An example on how to solve for A1 and A2

Slides:



Advertisements
Similar presentations
Boundary Conditions. Objective of Lecture Demonstrate how to determine the boundary conditions on the voltages and currents in a 2 nd order circuit. These.
Advertisements

Lecture - 9 Second order circuits
Reading Assignment: Chapter 8 in Electric Circuits, 9th Ed. by Nilsson
AP Electricity Quiz Review
Series RLC Network An example on how to solve for A 1 and A 2.
E E 2315 Lecture 10 Natural and Step Responses of RL and RC Circuits.
Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Transients Analysis.
ECEN 5817 Resonant and Soft-Switching Techniques in Power Electronics 1 Lecture 23 General Solution for the Steady-State Characteristics of the Series.
First Order Circuit Capacitors and inductors RC and RL circuits.
Lecture 10: RL & RC Circuits Nilsson 7.1 – 7.4
Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Transients Analysis.
Capacitance (II) Capacitors in circuits Electrostatic potential energy.
Series RLC Network. Objective of Lecture Derive the equations that relate the voltages across a resistor, an inductor, and a capacitor in series as: the.
First Order Circuits. Objective of Lecