related to photodiode parameters like NEP, detectivity, quantum efficiency. The document tests knowledge of fundamental electronic devices and circuits.
This doctoral thesis models pulse propagation in optical fibers using finite-difference methods. It summarizes the numerical model, which solves the nonlinear Schrödinger equation describing pulse propagation using Crank-Nicholson and split-step Fourier methods. It tests the model's accuracy by comparing results to analytic solutions and a commercial simulation program. Effects like dispersion, loss, self-phase modulation and polarization mode dispersion are modeled. Additional models are presented for optical amplifiers and filters used in the fiber links.
Solutions manual for microelectronic circuits analysis and design 3rd edition...Gallian394
Solutions Manual for Microelectronic Circuits Analysis and Design 3rd Edition by Rashid IBSN 9781305635166
Download at: https://goo.gl/ShMdzK
People also search:
microelectronic circuits analysis and design rashid pdf
microelectronics circuit analysis and design 3rd edition pdf
microelectronic circuits analysis and design pdf
microelectronic circuits analysis and design rashid solution manual
electronic circuit analysis and design
1) The document contains mathematical equations and symbols relating to electrical circuits, energy, and particle physics.
2) Variables include resistance, current, time, capacitance, and energy levels with subscripts and primes denoting separate but related quantities.
3) Equations appear to relate these variables mathematically, such as relating current over time to resistance and voltage in a circuit.
This document contains solutions to 10 exercises involving bipolar junction transistors and MOSFET circuits. The key points summarized are:
1) Exercise 10.1 calculates the reference and base currents in a BJT circuit. Exercise 10.2 calculates the reference current, zero bias current, and small signal output resistance.
2) Exercise 10.3 calculates the various currents in a multi-emitter BJT. Exercise 10.4 calculates the early effect resistance and voltage difference between two BJTs.
3) Exercise 10.5 again calculates reference current and zero bias current. Exercises 10.6 through 10.11 involve various MOSFET and BJT circuit calculations.
This document contains solutions to problems from Chapter 11. Problem 11.1 solves for resistances in a common emitter amplifier circuit. Part c) finds the maximum and minimum voltages. Problem 11.2 calculates the common-mode gain of a differential amplifier as -0.01012. Problem 11.3a) solves for currents and voltages in a multi-emitter transistor. Part b) solves for currents when the collector voltage is changed.
This document provides information about digital electronics and logic gates:
- It defines analog and digital signals and gives examples of each.
- It provides truth tables for basic logic gates like AND, OR, NAND, NOR and explains how digital inputs can represent 1s and 0s.
- It discusses logic expressions and how they can be simplified using Boolean algebra identities. Diagrams show how logic expressions can be implemented using logic gates.
- It provides details about common logic IC packages like their pin configurations and what types of gates they contain.
- Threshold voltages are discussed for interpreting 1s and 0s between different logic circuits.
- A truth table is given as an example to summarize the
This document appears to be an exam paper for a basic thermodynamics course consisting of multiple choice and numerical problems. Some key points:
1) It asks students to differentiate between control mass/volume and intensive/extensive properties and classify some examples.
2) It includes problems on gas thermometry, gas expansion processes, calculating work done by gases, and the steady flow energy equation as applied to systems like turbines and nozzles.
3) Questions cover concepts like the zeroth law of thermodynamics, definitions of work, the first law of thermodynamics as an equation, and analyzing compressor processes.
This document contains 25 multiple choice questions from a GATE EE exam, along with explanations for each answer. The questions cover topics such as probability, complex variables, circuit analysis, differential equations, and electric machines. For each question, the correct answer is identified and a brief explanation of the solution steps is provided.
1. The question document contains details about an engineering mathematics examination including 5 questions from Part A and 3 questions from Part B.
2. The questions cover topics such as Fourier series, numerical methods, differential equations, and Laplace transforms.
3. Students are required to answer 5 full questions by selecting at least 2 questions from each part.
This document contains the instructions and questions for a mathematics examination. It includes 12 questions split into two sections. The questions cover topics like differential equations, derivatives, integrals, and vector calculus. Students are instructed to answer 1 question from each section, show working, use diagrams where needed, and assume suitable data if questions don't provide enough information. Calculators, tables and other aids are permitted.
The document describes a proposed social website project that would allow people to connect with others and share information. It discusses the objectives of allowing existing members to view new members and increase social connections. It outlines some potential issues with security, portability and compatibility. The benefits are seen as improved accessibility to information and keeping in touch with low cost. It provides an overview of the key modules to be included and proposes future enhancements around privacy features and administrative functions to prepare for a successful launch.
Dealing with debt presentation - Katie Blacklock and Matthew WhittakerResolutionFoundation
This is the slide presentation by Katie Blacklock and Matthew Whittaker at the Resolution Foundation event, Dealing with debt. It covers Resolution Foundation's latest analysis of which households are most at risk and what policy responses might be needed to help ease the adjustment back to a more normal era of monetary policy.
This event was held on 3 June 2014.
For more details go to: res-fdn.org/dealwithukdebt
This event was live tweeted at @ukdebt
The document discusses signs that it may be time for a worker to consider changing jobs or "trading up" to a new position. It lists potential motivations for seeking a new role such as feeling overworked, unappreciated, or that one's skills have become stale. It cautions that job transitions can fail if a person is burnt out or makes a rushed decision without fully understanding the new role and market demands. The document provides advice on assessing one's options in the current role and market before deciding to change positions and grabbing opportunities by using recruiters, job boards, and networking to facilitate a successful transition.
The document describes a whiplash rehabilitation device that has a hard shell at the front for structure and straps at the back allowing it to fit all users. The device is designed to cater for all users by providing a rigid front shell for support combined with an adjustable strap system to ensure a customized fit.
A Halloween-inspired discussion of the 10 scariest intranet mistakes and how to avoid them. Presented by Carmine Porco, Prescient's GM and VP of Client Deliverables and Chris Chambers, Prescient's VP of Client Development.
View the webinar video here: http://bit.ly/tuO5KU
From the UK\'s richest man, billionaire Lakshmi Mittal, to the humble cornershop, Asian business has made an indelible mark on Britain\'s landscape. It is unquestionably one of the great success stories of recent times and has made our nation both rich and proud.
It\'s a powerful story, often of turmoil and trauma in the beginning followed by success and achievement only in the later years through endeavour, passion, risk and great self-belief. Qualities that stand any business in good stead, especially through tough and challenging times.
The Asian Business Awards is a remarkable institution in itself. Last year, the fortunes of guests attending topped £12bn - making the Asian Business Awards one of the wealthiest gatherings the UK has ever seen. From pharmacy to food, hotels to wholesale and travel to trading, no other event bring this wealth of wealth under one roof. The awards also launch the much anticipated annual Asian Rich List which is regarded as the authoritative source of Asian wealth in the UK.
Recognising this remarkable story is important and critical in passing the torch of entrepreneurship and business flair from one generation to the next; of keeping this story, alive and very much kicking.
We urge you to join us on the evening and become part of the success.
Web delivery of giant climate data sets to facilitate open scienceJames Hiebert
The world's best climate scientists are busy churning out computed projections of our planet's future atmosphere and ocean. However, major barriers exist to publishing geospatiotemporal climate data in an open and transparent manner. This is primarily due to the sheer volume of information that these data represent. A handful of variables for a handful of model realizations by dozens of models for scores of scenarios over centuries of time on tens to hundreds of thousands of grid points create big data. Secondary challenges include the novelty of high-resolution climate data and the complexity and sometimes enigmatic nature of climate data interpretation. Finally, scientific teams often lack either the background, the resources, or the mandate required to facilitate strategic data sharing, aside from dropping their output files onto an FTP site.
In the spirit of open data and open science, the Pacific Climate Impacts Consortium (PCIC), a regional climate services provider in British Columbia, Canada, has been making a concerted effort to use geospatial FOSS in order to expand the availability, comprehensibility and transparency of big climate data sets from the Coupled Model Intercomparison Project (CMIP5) experiment. This presentation will describe some of the technical challenges to serving large geospatiotemporal climate data sets over the web, including some of the nuances of climate data. Additionally, I will explain the requirements of the geospatial climate community, and outline the FOSS solutions that we have employed to serve big climate data sets over the web.
This document summarizes Daniel Mumby's background as an experienced entrepreneur and founder of StartUp Foundation. He has over 20 years of corporate experience and has successfully founded 15 ventures across various industries. StartUp Foundation aims to inspire, train, and invest in aspiring entrepreneurs through programs that connect, collaborate, and share knowledge to help build successful ventures. The organization works with experienced professionals who have significant success factors like expertise, networks, and capital, but may face personal and career risks in starting ventures, and helps mitigate those risks.
This document provides guidance for conducting usability testing and quality assurance. It discusses identifying websites to test and key tasks for each site. It encourages identifying sites that may be new to classmates and, ideally, sites that those conducting the test work on. For each chosen site, it recommends identifying 3-5 key tasks that a user should be able to accomplish on the site.
This document appears to be a resume or profile for an individual named Vishvas Yadav. It includes sections about education and experience. Vishvas Yadav worked at Company 1 from 2006-2009 and Company 2 from 2009-2011. It also mentions that Vishvas Yadav is the Program Director of C.M.M.A.A.O. Pvt. Ltd. Project Management Institute and is involved with fundraising campaigns for orphanages, NGOs, and charitable trusts.
This document contains contact information, educational background, work experience, skills, and details about a fundraising campaign for orphanages, NGOs and charitable trusts run by CODOCA MTVCOLA MARKETING ADVERTISING AND OUTSOURCING PRIVATE LIMITED and its program director Vishvas Yadav. The campaign called "Mission Vishvas" aims to provide funds to over 500 social organizations and will donate 50% of profits from sales of research papers to these causes after tax deductions.
This document provides model solutions to questions from the JEE Advanced 2013 exam. It includes solutions to multiple choice and numerical questions across physics, chemistry and mathematics. Some key details include:
- Solutions to 20 multiple choice questions from Paper I Section I and II of the exam.
- Solutions involving concepts like circular motion, momentum, electrostatics, thermodynamics and more.
- Solutions to 4 numerical questions involving calculations related to decay of particles, fundamental frequency of a vibrating string, and more.
- Solutions to 20 multiple choice questions from Paper II Section I, II and III covering topics in chemistry, organic chemistry, coordination chemistry and biochemistry.
- Solutions to 4 numerical questions related
1) The document provides model solutions to questions from JEE Advanced 2013 Paper II.
2) It includes the answers to multiple choice and numerical questions across various sections - mechanics, electricity & magnetism, atomic structure, thermodynamics etc.
3) The solutions show the step-by-step working for calculating numerical values or explaining concepts to arrive at the correct answer option.
1. The document discusses various topics related to waves including reflection, refraction, interference, diffraction, and standing waves.
2. Reflection can be of two types - closed or fixed end reflection where the phase changes by 180 degrees, and open or free end reflection where the phase does not change.
3. Refraction follows Snell's law where the ratio of sines of the angles of incidence and refraction is equal to the ratio of phase velocities in the two media. Refraction occurs when waves move from a deep to shallow region or vice versa.
1. The document discusses various topics related to waves including reflection, refraction, interference, diffraction, and standing waves.
2. Reflection can be of two types - closed or fixed end reflection where the phase changes by 180 degrees, and open or free end reflection where the phase does not change.
3. Refraction follows Snell's law where the ratio of sines of the angles of incidence and refraction is equal to the ratio of phase velocities in the two media. Refraction occurs when waves move from a deep to shallow region or vice versa.
This document contains formulas for calculus including differentiation, integration, and trigonometry. It includes:
- Rules for differentiation of basic functions like polynomials, exponentials, logarithms, trigonometric functions and their inverses
- Trigonometric identities like sum/difference formulas, double angle formulas, and trigonometric function definitions
- Rules for indefinite integration of basic functions and trigonometric functions
- Trigonometric integrals involving trigonometric functions of x
This document contains equations related to fluid mechanics and wave theory. Some key equations include:
1) Equations for velocity profiles in boundary layers and free surface waves.
2) Equations describing linear wave theory including wave celerity, wave frequency, and potential functions for wave propagation.
3) Equations for hydrostatic pressure and continuity.
4) Equations describing forces on structures like drag and inertia.
1. This document is a marking scheme for Additional Mathematics paper 1 containing solutions and marks for various questions.
2. It provides the correct solutions, working and marks awarded for 16 questions ranging from simple calculations to complex problem solving.
3. The marking scheme acts as a guide for examiners to consistently and fairly award marks for student responses based on the level of accuracy and working shown.
8 Continuous-Time Fourier Transform Solutions To Recommended ProblemsSara Alvarez
This document provides solutions to problems related to continuous-time Fourier transforms.
1) It calculates the Fourier transform of several signals using the definition and properties of the Fourier transform.
2) It demonstrates how scaling and shifting properties in the time domain relate to scaling and shifting in the frequency domain.
3) It shows how periodic signals can be represented using Fourier series coefficients that are samples of the continuous Fourier transform.
TMUA 2021 Paper 1 Solutions (Handwritten).pdfssuser625c41
(1) The document provides instructions for a test of mathematics paper with 20 questions and a time limit of 75 minutes. No calculators or additional materials are allowed.
(2) Candidates must fill out personal information on the answer sheet and choose one answer for each question, recording their choice on the answer sheet. There are no penalties for incorrect answers.
(3) The test consists of 20 multiple choice questions about mathematics, each worth one mark. Candidates should attempt all questions within the time limit.
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3. 1
Câu 1.
LR = 150sos(100 t +
/3) (V); uRC = 50 6 sos(100 t - /12) (V). Cho R = 25 . C
A. 3 (A). B. 3 2 (A) . C.
2
23
(A). D. 3,3 (A)
MON =
12
5
)
12
(
3
MN = UL + UC
OM = URL = 75 2 (V)
ON = URC = 50 3 (V)
MN = UL + UC =
12
5
cos.222
RCRLRCRL UUUU 118 (V)
UR
2
= ULR
2
UL
2
= URC
2
UC
2
- UL
2
UC
2
= ULR
2
URC
2
= 3750
(UL + UC )(UL - UC ) = 3750 UL + UC = 3750/118 = 32 (V)
UL - UC =118 (V)
UL + UC = 32 (V)
Suy ra UL = 75 (V) UR = 222
75LRL UU = 75 (V)
R
Câu 2.
1 R =
40V, UL = 40V, UC = 70V.Khi C = C2 C = 50 2
A. 25 2 (V). B. 25 (V). C. 25 3 (V). D. 50 (V).
Khi C = C1 UR = UL ZL = R
22
)( CLR UUU = 50 (V)
O
UR
N
UCR
UL M
4. 2
Khi C = C2 R L
U = 2
2
2
)'(' CLR UUU = 50 (V) R = 25 2
Câu 3.
4
1
(H) và
t + /6) (V). Khi = 1 thì
2 cos(240 t -
góc
A. uC = 45 2 cos(100 t - /3) (V); B. uC = 45 2 cos(120 t - /3) (V);
C uC = 60cos(100 t - /3) (V); D. uC = 60cos(120 t - /3) (V);
= 1 ta có 1
ZL1
4
1
= 60
= u - i =
4
)
12
(
6
tan = 1
R = ZL1 ZC1; Z1 = 245
1
245
I
U
Z1
2
= R2
+ (ZL ZC)2
= 2R2
R = 45
R = ZL1 ZC1 ZC1 = ZL1 R = 15
ZC1 =
C1
1
C =
3600
1
15.240
11
11 CZ
(F)
22
2 )120(
3600
1
.
4
1
11
LC
2
C2 = ZL2 = 2 L = 30 ( )
I2 = 2
45
245
R
U
(A); uc 2
C2 =
326
UC2 = I2,ZC2 = 30 2 (V)
C = 60cos(
Câu 4 x 63,8C F
1
L H
x
5. 3
A. 0 ;378,4W B. 20 ;378,4W
C. 10 ;78,4W D. 30 ;100W
P = I2
R=
R
ZZ
R
U
ZZR
RU
CLCL
2
2
22
2
)()(
x + r = Rx
ZL ; ZC = 6
10.8,63.314
1
2
1
fC
50
P = Pmax
R
3500
- 2
3500
R
R = 50
Khi R < 50
Khi R > 50
.
x = R r = 0
P = 4,378
)( 22
2
CL ZZr
rU
W
x = 0, P = 378 W
Câu 5.
AN và uAB AM
3 120 3( )AB AN MNU U U V 2 2I A L là
A. 30 3 B. 15 6 C. 60 D. 30 2
AB = UAB UAB = 120 3 (V)
AM = UAM = Ur + UL
NR
A B
M CL,r
A
Ur E UR F
I
UAN
N
UAM
M
UAB B
6. 4
AN = UAN UAN = 120 3 (V)
AE = Ur
EF = MN = UMN = UR UMN = UR = 120 (V)
AF = Ur + UR ; EM = FN = UL ; NB = UC NAB = MAF suy ra MAN = FAB
AB = UMN suy ra UL
2
= (UL UC)2
UC = 2UL suy ra NAF = FAB
MAN = ANM tam giác AMN cân MN = AM hay UAM = UR = 120(V)
Ur
2
+ UL
2
= UAM
2
= 1202
(1)
(Ur + UR)2
+ (UL UC)2
= UAB
2
hay (Ur + 120)2
+ UL
2
= 1202
(2)
r = 60 (V); UL = 60 3 (V)
L = 615
22
360
I
UL
( ),
Câu 6.
1 2
0cos t (U0 và
LC
12
AM và uMB là 900
A. 85 W B. 135 W. C. 110 W. D. 170 W.
Khi
LC
12
L = ZC
P =
21
2
RR
U
(1). Ta có: tan 1 =
1R
ZC
; tan 2 =
1R
ZL
2 - 1 = 900
tan 1. tan 2 = -1
1R
ZC
1R
ZL
= -1 ZL = ZC = 21RR (2)
P2 = I2
2
R2 = 22
2
2
2
LZR
RU
=
21
2
2
2
2
RRR
RU
21
2
RR
U
= P = 85W.
Câu 7:
6 cos(100 ,
LR1
C M R2
BA
L,
r
R
A B
C NM
7. 5
AN và uMB
là 900
, uAN và uAB là 600
. Tìm R và r
A. R=120 ; r=60 B. R=60 ; r=30 ;
C. R=60 ; r=120 D. R=30 ; r=60
OO1 = Ur
UR = OO2 = O1O2 = EF
UMB = OE UMB = 120V (1)
UAN = OQ
UAB = OF UAB = 120 3 (V) (2)
EOQ = 900
FOQ = 600
Suy ra = EOF = 900
600
= 300
.
Xét tam giác OEF: EF2
= OE2
+ OF2
2.OE.OFcos300
EF = OE = 120 (V) Suy ra UR = 120(V) (3)
UAB
2
= (UR + Ur)2
+ (UL UC)2
L UC)2
= UMB
2
Ur
2
( xét tam giác vuông OO1E)
UAB
2
= UR
2
+2UR.Ur + UMB
2
Ur = 60 (V) (4)
= FOO3 = 300 0
)
I = P / U cos 360/(120 3 cos300
) = 2 (A): I = 2A (5)
R/I = 60 ; r = Ur /I = 30
Câu 8. 2 cos t (có 200; ] )
, L =
1
(H); C =
4
10
(F).
A.100 V; 50V. B.50 2 V; 50V. C.50V;
3
100
v. D. .
3
100
;
53
400
VV
:
UAN Q
O3
UL
UL + UC
O
UC
Ur O1 UR
O
UAB FUMB E
UR + Ur
8. 6
Ta có UL = IZL;
UL=
22
4
4
282
2
2
42
22 11
10.7
1
10
1
)2(
11
)
1
(
U
L
C
L
R
C
UL
C
LR
LU
y = 2
4228 1
10.710 XX
2
1
2
hay L
UL = ULmax khi
ULmax =
22
4
4
28 11
10.7
1
10
U
53
400
1
4
7
16
1
100
1
.4
1
10.7
10.16
1
10 22
4
48
28
U
(V)
UL = ULmin khi
ULmin =
22
4
4
28 11
10.7
1
10
U
3
100
171
100
11
10.7
10
1
10 22
4
48
28
U
Câu 9.
3
5
2
AD AM + UMD
A. 240 (V). B. 240 2 (V). C. 120V. D. 120 2 (V)
:
Ta có ZL ZAM = 8022
LZR
AM + UMD)2
.
AM + UMD
Y = (UAM + UMD)2
= I2
( ZAM
2
+ZC
2
+ 2ZAM.ZC) = 22
222
)(
)2(
CL
CAMCAM
ZZR
ZZZZU
9. 7
Y =
640080
)6400160(
)40(40.3
)16080(
2
22
22
222
CC
CC
C
CC
ZZ
ZZU
Z
ZZU
Y = Ymax
640080
)6400160(
2
2
CC
CC
ZZ
ZZ
= 1+
640080
240
2
CC
C
ZZ
Z
X =
640080
240
2
CC
C
ZZ
Z
=
80
6400
240
C
C
Z
Z
X = Xmax ZC
2
= 6400 ZC = 80
AM + UMD C = 80
(UAM + UMD)max = )( CAM ZZ
Z
U
= 2240
80
160.2120
)8040(40.3
)8080(2120
22
(V)
AM + UMD)max = 240 2 (V)
Câu 10
0
1
1=3C thì
2 = 900
- 1
0.
Ta có ZC2 = ZC1/3 = ZC/3
Do Ud = IZd = I 22
LZR : Ud1 = 30V; Ud2 = 90V
Ud2 = 3Ud1 I2 = 3I1
UC1 = I1ZC
UC2 = I2ZC2 = 3I1ZC/3 = I1ZC = UC1 =UC
C; Ud1U1; Ud2U2 C
U1 = U2
2=900
- 1 .
Tam giác OU1U2
C; U1 và U2
CU1 U2 d1Ud2
Suy ra U1U2 = Ud1Ud2 = 90 30 = 60V
1 = OU2 = U1U2/ 2
UR2 IO 1 UR1
UC
U1
U2
Ud2
UL2
Ud1UL1
10. 8
Suy ra U = 60/ 2 = 30 2 U0 = 60V
Câu 11:
0u U cos t 1 2
( 2 < 1
?
A. R = 1 2
2
( )
L n 1
B. R = 1 2
2
L( )
n 1
C. R = 1 2
2
L( )
n 1
D. R = 1 2
2
L
n 1
: I1 = I2 =Imax/n Z1 = Z2 1 L -
C1
1
= - 2 L +
C2
1
2 L-=
C1
1
mà I1 = Imax/n
)
1
(
1
1
2
C
LR
U
=
R
U
n
1
n2
R2
= R2
+( 1 L -
C1
1
)2
= R2
+ ( 1 L - 2 L )2
(n2
1)R2
= ( 1 - 2 )2
L2
R = 1 2
2
L( )
n 1
Câu 12. 0 cos t ( U0
2
1,V2, V3
A. V1, V2, V3. B. V3, V2, V1. C. V3, V1, V2. D. V1, V3,V2.
1,2,3
U1=IR =
22
)
1
(
C
LR
UR
U1 = U1max 1
2
=
LC
1
(1)
U2 = IZL = 2
2
2
22
22222
2
1
)
1
(
y
U
C
L
C
LR
UL
C
LR
LU
U2 = U2max khi y2 = 2
2
2
42
2
11
LC
L
R
C
2min
2
1
2 theo x, cho y2 = 0 x = 2
1
= )2(
2
2
CR
C
LC
11. 9
)2(
2
22
2
2
R
C
L
C
=
)2(
2
2
CRLC
(2)
U3 = IZC = 2
3
22
222222
)2
1
()
1
(
y
U
C
L
C
LRC
U
C
LRC
U
U3 = U3max khi y3 = L2 4
+(R2
-2
C
L
) 2
+ 2
1
C
3min
2
3 3 = 0
y = 2
= 2
2
2
2
2
1
2
2
L
R
LCL
R
C
L
3
2
= 2
2
2
1
L
R
LC
(3)
So sánh (1); (2), (3):
3
2
= 2
2
2
1
L
R
LC
< 1
2
=
LC
1
2
2
- 1
2
=
)2(
2
2
CRLC
-
LC
1
=
)2()2(
)2(2
2
2
2
2
CRLLC
CR
CRLLC
CRLL
>0
(Vì CR2
< 2L nên 2L CR2
> 0 )
2
2
=
)2(
2
2
CRLC
> 1
2
=
LC
1
Tóm lai ta có 3
2
= 2
2
2
1
L
R
LC
< 1
2
=
LC
1
< 2
2
=
)2(
2
2
CRLC
3, V1 , V2
Câu 13 .
AB = 100 2 cos1 1
1 = 0,5A, UMB AB
0
2
AM 2:
A.
21
(H). B.
31
(H). C.
32
(H). D.
5,2
(H).
Ta có ZC =100/0,5 = 200 , 360tantan 0
R
ZZ CL
(ZL ZC) = R 3
Z = U/I = 100/0,5 = 200
Z = RZZR CL 2)( 22
R = 100
12. 10
UAM = I.ZAM =
2212
22222
22
100
)100(400
1
2)(
L
L
L
CLCLCL
L
Z
Z
U
ZR
ZZZZR
U
ZZR
ZRU
UAM =UAMmin khi y = 22
100
100
L
L
Z
Z
= ymax
y = ymax ZL
2
200ZL -100 = 0
ZL = 100(1 + 2 )
L =
21
Câu 14.
R=100
1L=L và khi 1
2
L
L=L =
2
L1
A.
-4
1
4 3.10
L = (H);C= (F) B.
-4
1
4 10
L = (H);C= (F)
C.
-4
1
2 10
L = (H);C= (F) D.
-4
1
1 3.10
L = (H);C= (F)
4
1 = P2 I1 = I2 Z1 = Z2
L1 ZC)2
= (ZL2 ZC)2
. Do ZL1 ZL2 nên ZL1 ZC = ZC ZL2 = ZC -
2
1LZ
1,5ZL1 = 2ZC (1)
tan 1 =
R
ZZ CL1
=
R
ZL
4
1
và tan 2 =
R
Z
Z
R
ZZ C
L
CL 2
1
2
=
R
ZL
4
1
1 + 2 =
2
tan 1. tan 1 = -1 ZL1
2
= 16R2
ZL1 = 4R = 400
L1 =
41LZ
(H)
ZC = 0,75ZL1 = 300 C =
3
10
.
1 4
CZ
(F)
Câu 15:
1= 2 cos 100
12
t (A) và i2=
13. 11
7
2 cos 100
12
t
A. 2 2 cos(100 t+
3
)(A) . B. 2 cos(100 t+
3
)(A).
C. 2 2 cos(100 t+
4
)(A) . D. 2cos(100 t+
4
)(A).
au suy ra ZL = ZC
pha 1 1 và 2 2 1= - tan 2
2 cos(100 t + ) (V).
1 = (- /12) = + /12 2 = 7 /12
tan 1 = tan( + /12) = - tan 2 = - tan( 7 /12)
tan( + /12) + tan( 7 /12) = 0 sin( + /12 + 7 /12) = 0
Suy ra = /4 - tan 1 = tan( + /12) = tan( /4 + /12) = tan /3 = ZL/R
ZL = R 3
U = I1
2 2
12 120LR Z RI (V)
L = ZC
2 cos(100 t + /4) .
i = 2 2
Câu 16. ( m
Cho 0
Imax 1 , 2 1 2 thì
là ax
2
mI
I .Cho
3
4
L
I1 = I2 Z1 = Z2 (ZL1 ZC1)2
= (ZL2 ZC2)2
ZL1 + ZL2 = ZC1 + ZC2
L( 1 + 2) =
21
21
21
)
11
(
1
CC
LC =
21
1
ZC1 = ZL2
Imax =
R
U 2
; I1 =
Z
U
=
2
11
2
)( CL ZZR
U
=
R
U
2
2
4R2
= 2R2
+ 2(ZL1 ZC1)2
R2
= (ZL1 ZL2)2
= L2
( 1 - 2)2
R = L ( 1 - 2) = 200
4
3
= 150(
Câu 17:
0cos( t 0, C, là các
R = 220V và uL = U0L
RC
A. 220V. B. 220 2 V. C. 110V. D. 110 2 .
14. 12
L và i: UL - i =
2
i =
3
-
2
= -
6
L = ZC và U = UR = 220 (V)
L = 2ZL
URC =
22
22
)'('
'
CL
C
ZZR
ZRU
=
22
22
)2('
'
CC
C
ZZR
ZRU
=
Câu 18: 0cos(100 t+
4
10
C F
2
L H 1 2 os(100 t /12)i I c A. Khi
4
L H thì
2 2 os(100 t / 4)i I c A
A. 100 3 B. C. D. 100 2
:
Ta có ZC = 100 ; ZL1 = 200 ; ZL2 = 400
tan 1 =
R
ZZ CL1
=
R
100
1 = +
12
tan 2 =
R
ZZ CL2
=
R
300
= 3tan 1 2 = +
4
2 - 1 =
4
-
12
=
6
tan( 2 - 1) = tan
6
=
3
1
tan( 2 - 1) =
3
1
tan31
tan2
tantan1
tantan
1
2
1
12
12
tan 1 =
3
1
R
100
=
3
1
R = 100 3 (
Câu 19
qIcos = I2
r.
cosqU
P
=
8,0.220
88
= 0,5 (A); r = 2
I
P
= 352
15. 13
Z =
I
Uq
= 22
LZr = 440
Khi mác vào U = 380V: I =
Z
U
=
22
)( LZrR
U
=
222
2 LZrRrR
U
R2
+ 2Rr + 2
quatZ = 2
)(
I
U
R2
+ 704R +4402
= 7602
R2
+ 704R 384000 = 0 R = 360,7
Câu 20
v
3
AB tính theo R là?
I =
Z
U
=
Z
E
2 N 0 = 2 2 fN 0 = U ( do
r = 0)
Z = 222
LR
Khi n1 = n thì 1 = ; ZL1 = ZZ
Khi n3 = 3n thì 3 = 3 ; ZL3 = 3ZZ ---->
3
1
I
I
=
3
1
E
E
1
3
Z
Z
=
3
1
1
3
Z
Z
3
1
22
22
9
L
L
ZR
ZR
=
3
1
I
I
=
3
1
R2
+ 9 2
LZ = 3R2
+3 2
LZ
6 2
LZ = 2R2 2
LZ = R2
/3 ZL =
3
R
- Khi n2 = 2n thì 2 = 2 ; ZL2 = 2ZZ =
3
2R
Câu 21:
0. osu U c t 1
'
C 3C
2 1
2
0 ?U
A. 60V . B. 30 2V C. 60 2V . D. 30V
Ud1 = 30 (V)
Ud2 = 90 (V)
1
2
d
d
U
U
= 3 I2 = 3I1 Z1 = 3Z2 .Z1
2
= 9Z2
2
16. 14
R2
+ (ZL ZC1)2
= 9R2
+ 9(ZL -
3
1CZ
)2
2(R2
+ZL
2
) = ZLZC1
ZC1 =
L
L
Z
ZR )(2 22
1
1
d
d
Z
U
=
1Z
U
U = Ud1
1
1
dZ
Z
= Ud1
22
2
1
2
)(
L
cL
ZR
ZZR
= Ud1 22
1
2
1
22
2
L
CLCL
ZR
ZZZZR
=
Ud1 22
22
2
222
22 )(2
2
)(4
L
L
L
L
L
L
L
ZR
Z
ZR
Z
Z
ZR
ZR
= Ud1 3
)(4
2
22
L
L
Z
ZR
= Ud1 1
4
2
2
LZ
R
tan 1 =
R
ZZ CL 1
; tan 1 =
R
ZZ CL 2
=
R
Z
Z C
L
3
1
2 1
2
1 + 2 =
2
tan 1 tan 2 = -1 ( vì 1 < 0)
R
ZZ CL 1
R
Z
Z C
L
3
1
= -1 (ZL ZC1)(ZL -
3
1CZ
) = - R2
R2
+ ZL
2
4ZL
3
1CZ
+
3
2
1CZ
= 0 (R2
+ ZL
2
) 4ZL
L
L
Z
ZR
3
)(2 22
+ 2
222
3
)(4
L
L
Z
ZR
= 0
(R2
+ ZL
2
)[1-
3
8
+ 2
22
3
)(4
L
L
Z
ZR
] = 0 2
22
3
)(4
L
L
Z
ZR
-
3
5
= 0 2
2
3
4
LZ
R
=
3
1
2
2
4
LZ
R
= 1 U = Ud1 1
4
2
2
LZ
R
= Ud1 2
0 = U 2 = 2Ud1 = 60V.
Câu 22
n0
1 vòng/phút và n2
1, n2 và n0 là
A. 2
0 1 2.n n n B. 2 2 2
0 1 2n n n C. 2
2
2
1
2
2
2
12
0
nn
nn
n D. 2
2
2
1
2
2
2
12
0
2
nn
nn
n
2 N 0 = 2 2 fN 0 = U ( do r = 0)
Do P1 = P2 I1
2
= I2
2
ta có:
2
1
1
2
2
1
)
1
(
C
LR
=
2
2
2
2
2
2
)
1
(
C
LR
])
1
([ 2
2
2
22
1
C
LR = ])
1
([ 2
1
1
22
2
C
LR
17. 15
C
L
C
LR 2
122
2
2
122
2
2
1
22
1 2 =
C
L
C
LR 2
222
1
2
222
2
2
1
22
2 2
)2)(( 22
2
2
1
C
L
R = )(
1
2
2
2
1
2
1
2
2
2
C
= 2
2
2
1
2
1
2
2
2
1
2
2
2
))((1
C
(2
C
L
- R2
)C2
= 2
2
2
1
11
(*)
I =
Z
E
Z
U
P = Pmac khi E2
/Z2
22
2
)
1
(
C
LR
y =
2
22
222
2
1
1
C
L
C
LR
=
2
2
2
42
2
11
1
LC
L
R
C
max thì
2
1
y = 22
2
2
)2( Lx
C
L
R
C
x
0 = 2
0
1
=
2
1
C2
(2 )2
R
C
L
(**)
2
2
2
1
11
= 2
0
2
2
0
2
2
2
1
211
fff
hay 2
0
2
2
2
1
211
nnn 2
2
2
1
2
2
2
12
0
2
nn
nn
n
Câu 23 : 1 =
4
10
F
và C= C2 =
4
10
2
F thì UC C
A. C =
4
3.10
4
F . B. C =
4
10
3
F C. C =
4
3.10
2
F. D. C =
4
2.10
3
F
UC1 = UC2
2
1
2
1
)( CL
C
ZZR
UZ
=
2
2
2
2
)( CL
C
ZZR
UZ
18. 16
2
1
22
C
L
Z
ZR
- 2
1C
L
Z
Z
+1 = 2
2
22
C
L
Z
ZR
- 2
2C
L
Z
Z
+1 (R2
+ 2
LZ )( 2
1
1
CZ
- 2
2
1
CZ
) = 2ZL(
1
1
CZ
-
1
1
CZ
)
1
1
CZ
+
1
1
CZ
= 22
2
L
L
ZR
Z
(1)
UC =
22
)( CL
C
ZZR
UZ
= UCmax khi y = 2
22
C
L
Z
ZR
- 2
C
L
Z
Z
+1 = ymin
y = ymin khi ZC =
L
L
Z
ZR 22
CZ
1
= 22
L
L
ZR
Z
(2)
1
1
CZ
+
1
1
CZ
=
CZ
2
C =
2
21 CC
=
4
3.10
4
(F).
Câu 24:
hi f 1 thì
1 2 6 os 100 ( )
4
i c t A
2
A. 2
5
2 3 os 100 ( )
12
i c t A B. 2
5
2 2 os 100 ( )
12
i c t A
C. 2 2 2 os 100 ( )
3
i c t A D. 2 2 3 os 100 ( )
3
i c t A
Khi C = C1 UD = UC = U Zd = ZC1 = Z1
Zd = Z1
2
1
2
)( CL ZZr = 22
LZr ZL ZC1 = ZL
ZL =
2
1CZ
(1)
Zd = ZC1 r2
+ZL
2
= ZC!
2
r2
=
4
3 2
1CZ
r =
2
3 2
1CZ
(2)
tan 1 =
3
1
2
3
2
1
1
1
1
C
C
C
CL
Z
Z
Z
r
ZZ
1 = -
6
Khi C = C2 UC = UCmax khi ZC2 = 1
1
2
1
22
2
2
C
C
C
L
L
Z
Z
Z
Z
Zr
19. 17
2 = 1
2
1
2
1
12
1
2
2
2
33)2
2
(
4
3
)( CCCCCL ZZZ
Zc
ZZZr
tan 2 = 3
2
3
2
2
1
1
1
2
C
C
C
CL
Z
Z
Z
r
ZZ
2 = -
3
U = I1Z1 = I2Z2 I2 = I1 2
3
32
3
1
2
1 I
Z
Z
(A)
i2 = I2 )
364
100cos(2 t = 2 )
12
5
100cos(2 t (A).
Câu 25.
t +
là: A. 50W. B. 200W. C. 25W, D, 150W
1chieu
2
< Pmax (do Z > R)
P = I2
R <
R
U 2
=
100
)250( 2
Câu 26
A.10 5 V B.28V C.12 5 V D.24V
r
P
= 2A;
Ud =
cosI
P
= 20V , I =
d
d
Z
U
=
dZ
20
Zd =
2
20
= 10
Zd = 22
LZr ZL = 22
rZL = 6
I =
Z
U
U = IZ = I 22
)( LZRr = 2 22
612 = 12 5 (V).
Câu 27
0
L=UL 2 cos(100 t + 1 L =U0L
cos( t+ 2 L=U0L / 2
A.160 (rad/s) B.130 (rad/s) C.144 (rad/s) D.20 30 (rad/s)
20. 18
UL = IZL =
22
)
1
(
C
LR
LU
UL =ULmax khi y = 2
22
)
1
(
C
LR
= ymin
2
0
1
=
2
2
C
(2
C
L
-R2
0 = 120 rad/s
0L = UL 2 Suy ra UL L
22
)
1
(
C
LR
=
22
)
'
1
'(
'
C
LR
2
[ 22
)
'
1
'(
C
LR ] = 2
[ 22
)
1
(
C
LR ]
( 2
- 2
)( 2
C
L
-R2
) = 2
1
C
( 2
2
'
- 2
2
'
) = 2
1
C
( 2
- 2
)( 2
'
1
+ 2
1
)
C2
( 2
C
L
-R2
) = 2
'
1
+ 2
1
= 100 rad/s
2
0
2
= 2
'
1
+ 2
1 2
= 2
0
2
2
0
2
2
2
0
2
0
2 2222
120100.2
120.100
= 160,36
Câu 28 6 cos(100 p
UC = UCmax = 200 (V) khi ZC =
L
L
Z
ZR 22
ULUC = UR
2
+ UL
2
UR
2
+ UL
2
=200UL
U2
= UR
2
+(UL UC)2
(100 3 )2
= UR
2
+ UL
2
+2002
400UL
30000 = 200UL + 40000 400UL UL = 50 (V)
Câu 29.
0. osu U c t 1
21. 19
'
C 3C
2 1
2
0 ?U
Ud1 = 30 (V)
Ud2 = 90 (V)
1
2
d
d
U
U
= 3 I2 = 3I1 Z1 = 3Z2 Z1
2
= 9Z2
2
R2
+ (ZL ZC1)2
= 9R2
+ 9(ZL -
3
1CZ
)2
2(R2
+ZL
2
) = ZLZC1 R2 + ZL
2
=
2
1CLZZ
1
1
d
d
Z
U
=
1Z
U
U = Ud1
1
1
dZ
Z
= Ud1 22
1
2
1
22
2
L
CLCL
ZR
ZZZZR
= Ud1 3
2
?
1
Z
ZC (*)
tan 1 =
R
ZZ CL 1
; tan 1 =
R
ZZ CL 2
=
R
Z
Z C
L
3
1
2 1
2
1 + 2 =
2
tan 1 tan 2 = -1 ( vì 1 < 0)
R
ZZ CL 1
R
Z
Z C
L
3
1
= -1 (ZL ZC1)(ZL -
3
1CZ
) = - R2
R2
+ ZL
2
4ZL
3
1CZ
+
3
2
1CZ
= 0
2
1CLZZ
4ZL
3
1CZ
+
3
2
1CZ
= 0
3
2
1CZ
-
6
5 1CLZZ
= 0
3
1CZ
-
6
5 LZ
= 0 ZC1 = 2,5ZL (**) U = Ud1 3
2
?
1
Z
ZC = Ud1 2
0 = U 2 = 2Ud1 = 60V.
Câu 30.
Khi rô to quay v 6
A. 4 5 ( ) B. 2 5 ( ) C. 16 5 ( ) D. 6 5 ( )
I =
Z
U
=
Z
E
2 N 0 = 2 2 fN 0 = U ( do
r = 0)
Z = 22
2 1
C
R
22. 20
Khi n1 = n thì 1 = ; I1 =
1Z
E
; ZC1 = ZC =
C
1
Khi n2 = 2n thì 2 = 2 ; ZC2 = ZC1 /2 = ZC /2 I2 =
2Z
E
2
1
I
I
=
3
1
E
E
1
2
Z
Z
=
2
1
1
2
Z
Z
2
1
22
2
2
4
C
C
ZR
Z
R
=
2
1
I
I
=
6
1
6R2
+ 1,5 2
CZ = 4R2
+4 2
CZ
2,5 2
CZ = 2R2 2
CZ = 2R2
/2,5 = ZC =
5
2R
= 12 5 ( )
- Khi n3 = 3n thì 3 = 3 ; ZC3 = ZC /3 = 4 5 (
Câu 31
A. 3 /(40 )(H) và 150 B. 3 /(2 )và 150
C. 3 /(40 ) (H) và 90 D. 3 /(2 )và 90
:
1 =
22
LZR
U
tan 1 =
R
ZL
= tan
6
=
3
1
ZL =
3
R
(1) và U = I1
22
LZR =
3
2,0 R
(V) (2)
C = UV = 20V
2 = -
2
- (-
6
) = -
3
tan 2 =
R
ZZ CL
= - tan
3
= - 3 ZC ZL = R 3
ZC = R 3 +
3
R
=
3
4R
; Z2 = 22
)( CL ZZR = 2R
UC =
2Z
UZC
=
3
2U
3
2U
= 20 U =
3
2,0 R
= 10 3 R = 150 ( )
ZL =
3
R
= 50 3 2 fL = 50 3 L =
1000.2
350
=
.40
3
(H)
L =
.40
3
(H) ; R = 150 ( )
Câu 32. AB = Uocos DH
AD
o
DH
o
DB
pha 90o
AB. Tính Uo?
23. 21
A. Uo = 136,6V. B. Uo = 139,3V. C. oU 100 2V. D. Uo = 193,2V.
UAD ; UDH ; UHB
UAB = UAD + UDH + UHB
Tam giác DHB vuông cân.
UHB = UDH = 100V
UDB = 100 2 (V)
có góc D = 750
UAB = UDB sin750
= 100 2 sin750
U0 = UAB 2 = 200sin750
= 193,18V
Hay U0 = 193,2 V
Câu 33: 2
4cos t (A)
Ta có i = 2
4cos t = 2cos2 t + 2 (A)
- 1 = 2cos2 1 = 2 (A)
- T 2 = 2 (A)
:
2
I = I1 = 2 (A)
P = P1 + P2 = I1
2
R + I2
2
R = I2
R I = 62
2
2
1 II (A)
Câu 34.
12
12
5
A. U 5 . B. U 7 . C. U 2 . D. U 3 .
1,u2
u1 = U 2 cos( t +
12
). ; u2 = 2U 2 cos( t +
12
5
).
A D H B
300
450
H
B
D
A
24. 22
2
ABU = U2
+ 4U2
- 2.2U2
cos 1200
= 7U2
UAB = U 7
Câu 35
ULmax= 2URmax Lmax Cmax?
A 2/ 3 . B. 3 /2. C. 1/ 3 . D. 1/2
Ta có UR = URmax = U và UC = UCmax =
R
UZC
L = ZC
UL = ULmax khi ZL =
C
C
Z
ZR 22
: (*)
ULmax =
122
22
L
C
L
C
Z
Z
Z
ZR
U
=
L
C
Z
Z
U
1
= 2URmax = 2U
1 -
L
C
Z
Z
=
4
1
ZL =
3
4
ZC (**)
C = R 3 Cmax =
R
UZC
= U 3
max
max
C
L
U
U
=
3
2
U
U
=
3
2
,
Câu 36
2 cos t = 1 = 40 rad/s và khi = 2
= 360 rad/s
A. 100 (rad/s). B. 110 (rad/s). C. 200 (rad/s). D. 120 (rad/s).
I1 = I1 Z1 = Z1 (ZL1 ZC1)2
= (ZL2 ZC2)2
Do 1 2 nên (ZL1 ZC1) = - (ZL2 ZC2) ZL1 + ZL2 = ZC1 + ZC2
( 1 + 2)L =
C
1
(
1
1
+
2
1
) LC =
21
1
(*)
Khi I = Imax 2
1
(**)
= 21 = 120
1200
25. 23
Câu 37
A. 0,685I B. I C. 2I/ 7 D. I/ 7
L
Ta có : tan d =
R
ZL
= tan
3
= 3 ZL = R 3 Zd = 2R
Ud = UC ZC = Zd = 2R. Z = 2R 32
322R
U
(*)
L =
2
LZ
=
2
3R
22
)'( CL ZZR
U
=
22
)2
2
3
( R
R
R
U
=
3823
2
R
U
(**)
I
I'
=
3823
324
= 0,6847 .
Câu 38 1 2
ULmax
A 45,21 B 23,12 C 74,76 D 65,78
: UL = IZL =
22
)
1
(
C
LR
LU
UL1 = UL2
2
1
1
2
1
)
1
(
C
LR
=
2
2
2
2
2
)
1
(
C
LR
2
1
1
+ 2
2
1
= 4 2
C2
(2
C
L
- R2
) (*)
UL = ULmax khi
22
)
1
(
C
LR
LU
=
2
22
)
1
(
C
LR
UL
26. 24
hay y = 2
22
)
1
(
C
LR
= ymin 2
2
= 4 2
C2
(2
C
L
- R2
) (**)
2
2
= 2
1
1
+ 2
2
1
hay 2
2
f
= 2
1
1
f
+ 2
2
1
f
f =
2
2
2
1
21 2
ff
ff
= 74,67 (Hz).
Câu 39:
1 1
2
f1 ?
A. 80Hz B. 50Hz C. 60Hz D. 70Hz
cos 1 = 0,6 tan 1 =
3
4
tan 1 =
rR
ZL
=
3
4
ZL =
3
4
(R + r) (*)
cos = 0,8 tan = ±
4
3
tan =
rR
ZZ CL
= ±
4
3
ZL ZC = ±
4
3
(R +r) (**)
C
L
Z
Z
= 2
1 LC và 2
2 LC = 1
C
L
Z
Z
= 2
2
2
1
= 2
2
2
1
f
f
f1 = f2
C
L
Z
Z
* Khi ZL ZC =
4
3
(R +r) ZC =
12
7
(R +r)
C
L
Z
Z
=
7
16
f1 =
7
4 2f
= 151,2 Hz
** Khi ZL ZC = -
4
3
(R +r) ZC =
12
25
(R +r)
C
L
Z
Z
=
25
16
f1 = f2
C
L
Z
Z
= f2.
5
4
Câu 40: 2 osu U c t
thì
NB L, ZC
C là:
A. 21 ; 120 . B. 128 ; 120 . C. 128 ; 200 . D. 21 ; 200 .
CL; rR
N BA M
27. 25
: PR = I2
R = 22
2
)()( CL ZZrR
RU
=
r
R
ZZr
R
U
CL
2
)( 22
2
PR = PRmax khi R2
= r2
+ (ZL ZC)2
. (1)
thì PR = PRmax C = UCmax
C =
L
L
Z
ZrR 22
)(
=
LZ
rR 2
)(
+ ZL (2)
L ZC
C nguyên thì (R+r)2
= nZL )
C = n + ZL ZC ZL = n (4)
Thay (4) vào (1) r2
+ n2
= R2
= 752
. (5)
r = 21 n = 72.
Thay R, r, n vào (3) ZL = 128 Thay vào (4) ZC = 200
: r = 21 ; ZC = 200 .
Câu 41
A.10 5 V B.28V C.12 5 V D.24V
=
dZ
r
=0,8 Zd = 10 và ZL = 6 ,
r
P
= 2 (A)
U = I 22
)( LZrR = 2 22
612 = 12 5 (V)
Câu 42 0 1, C2 1//C2
1 1 2
2 1
C// = C1 + C2; Cnt =
21
21
CC
CC
; =
LC
1
C =
L2
1
C// =
L2
1
1
C1 + C2 =
L2
1
1
(*)
Cnt =
L2
2
1
21
21
CC
CC
=
L2
2
1
C1C2 =
L2
2
1
L2
1
1
= 22
2
2
1
1
L
(**)
C1 + 22
2
2
1
1
L 1
1
C
=
L2
1
1
(***)
28. 26
C1 =
L2
1
(****)
Thay (****) vào (***)
L2
1
+ 22
2
2
1
2
L
L
=
L2
1
1
2
1
+ 2
2
2
1
2
= 2
1
1
2
2
2
1 + 4
= 2
2
2 4
- 2
2
2
+ 2
2
2
1 = 0 (*****)
và =
Câu 43 : , L, . 2 = U0cos t (V),
. Cmax. Cmax
A. UCmax =
2
C
2
L
U
Z
1
Z
C. UCmax =
2
L
2
C
U
.
Z
1
Z
B. UCmax =
2 2
2U.L
4LC R C
D. UCmax =
2 2
2U
R 4LC R C
UC =
22
)_( CL
C
ZZR
UZ
=
C
1
22
)
1
(
C
LR
U
=
C
1
2
2242 1
)2(
CC
L
RL
U
UC = UCmax khi 2
= 2
2
2
2
L
R
C
L
và UCmax =
C
1
2
42
4
4
L
R
C
L
R
U
=
22
4
2
CRLCR
LU
UCmax =
22
4
2
CRLC
L
R
U
=
)4(
4
22
2
2
CRLC
L
R
U
=
)
4 2
242
L
CR
L
CR
U
=
)
4
1(1 2
242
L
CR
L
CR
U
=
2
2
)
2
1(1
L
CR
U
=
2
222
4
)2(
1
L
CR
C
L
U
=
22
4
22
4
)2(
1 CL
L
R
C
L
U
=
224
1 CL
U
=
2
2
1
C
L
Z
Z
U
Câu 44: -
29. 27
ch r-L-
A. B. C. D.
:
0 , ZL , ZC
t 2
1 = 70 thì I1 = 0,75A, P1 = 0,928P = 111,36W
P1 = I1
2
R0 (1) R0 = P1/I1
2
198 (2)
I1 =
2222
101 )(268
220
)()( CLCL ZZZZRR
U
Z
U
Suy ra
(ZL ZC )2
= (220/0,75)2
2682
ZL ZC 119 (3)
Ta có P = I2
R0 (4)
22
20 )()( CL ZZRR
U
Z
U
(5)
P = 22
20
0
2
)()( CL ZZRR
RU
R0 + R2 256 R2 58
R2 < R1 2 R1 = - 12
Câu 45
uR = 50 2
2 V và uR = -25 2
60 3 V. B. 100 V. C. 50 2 V. D. 50 3 V
uR = 50 2 ) (V). UR = 50 (V)
2 ;(V) uR = -25 2 (V) u = 2 uR Z = 2R
Z2
= R2
+ ZC
2
ZC
2
= 3R2
ZC = R 3
UC = UR 3 = 50 3
Câu 46 :
R =
ULr = 25V; UC
A. B. C. D.
30. 28
Ta có Ur
2
+ UL
2
= ULr
2
(UR + Ur)2
+ (UL UC)2
= U2
2 (V)
Ur
2
+ UL
2
= 252
(*)
(25+ Ur)2
+ (UL 60)2
= U2
= 3200
625 + 50Ur + Ur
2
+ UL
2
-120UL + 3600 = 3200
12UL 5Ur = 165 (**)
* UL1 = 3,43 (V) Ur1 = 24,76 (V)
2
* UL = 20 (V) Ur = 15 (V)
Lúc này cos =
U
UU rR
=
2
1
P = UIcos I = 1 (A)
Câu 46
1 = R2 = 20 ; R3 = 40 .
A. 6 A B. 3 A C. 0 A D. 2 3 A
R
UP
I1 = I2 = 6A; I3 = 3 A
1 + i2 + i3
I = I1 + I2 + I3
1, i2., i3 là 2 /3
3 = 3 A
Câu 47
6
6 :
A. 75 10 V. B. 75 3 V C. 150 V. D. 150 2 V
ULr
U
UC
UrUR
I3
I2
I1
I3
I
I1
I2
I
I3
I2
I1
31. 29
UC = UCmax khi = 900
RL
2
maxCU = U2
+ 2
RLU
Khi u = 75 6 V thì uRL = 25 6 V Z = 3ZRL hay U = 3URL
2
maxCU = U2
+ 2
RLU = 10 2
RLU .
RL C
R ta có: U.URL = URUC
3 2
RLU = 10 URLUR 3URL = 10 UR = 75 10
URL = 25 10 (V). 10
Câu 48
A.10 5 V B.28V C.12 5 V D.24V
=
dZ
r
=0,8 Zd = 10 và ZL = 6 ,
r
P
= 2 (A)
U = I 22
)( LZrR = 2 22
612 = 12 5
Câu 49
0 cos C d
C =
3
10.125,0
A. 80 rad/s. B. 100 rad/s. C. 40 rad/s. . D.50 rad/s.
Do ZC = Zd = Z. UC = Ud = U. = 100I
L = Ud/2 = 50I
2ZL = ZL = 50
ZL = L; ZC =
C
1
C
L
= CLZZ = 5000 (*)
)(
1
CCL
= 80 L(C+ C) = 2
)80(
1
(**)
O
UC
U
UR
UR
UC
Ud
U
UL
32. 30
5000C(C+ C) = 2
)80(
1
C2
+( C)C -
5000.)80(
1
2
= 0 C2
+
3
10.125,0
C -
5000.)80(
1
2
= 0
C2
+
8
10. 3
C -
4.8
10
2
6
= 0 C =
8
10. 3
F
ZC =
C
1
= 100 =
CZC
1
= 80
Câu 50
0cos C Lr
C = 0,125.10-3
/
A. 80 rad/s B. 100 rad/s C. 40 rad/s D. 50 rad/s
ZC = ZLr = Z = 100 ZL =
2
CZ
= 50
ZL.ZC =
C
L
= 5000 ( 2
) L = 5.103
C (*)
2
0 =
)(
1
CCL
5.103
C2
+ 5.103
C.C - 2
0
1
= 0
5.103
C2
+ 5.103
3
10.125,0
.C - 22
80
1
= 0
5.103
C2
+
.625,0
C -
6400
1
= 0
C =
.25,1
.10-4
(F);
. =
CZC
1
=
4
10.
25,1
..100
1
= 80 rad/s.
Cách 2 ; ZLr = Z r2
+ ZL
2
= r2
+ (ZL ZC)2
ZC = 2ZL ZL =
2
CZ
= 50
C thì Z = ZL =
2
CZ
C = C =
.25,1
10-4
F
UL
Ud
UC
U
33. 31
=
CZC
1
=
4
10.
25,1
..100
1
= 80
Câu 51 : u U 2 cos(100 t)V R 100 2 ,
25
1C (µF) và
3
125
2C (µF)
A.
50
C (µF). B.
3
200
C (µF)., C.
20
C (µF). D.
3
100
C (µF)
Ta có 1
1 2 2
1( )
C
C
L C
UZ
U
R Z Z
2
2 2 2
2( )
C
C
L C
UZ
U
R Z Z
UC1 = UC2
2 2
1 2
2 2 2 2
1 2( ) ( )
C C
L C L C
Z Z
R Z Z R Z Z
ZC1 C2
R2
+ ZL
2
=
21
212
CC
CCL
ZZ
ZZZ
=
240400
240.400.2 LZ
= 300ZL
L = ZC
Thay R =100 2
ZC
2
- 300ZC +20000 = 0
: ZC C
Khi ZC C =
4
10 50
2
F F
Khi ZC C =
4
10 100
F F
Câu 52
A. 10V. B. 10 2 V. C. 20V. D. 20 2 V.
: Do UR = UL = UC trong R = 20V
L R =
2
U
= 10 2
34. 32
Câu 53 0 ;f1 ;f2
:
A.f0 =
2
1
f
f
B. f0 =
1
2
f
f
C.f1.f2 = f0
2
D. f0 = f1 + f2
UR = Urmax L = ZC f0
2
=
LC2
4
1
(1)
UC = UCmax khi ZC2 =
2
2
2
2
L
L
Z
ZR
R2
= ZL2ZC2 ZL2
2
(*)
UL = ULmax khi ZL1 =
1
2
1
2
C
C
Z
ZR
R2
= ZL1ZC1 ZC1
2
(**)
L1ZC1 ZC1
2
= ZL2ZC2 ZL2
2
ZL.ZC =
C
L
suy ra ZC1 = ZL2
Cf12
1
= 2 f2L f1f2 =
LC2
4
1
(2)
2) ta có f1f2 = f0
2
Câu 54 1
1 1
1 = 20 AB = 200 2 m (
AB
A. 266,4W B. 120W C. 320W D. 400W
2 cos t = 200 2 cos100
2 cos(100 t -
2 (V) cos
t+1/600)s
i = 0 i = 2 2 cos[100 (t +
600
1
) - ] = 0 cos(100 t +
6
- ) = 0
cos100 t.cos(
6
- ) - sin100 t.sin(
6
- ) = 0 cos(
6
- ) = 0 (vì sin100 t = 0 )
=
6
-
2
= -
3
PMB = UIcos - I2
R1 = 200.2.0,5 4. 20 = 120W.
Câu 55 1 =
220 2 cos(100 t) (V) , u2 = 220 2 cos(100 t +
3
2
) (V), u3 = 220 2 cos(100 t -
3
2
) (V), .
1=R2=R3
35. 33
A. i = 50 2 cos(100 t +
3
) (A) B. i = 50 2 cos(100 t + ) (A)
C. i = 50 2 cos(100 t +
3
2
) (A) D. i = 50 2 cos(100 t -
3
) (A)
I1 = I3 =
2,2
220
= 100 (A)
I2 =
4,4
220
I0 = I1 + I2 + I3 = I13 + I2
I13 = I1 = I3 = 100A
I0 = I13 I2 = 50 (A)
0 = -
3
i = 50 2 cos(100 t -
3
Câu 56: A và
A , B
có uMB AB
A. UAM B. UAM C. UAM D. UAM
sin
AMU
=
sin
ABU
UAM =
sin
sinABU
Do góc , UAB AM = 900
MB AB thì UAM
AM
I =
22
)
1
(
C
LR
U AB
UAM
Câu 57 0cos100
I1
-2 /3
2 /3
- /3
I0
I3
I2
I1
UAM
UM
UA
36. 34
A. 2I B.I 2 C.I D. I/ 2
2
R =
2
2
0 RI
2
R =
2
P
=
4
2
0 RI
I
I'
=
2
1
2
I
.
Câu 58:
1R
là 1 1 1, , osR CU U c 2R
2 2 2, , osR CU U c 1
2
0,75
R
R
U
U
và 2
1
0,75
C
C
U
U
1osc là:
A. 1 B.
1
2
C. 0,49 D.
3
2
2
1
R
R
U
U
=
4
3
UR2 =
9
16
UR1 (*)
1
2
C
C
U
U
=
4
3
UC2 =
16
9
UC1 (**)
U2
= 2
1RU + 2
1CU = 2
2RU + 2
2CU = (
9
16
)2 2
1RU + (
16
9
)2 2
1CU
(
9
16
)2 2
1RU - 2
1RU = 2
1CU - (
16
9
)2 2
1CU 2
1CU = (
9
16
)2 2
1RU
U2
= 2
1RU + 2
1CU = [(1 + (
9
16
)2
] 2
1RU U =
9
169 22
UR1
cos 1 =
U
UR1
=
22
169
9
=
Câu 59 2 cos(110 t
A. 26,9 µF. B. 27,9 µF. C. 33,77 µF. D. 23,5 µF
37. 35
0 cos( t + )
- Q0sin( t + ) = I0cos(( t + +
2
)
0 = Q0 Q0 0
I = I ZC = ZL
L2
1
=
3,0.)110(
1
2
=
Câu 60 C =600 ôm;
ZL
:
A. 400 2 V. B. 471,4 V. C. 666,67 V. D. 942,8 V.
22
)( CL ZZR = 215200 = 464 ( )
UC =
Z
U
ZC =
464
600
Cmax = 400 2 (V)
600
464
400 2 = 437,5 (V).
Câu 61.
,
6
41
3
10 4
A. 10vòng/s B. 15 vòng/s C. 20 vòng/s D. 5vòng/s
0 = N 0 = 2 fN 0 U = E =
2
0E
Z
U
Do I1 = I2 ta có:
2
1
1
2
2
1
)
1
(
C
LR
=
2
2
2
2
2
2
)
1
(
C
LR
])
1
([ 2
2
2
22
1
C
LR = ])
1
([ 2
1
1
22
2
C
LR
C
L
C
LR 2
122
2
2
122
2
2
1
22
1 2 =
C
L
C
LR 2
222
1
2
222
2
2
1
22
2 2
)2)(( 22
2
2
1
C
L
R = )(
1
2
2
2
1
2
1
2
2
2
C
= 2
2
2
1
2
1
2
2
2
1
2
2
2
))((1
C
38. 36
2
2
2
1
11
= (2
C
L
- R2
)C2
= 2
3
9
10.4
(*)
= 2 f = 2 np
2
2
2
1
11
= 22
4
1
p
( 2
2
2
1
11
nn
) = 22
4
1
p
( 2
1
n
+ 2
9
1
n
) = 222
36
10
np
= 222
536
10
n
(**)
222
536
10
n
= 2
3
9
10.4
n2
= 22
536
10
3
2
10.4
9
= 25
n = 5 v
Câu 62
u = 220 2
X X
X X bao nhiêu ?
UC =
22
)_( CL
C
ZZR
UZ
=
C
1
22
)
1
(
C
LR
U
=
C
1
2
2242 1
)2(
CC
L
RL
U
UC = UCmax khi 2
= 2
2
2
2
L
R
C
L
và UCmax =
C
1
2
42
4
4
L
R
C
L
R
U
=
22
4
2
CRLCR
LU
=
3
5U
6L = 5R 22
4 CRLC R2
C2
4LC + 2
2
25
36
R
L
C = 2
6,12
R
LL
= (2±1,6).
4
10
F 1 =
4
10..6,3
F và C2 =
5
10.4
F
2
= 2
2
2
2
L
R
C
L
=
LC
1
- 2
2
2L
R
> 0 C < 2
2
R
L
=
4
10.2
F 1
CX = C2 =
5
10.4
F 2
=
2
1
LC
- 2
2
2L
R
=
4
10 25
-
2
100 22
= 2.104 2
= 100 2 rad/s
X = 50 2 Hz X =
5
10.4
F và fX = 50 2 Hz
Câu 63: u
2 0
C
A. C = 03C
4
. B. C = 0C
2
. C. C = 0C
4
. D. C = 0C
3
.
39. 37
Ta có Ud = I 22
LZR ; Ud = Udmax khi I = Imax L = ZC0 (*)
Udmax = 2U Zd = 2Z = 2R ( vì ZL = ZC0) R2
+ ZL
2
= 4R2
R =
3
LZ
=
3
0CZ
(**)
UC = UCmax khi ZC =
L
L
Z
ZR 22
=
0
2
0
2
0
3
C
C
C
Z
Z
Z
=
3
4 0CZ
ZC =
3
4 0CZ
C =
4
3 0C
Câu 64: )(cos0 VtUu
rR
AM và uNB V530 0
A. 2120 V. B. 120V. C. 260 V. D. 60 V.
Do R = r UR = Ur
Ta có :(UR + Ur)2
+ 2
LU = 2
AMU
4 2
RU + 2
LU = 2
AMU (1)
2
RU + (UL UC)2
= 2
NBU (2)
UAM = UNB ZAM = ZNB 4R2
+ ZL
2
= R2
+ (ZL ZC)2
3R2
+ ZL
2
= (ZL ZC)2
(*)
uAM và uBN vuông pha tan AM.tan NB = -1
R
ZL
2 R
ZZ CL
= -1 (ZL ZC)2
= 2
2
4
LZ
R
(**)
2
+ ZL
2
= 2
2
4
LZ
R
ZL
4
+ 3R2
ZL
2
4R2
= 0 ZL
2
= R2
L
2
= UR
2
5UR
2
= 2
AMU = (30 5 )2
UR = 30 (V)
UR = UL =30 (V) (4)
2
RU + (UL UC)2
= 2
NBU (UL UC)2
= (30 5 )2
302
= 4.302
UAB
2
= :(UR + Ur)2
+ (UL UC)2
= 4UR
2
+ (UL UC)2
= 2.4.302
UAB = 60 2 (V) U0 = UAB 2
Câu 65
UC
UA
UL
UA
Ur
UR
2U
UN
40. 38
R
U
22
LZR
U
2
0
2
U
u
+ 2
0
2
I
i
=1
Ta có I =
22
LZR
U
erbol
22
LZR
U
Câu 66.
0vòng/min. B. 900vòng/min. C. 3000vòng/min. D. 1500vòng/min.
n =
p
f
=
3
50
vòng/s =
3
50
.60 vòng/min = 1000 vòng/min.
Câu 67 0cos t (U0
CR2
< 2L.Khi = 1 = 2
= 0
1, 2 và 0 là :
A. )(
2
1 2
2
2
1
2
0 B. )(
2
1
210
C. 2
0
1
=
2
1
( 2
1
1
+ 2
2
1
) D. 0 = 21
: UL =
22
)( CL
L
ZZR
UZ
. Do UL1 = UL2
2
1
1
2
2
1
)
1
(
C
LR
=
2
2
2
2
2
2
)
1
(
C
LR
2
1
2
2
C
L
R
+ 24
1
1
C
= 2
2
2
2
C
L
R
+ 24
2
1
C
(2
C
L
- R2
)( 2
2
1
- 2
1
1
) = 24
2
1
C
- 24
1
1
C
(2
C
L
- R2
) = 2
1
C 2
2
2
1
2
2
2
1
41. 39
2
1
1
+ 2
2
1
= C2
(2
C
L
- R2
) (*)
UL = ULmax khi 2
2
2
C
L
R
+ 24
1
C
+ L2
2
0
1
=
2
2
C
(2
C
L
- R2
) (**)
2
0
1
=
2
1
( 2
1
1
+ 2
2
1
) . 2
< 2L
Câu 68:
1, U2 , U3
1 = 100V, U2 = 200V, U3 1 2
A. 233,2V. B. 100 2 V. C. 50 2 V. D. 50V.
U = 2
32
2
1 )( UUU = 2
32
2
1 )''(' UUU = 100 2 (V)
2 3)2
= U2
1
2
= 13600
U2 U3 = I(Z2 Z3) =100 (V) (*)
2 3 2 Z3) = 13600 (V) (**)
I
I'
=
100
13600
2
2'
U
U
=
2
2'
IZ
ZI
=
I
I'
=
100
13600
2 =
100
13600
U2
Câu 69
1 60f Hz cos 1 2 120f Hz,
cos 0,707 3 90f Hz
A. 0,872 B.0,486 C. 0,625 D. 0,781
Ta có ZL1 = ZC1 1L =
C1
1
LC = 2
1
1
(1)
cos 2 = 0,707 2 = 450
tan 2 =
R
ZZ CL 22
=1 R = ZL2 - ZC2
tan 3 =
R
ZZ CL 33
=
22
33
CL
CL
ZZ
ZZ
=
C
L
C
L
2
2
3
3
1
1
=
3
2
1
1
2
1
2
2
2
1
2
3
=
3
2
2
1
2
2
2
1
2
3
=
3
2
f
f
2
1
2
2
2
1
2
3
ff
ff
tan 3 =
3
2
f
f
2
1
2
2
2
1
2
3
ff
ff
=
3
4
22
22
60120
6090
=
3
4
12
5
=
9
5
(tan 3)2
= 25/91
42. 40
81
106
81
25
1
cos
1
3
2
cos2
3 = 81/106 cos 3
Câu 70
R = 100 2 V,
UL
A. UC = 100 3 V B. UC = 100 2 V C. UC = 200 V D. UC = 100V
Ta có UAM =
22
22
)( CL
C
ZZR
ZRU
=
22
22
)(
1
C
CL
ZR
ZZR
=
22
2
2
1
1
C
CLL
ZR
ZZZ
AM = UAMmax 22
2
2
C
CLL
ZR
ZZZ
= ymin
( 22
CZR )(-2ZL) ( CLL ZZZ 22
)2ZC = 0 ZC
2
ZLZC R2
= 0
Hay UC
2
ULUC UR
2
= 0 UC
2
100UC 20000 = 0
UC = 200(V)
Câu 71 1L1C1 1 2L2C2 2 ,
1= 2 .
1và 2
A. =2 1. B. = 3 1. C. = 0. D. = 1.
2
=
LC
1
=
21
21
21 )(
1
CC
CC
LL
2
1 =
11
1
CL
L1 =
1
2
1
1
C
; 2
2 =
22
1
CL
L2 =
2
2
2
1
C
L1 + L2 =
1
2
1
1
C
+
2
2
2
1
C
= 2
1
1
(
1
1
C
+
2
1
C
) = 2
1
1
21
21
CC
CC
( vì 1= 2.)
2
1 =
21
21
21 )(
1
CC
CC
LL
= 2
= 1
Câu 72
3
A. 4 (A). B. 3 2 (A). C. 3 (A). D. 5(A).
t1 = T/3: Q1 = I1
2
Rt1 = 9RT/3 = 3RT
t2 = T/4: Q2 = I2
2
Rt2 = 4RT/4 = RT
43. 41
t3 = 5T/12: Q3 = I3
2
Rt3 = 12R.5T/12 = 5RT
t = t1 + t2 + t3 = T là Q = I2
Rt = I2
RT
Mà Q = Q1 + Q2 + Q3 = 9RT I2
= 9 án C
Câu 73
2
2
A. 100 V. B. 100 2 V. C. 100 3 V. D. 120 V.
tan 1 =
1
11
R
CL
U
UU
; tan 2 =
2
22
R
CL
U
UU
1 - 2 = /2 tan 1 tan 2 =
1
11
R
CL
U
UU
2
22
R
CL
U
UU
= -1
(UL1 UC1)2
.(UL2 UC2)2
= 2
1RU 2
2RU . 2
1MBU 2
2MBU = 2
1RU 2
2RU
8 4
1MBU = 2
1RU 2
2RU .(*) (vì UMB2 = 2 2 UMB1)
2
1RU + 2
1MBU = 2
2RU + 2
2MBU (= U2
) 2
2RU = 2
1RU - 7 2
1MBU (**)
4
1MBU = 2
1RU 2
2RU = 2
1RU ( 2
1RU - 7 2
1MBU )
4
1RU - 7 2
1MBU 2
1RU - 8 4
1MBU = 0 2
1RU = 8 2
1MBU
2
1RU + 2
1MBU = U2 2
1RU +
8
2
1RU
= U2
UR1 =
3
22
U = 100 2
Câu 74:
1 = 0I cos(100 t )
4
2 0i I cos(100 t )
12
A. u 60 2 cos(100 t )
12
(V). B. u 60 2 cos(100 t )
6
(V)
C. u 60 2 cos(100 t )
12
(V). D. u 60 2 cos(100 t )
6
(V).
1 = I2 (ZL ZC)2
= ZL
2
ZC = 2ZL
tan 1 =
R
ZZ CL
= -
R
ZL
(*) tan 1 =
R
ZL
(**) 1 + 2 = 0
44. 42
1 = u -
4
; 2 = u +
12
2 u -
4
+
12
= 0 u =
12
u 60 2 cos(100 t )
12
,
Câu 75:
A. 3 B. 3 C. 3 D. 3
d = Ip =
R
U p
d = 3 p = 3
R
U p'
= 3
R
Ud
= 3
R
U
R
U Pp
3
3
= 3I
án A
Câu 76 :
100cos(100 / 3)( )u t V -3
n
A. 4
( 3 2).10 ( )C B. 4
(1 3).10 ( )C
C. 4
( 3 2).10 ( )C D. 4
( 3 1).10 ( )C
-3
(s) = T/4 ( chu kì T = 0,02s)
-4
cos(100 t +
3
) (C)
Khi t1 1 = 2.10-4
cos(
3
) (C) = 10-4
1
2 = T/4 = 5.10-3
(s); q2 = 2.10-4
cos(
2
+
3
) = - 2.10-4
2
3
(C)
-3
(s) k
Q = q1 + q2 = 4
(1 3).10 ( )C
Câu 77: 2
A. 60V B. 120V C. 30 2 V D. 60 2 V
45. 43
: Lmax khi ZL =
2 2
C
C
R Z
Z
(1)và ULmax =
2 2
CU R Z
R
Ta có: 2 2 2
2 2
30 2 30
2 ( )
( )
C
C L C
C CL C
UU
Z R Z Z
Z Z ZR Z Z
(2)
4 2 2 4 2 2
2 0C C C CR Z R Z R Z R Z
Lmax =
2
2 60
UR
U
R
V.
Câu 78
R=100
1L=L và khi 1
2
L
L=L =
2
L1
A.
-4
1
4 3.10
L = (H);C= (F) B.
-4
1
4 10
L = (H);C= (F)
C.
-4
1
2 10
L = (H);C= (F) D.
-4
1
1 3.10
L = (H);C= (F)
4
Do công suát P1 = P2 I1 = I2 Z1 = Z2
L1 ZC)2
= (ZL2 ZC)2
. Do ZL1 ZL2 nên ZL1 ZC = ZC ZL2 = ZC -
2
1LZ
1,5ZL1 = 2ZC (1)
tan 1 =
R
ZZ CL1
=
R
ZL
4
1
và tan 2 =
R
Z
Z
R
ZZ C
L
CL 2
1
2
=
R
ZL
4
1
1 + 2 =
2
tan 1. tan 1 = -1 ZL1
2
= 16R2
ZL1 = 4R = 400
L1 =
41LZ
(H)
ZC = 0,75ZL1 = 300 C =
3
10
.
1 4
CZ
(F)
Câu 79.
200 2 cos(100 1 2
?
ZL1 = 400 ; ZL2 = 200 ;
46. 44
P1 = P2 I1 = I2 (ZL1 ZC) = -((ZL2 ZC) > ZC = (ZL1 + ZL2)/2 = 300
P1 = 2
1
2
2
)( CL ZZR
RU
200 = 22
2
100
)200(
R
R
R2
+ 1002
= 200R R = 100
Câu 80
u = 120 2 cos(100 nh R, khi R = R1
R = R2 2 1 = P2 và ZC > ZL. Khi R = R3
3?
P1 = P2 22
1
1
)( CL ZZR
R
= 22
2
2
)( CL ZZR
R
(ZL ZC)2
= 144
hay ZC ZL = 12 ( vì ZC > ZL)
Khi R = R3 P = Pmax khi R = R3 = ZC ZL =12
Z3 = 212)( 22
3 CL ZZR ( ) I3 = U/Z3 = 5 2 (A)
tan 3 =
3R
ZZ CL
= - 1 3 = -
4
3 = 10cos(100 t +
4
)
Câu 81
góc 0 1 2 1 và 2
A. 1/ 1 + 1/ 2 = 2/ 0 B. 1 + 2 = /2
C. 1 + 2 = 2 0 D. 2 - 1 = /2
tan 1 =
R
ZZ CL 1
ZC1 = ZL - Rtan 1
tan 2 =
R
ZZ CL 2
ZC2 = ZL - Rtan 2
ZC1 + ZC2 = 2ZL R(tan 1 +tan 2)
ZC1 ZC2 = ZL
2
RZL(tan 1 +tan 2) + R2
tan 1.tan 2
tan 0 =
R
ZZ CL 0
=
LZ
R
UC1 = UC2
1
1
CZ
+
2
1
CZ
=
0
2
CZ
= 22
2
L
L
ZR
Z
21
21
CC
CC
ZZ
ZZ
= 22
2
L
L
ZR
Z
21
2
21
2
21
tan.tan)tan(tanZ
)tan(tan2
RRZ
RZ
LL
L
= 22
2
L
L
ZR
Z
47. 45
21
21
tan.tan-1
tantan
= 22
2
L
L
ZR
RZ
=
1
2
2
2
L
L
Z
R
Z
R
=
0
2
0
tan-1
tan2
tan( 1+ 2)) = tan2 1+ 2) = 2 0 .
Câu 82.
u=125 2 cos100 t,
AM MB 1= 100 và
2= 56,25
A. 0,96 B. 0,85 C. 0,91 D. 0,82
cos 1 =cos 2
1Z
rR
=
2Z
rR
Z1 = Z2 1L -
C1
1
=
C2
1
- 2L
( 1+ 2 )L =
C
1
(
1
1
)
1
2
LC =
21
1
hay ZC1 = ZL2. (1)
tan AM =
R
ZL1
; tan MB =
r
ZC1
uAM MB và r = R
ZL1ZC1 = R2
ZL1.ZL2 = R2
L =
21
R
cos 1 =
1Z
rR
=
2
11
2
)(4
2
CL ZZR
R
=
2
21
2
)(4
2
LL ZZR
R
=
22
21
2
)(4
2
LR
R
cos 1 =
21
2
2
21
2
)(4
2
R
R
R
=
21
2
21 )(
4
2
=
Câu 83
2x = z
AM
hd
hd
hd
48. 46
: Ta có: ZAM =
x
ZL
; ZMN =
y
R
=
4
120
= 30 ( ) (1) và ZNB =
z
ZC
Khi UMB = UAB = 100 (V) thì IMB = IAB = 2 (A) ZMB = ZAB = 50 ( )
2
MBZ = (
y
R
)2
+(
z
ZC
)2
; 2
ABZ = (
y
R
)2
+ (
x
ZL
-
z
ZC
)2
(
y
R
)2
+(
z
ZC
)2
= 502
(
z
ZC
)2
= 502
302
= 402
z
ZC
= 40 ( ) (2)
ZMB = ZAB (
z
ZC
)2
= (
x
ZL
-
z
ZC
)2
x
ZL
= 2
z
ZC
x
ZL
= 80 ( ) (3)
x =
80
LZ
; y =
30
R
; z =
40
CZ
Theo bài ra 2x = z y 2
80
LZ
=
40
CZ
-
30
R
40
CZ
-
40
LZ
=
30
R
(ZC ZL) =
3
4R
.
22
)( CL ZZR = 100V/1A = 100 ( )
R2
+ (ZL ZC)2
= 1002
R2
+ (
3
4R
)2
= 1002
R = 60 (
Câu 84
200 2 cos(100 t) V 1 2
ZL1 = 400 ; ZL2 = 200 ;
P1 = P2 I1 = I2 (ZL1 ZC) = -((ZL2 ZC) ZC = (ZL1 + ZL2)/2 = 300
P1 = 2
1
2
2
)( CL ZZR
RU
200 = 22
2
100
)200(
R
R
R2
+ 1002
= 200R R = 100
Câu 85: .
= 240 2 cos( ,
I = 3 A, UCL= 80 3 V, RC CL. Tính L?
A. 0,37H B. 0,58H C. 0,68H D. 0,47H
:
49. 47
M N
C
A B
R L, r
Ta có U = 240 (V); UR = IR = 80 3 (V)
UR = ULC = 80 V. Xét tam giác cân OME
U2
= UR
2
+ UCL
2
2URULcos =
3
2
=
3
=
6
Xét tam giác OMN UC = URtan = 80(V) (*)
Xét tam giác OFE : EF = OE sin
UL UC = Usin
6
L = 200 (V)
L =
I
UL
=
3
200
L =
100
LZ
=
3100
200
= 0,3677 H
Câu 86.
AB = U 2 cos(100 t ) V.
0 , UAN = 300V ,
UMB = 60 3 V và uAN MB
0
.
:
A. 200V B. 125V C. 275V D. 180V
R = 4r UR = 4Ur
(UR + Ur)2
+ UL
2
= UAN
2
=> 25Ur
2
+ UL
2
= 90000 (1)
Ur
2
+ (UL UC)2
= UMB
2
= 10800 (2)
tan AM =
rR
L
UU
U
=
r
L
U
U
5
; tan MB =
r
CL
U
UU
uAN MB
0
tan AM tan MB =
r
L
U
U
5 r
CL
U
UU
= - 1 UL UC = -
L
r
U
U5
(UL UC )2
= 2
2
25
L
r
U
U
(3)
r
2
+ 2
2
2590000
25
r
r
U
U
= 10800 Ur
2
= 2700 (*) Ur = 30 3
UL
2
= 90000 25Ur
2
= 22500 UL = 150 (V) (**) và UC = UL +
L
r
U
U5
= 240 (V) (***)
UR + Ur = 150 3
U2
= (UR + Ur)2
+(UL UC)2
= 75600
UR+r+ = 5Ur
uAN MB
0
nên hai tam giác
CD
OE
=
CO
EF
=
DO
OF
L
r
U
U
=
r
LC
U
UU
5
=
AN
MBr
U
U
=
300
360
=
5
3
UC
M
UC
N
O /6
UR
UL
UR
Ur
U
UC
UL
UC
E
F
50. 48
UL =
3
5
Ur
(UR + Ur)2
+ UL
2
= UAN
2
=> 25Ur
2
+ UL
2
= 90000
25Ur
2
+
3
25
Ur
2
= 90000 Ur
2
= 2700 Ur = 30 3
UL = 150 (V); UC = 240 (V)
UR + Ur = 150 3
U2
= (UR + Ur)2
+(UL UC)2
= 75600
Câu 87.
u = 50cos(100 t + t + 2
2 cos(200 t + 2
i = 2 cos(200 t +
A. R = 25 ( ), L = 2,5/ (H), C = 10-4
/ (F). B. L = 5/12 (H), C = 1,5.1z0-4
/ (F).
C. L = 1,5/ (H), C = 1,5.10-4
/ (F). D. R = 25 ( ), L = 5/12 (H).
1 = u1 - i1 = -
2
Z1 = ZC1 ZL1 ( ZL1 < ZC1)
2 = u2 - i2 =
2
Z2 = ZL2 ZC2 = 2ZL1 -
2
1CZ
2 = 2f1)
Z1 =
2
225
1
1
I
U
= 25 ; Z2 =
1
50
2
2
I
U
= 50 ;
Ta có ZC1 ZL1 = 25 ;
2ZL1 -
2
1CZ
= 50 ;
Suy ra ZL1 = 125/3 ( ) L =
12
5
300
125
(H)
ZC1 = 200/3 ( ) C = 4
10.5,1
100.200
3
(F)
Câu 88.
6 cos100
UC U UC
D
F
O
UC-
U
E
UR+
UL
UM
UA
C
UL
51. 49
UAM = 50 3 (V) và uAN AB, uMN AB.
UAM = UC = 50 3 (V)
UAB = 50 3 (V)
-
3
UC UL = UAB sin
3
= 75 (V)
UL = 50 3 - 75 (V)
MN và i là
2
-
3
=
6
Ur = UL/tan
6
= UL 3
r =
I
Ur
= 75 37,5 3 = 10
Pd = I2
r = 40W
Câu 89. 0cos(120 t +
2
ZL = 20
i = I0cos(120 t + /3 - /2 ) = I0cos(120 t - /6 )
2
0
2
I
i
+ 2
0
2
U
u
= 1 I0
2
= 22
0
22
0
uU
iU
= 22
0
2
22
0
2
uIZ
iIZ
L
L
300I0
2
3200 = 400 I0 = 3 (A) t - /6 ) (A)
Câu 90
2
L
L.> U L.> U
C. R >
L
C
D .R<
C
L
UR = U.
L > U = UR thì ZL > R
R <
LC
L
=
C
L
.
CL; rR
M AB N
UM
UA
O /6
E
/3 Ur
U
UA
U
UC
UL
UR
52. 50
Câu 91.M
-3
: ZL = 60 ; ZC = 140 .; U = 80 2 (V)
P = I2
R = 22
2
)( CL ZZR
RU
R2
160R + 802
= 0 > R = 80
I =
R
P
= 1 (A). P = UIcos cos =
UI
P
=
2
1
=
2
2
=
4
.
4
Câu 92.
2
Ta có u = 400.cos2
U1 = 100 2 2 = 200 (V)
1 + P2 P = I2
R ; P1 = I1
2
R; P2 = I2
2
R
1 =
Z
U1
= 1(A) vì ZL
22
LZR = 100 2 2 =
R
U2
= 2(A)
I = 2
2
2
1 II = 5 (A)
Câu 93. 2 cos(100
R=50 L=100 C = 50
A) 12,5 ms B) 17,5 ms C) 15 ms D) 5 ms
2
=
R
ZZ CL
= 1 =
4
t -
4
) (A)
2 cos100 t cos(100 t -
4
)
t cos(100 t -
4
) < 0
.cos( -
4
cos > 0 khi -
2
< <
2
:
53. 51
cos( -
4
) > 0 khi -
2
< -
4
<
2
hay khi -
4
< <
4
3
:
( -
4
)
khi cos và cos( -
4
8
T
=
4
T
4
20
= 5 ms.
Câu 94
A. 12,5ms B. 15ms C. 17,5ms D. 20ms
f
1
= 0,02 (s) = 20 (ms)
ZL = 314. 318,3. 10-3
= 100 ; ZC = 6
10.92,15.314
1
= 200 ; Z = 100 2
=
R
ZZ CL
= - 1 = -
4
:
u = U0cos100 t (V)
0 cos(100 t +
4
) (A)
0I0 cos100 t cos(100 t +
4
) (W)
t cos(100 t +
4
) > 0
.cos( +
4
cos > 0 khi -
2
< <
2
:
+
+
+
+ ++
N
M
54. 52
cos( +
4
) > 0 khi -
2
< +
4
<
2
hay khi -
4
3
< <
4
:
+
4
)
khi cos và cos( +
4
(Trên hình
và cos( +
4
8
T
=
4
T
4
T
Do
4
20
= 15 ms.
Câu 95
1
H, C =
50
AB = U0cos(100 t +
12
AN
2
0, R, r
A. 200 2 V;
3
200
; 100 ; B. 400V;
3
200
;
3
100
;
C. 100 2 V;
3
200
; 100 ; D. 200 2 V;
3
200
;
3
100
;
A. i = 2 sin(100 t +
3
) A B. i = 2sin(100 t -
3
) A
C. i = cos(100 t +
3
) A D. i = 2 cos(100 t +
3
) A
:
a. Ta có: ZL = 100 ; ZC = 200 ;
tan AB =
rR
ZZ CL
= -
r3
100
+
+
+ +
+
+ N
M
L;
r
R C
A BNM
55. 53
tan MN =
r
ZL
=
r
100
uMN AB góc
2
tan AB tan MN = - 1
2
2
3
100
r
= 1 r =
3
100
. R = 2r =
3
200
MN
AB
U
U
=
ANZ
Z
= 1 ( Vì Z = ZAN = 200 ) UAB = UMN U0 = 200 2 (V)
b. tan AB =
rR
ZZ CL
= -
r3
100
= -
3
1
AB = -
6
: uAB
6
I =
Z
U AB
= 1 A
i = 2 cos(100 t +
12
+
6
) = 2 cos(100 t +
3
Câu 96 3 r = 30 3
AB = U0cos(100 t +
12
AN = 300V, UMB = 60 3
uAN MB là
2
0, L, C?
A.60 42 V;
5,1
H;
24
10 3
F; B. 120V;
5,1
H;
24
10 3
F;
C. 120V;
5,1
H;
3
10
F; D. 60 42 V;
5,1
H;
3
10
F;
tan AN =
rR
ZL
tan MB =
r
ZZ CL
u n uNB góc
2
nên
tan AN .tan MB = -1
rR
ZL
.
r
ZZ CL
= - 1 ZL(ZL ZC) = - 13500 (*)
MB
AN
U
U
=
MB
AN
Z
Z
=
22
22
)(
)(
CL
L
ZZr
ZrR
=
360
300
2
2
)(2700
67500
CL
L
ZZ
Z
=
3
25
(**)
L = 150 và ZC = 240
L;
r
R C
A BNM
56. 54
L =
5,1
H; C =
24
10 3
F;
ZMB = 22
)( CL ZZr = 10800 = 60 3 ( ) I =
MB
MB
Z
U
= 1A
Z = 22
)()( CL ZZrR = 75600 =30 84 ( )
U0 = I0Z = 2 .30. 84 = 60 42 (V)
0 = 60 42 V; L =
5,1
H; C =
24
10 3
Câu 97
L(t1) = -10 3 V, uC(t1) = 30 3 V, uR(t1)
L(t2) = 20V, uC(t2) = - 60V, uR(t2
A. 60 V. B. 50V. C. 40 V. D. 40 3 V.
Ta có uR = U0R cos t ; uL = U0L cos( t +
2
) = - U0L sin t; uC = U0C cos( t -
2
) = U0C sin t
2: uR(t2) = U0R cos t2 = 0V cos t2 = 0 sin t2 = ±1
uL(t2) = - U0L sin t2 = 20V U0L = 20V (*)
uC(t2) = U0C sin t2 = -60V U0C = 60V (**)
1: uR(t1) = U0R cos t1 = 15V.
uL(t1) = - 20 sin t1 = -10 3 V ; uC(t1) = 60 sin t1 = 30 3 V
sin t1 =
2
3
cos t1 = ±
2
1
. 0R = 30 V (***)
U0
2
= U0R
2
+ ( U0L U0C)2
= 302
+ 402
U0
Câu 98
L(t1) = -30 3 V, uR(t1
L(t2) = 60V, uC(t2) = -120V, uR(t2
A. 50V B. 100 V C. 60 V D. 50 3 V
Ta có uR = U0R cos t ; uL = U0L cos( t +
2
) = - U0L sin t; uC = U0C cos( t -
2
) = U0C sin t
2: uR(t2) = U0R cos t2 = 0V cos t2 = 0 sin t2 = ±1
uL(t2) = - U0L sin t2 = 60V U0L = 60V (*)
uC(t2) = U0C sin t2 = -120V U0C = 120V (**)
1: uR(t1) = U0R cos t1 = 40V.
uL(t1) = - 60 sin t1 = -30 3 V ;
57. 55
sin t1 =
2
3
cos t1 = ±
2
1
. 0R = 80 V (***)
U0
2
= U0R
2
+ ( U0L U0C)2
= 802
+ 602
U0
Câu 99
u = 220 2 cos100
300
AM + UMB
A. 440 V. B. 220 3 V. C. 220 V. D. 220 2 V.
:
tan AM =
R
ZL
= tan300
=
3
1
ZL =
3
R
ZAM = 22
LZR =
3
2R
(*)
AM + UMB)2
.
AM + UMB
Y = (UAM + UMB)2
= I2
( ZAM +ZC)2
= 22
22
)(
)(
CL
CAM
ZZR
ZZU
=
CLCL
CAM
ZZZZR
ZZU
2
)(
222
22
max C
( CLCL ZZZZR 2222
)2(ZAM + ZC) - (ZAM + ZC)2
2(ZC ZL) = 0. Do (ZAM + ZC) 0 nên
( CLCL ZZZZR 2222
) - (ZAM + ZC)(ZC ZL) = 0
(ZAM + ZL)ZC = R2
+ ZL
2
+ ZAMZL ZC =
3
2R
(***)
Z2
= 22
)( CL ZZR Z =
3
2R
(****)
AM = ZMB = ZAB nên UMB = UC = UAB
Câu 100 2 cos t
3
0
A. 20 3 B. 40 C. 40 3 D. 60
I1 =
22
CZR
U
ZRC = 40 3
ud = 60 2 cos( t +
3
) (V)
M
BA
A B
R C L ,r
58. 56
u = uRC + ud uRC = u ud
2
RCU = 1202
+ 602
2.120.60 cos600
= 10800
URC = 60 3 (V)
I =
RC
RC
Z
U
=
340
360
= 1,5 (A)
Suy ra Zd =
I
Ud
=
5,1
60
= 40
Câu 101:
ocos t. Các
R, uL, uC
A. (1), (3), (2). B. (3), (1), (2). C. (2), (1), (3). D. (3), (2), (1).
R; uL; uC
uR = U0Rcos
uL = U0Lcos( t +
2
uC = U0Ccos( t -
2
Câu 102: 0E
A. 0 0;E E . B. 0 0/2; 3 /2E E .
C. 0 0/2; /2E E . D. 0 03 /2; 3 /2E E .
Ta có
1 0 os te E c
2 0
2
os( t- )
3
e E c
3 0
2
os( t+ )
3
e E c
Khi e1 = 0 0
2 0 0 0
32 2 2
os( t- ) os cos sin sin
3 3 3 2
E
e E c E c t E t
0
3 0 0 0
32 2 2
os( t+ ) os cos sin sin
3 3 3 2
E
e E c E c t E t
UR
U
Ud
-Ud
59. 57
Câu 103 :
R =
ULr = 25V; UC
A. B. C. D.
Ta có Ur
2
+ UL
2
= ULr
2
(UR + Ur)2
+ (UL UC)2
= U2
2 (V)
Ur
2
+ UL
2
= 252
(*)
(25+ Ur)2
+ (UL 60)2
= U2
= 3200
625 + 50Ur + Ur
2
+ UL
2
-120UL + 3600 = 3200
12UL 5Ur = 165 (**)
* UL1 = 3,43 (V) Ur1 = 24,76 (V)
2
* UL = 20 (V) Ur = 15 (V)
Lúc này cos =
U
UU rR
=
2
1
P = UIcos I = 1 (A)
Câu 104
u=Uocos 1 2 ( 2 < 1)
là
A. R =
12
21
nL
. B. R =
1
)(
2
21
n
L
. C. R =
1
)(
2
21
n
L
. D. R =
12
21
n
L
.
:
Ta có: I1 =
2
1
1
2
)
1
(
C
LR
U
; I2 =
2
2
2
2
)
1
(
C
LR
U
I1 = I2 1L -
C1
1
= - ( 2L -
C2
1
) hay : ( 1 + 2 )L =
C
1
(
1
1
+
2
1
)
LC =
21
1
C 1 =
2
1
L
(*)
Khi I = I =
R
U
I1 = I2 =
n
Ic
=
nR
U
R2
+ ( 1L -
C1
1
)2
= n2
R2
( 1L -
C1
1
)2
=(n2
1)R2
(**)
2
1)R2
= ( 1L - 2L )2
= L2
( 1- 2)2
ULr
U
UC
UL
UrUR
60. 58
1
)(
2
21
n
L
Câu 105: 2
t (A)
- Gi ?
- Giá ?
- Gi ?
: Ta có i = 4cos2
t (A) = 2 (cos2 t + 2) = 2cos2 t + 2 (A)
- 1 = 2cos2 1 = 2 (A)
- 2 = 2 (A)
a. C
:
2 không
I = I1 = 2 (A)
P = P1 + P2 = I1
2
R + I2
2
R = I2
R I = 62
2
2
1 II (A)
b. Có giá t
I = 2cos2 t + 2 = 0 + 2 (A)
c. C
:
2
Imax = I1max = 2 (A)
Imax = I1max + 2 = 4 (A)
61. NHÀ XU T B NG H C S
***********
BÀI T P
VÀ
I
N XOAY CHI U
BIÊN SO N: TR N DUY KHOA
Ch u trách nhi m xu t b n:
NG H C S
HÀ N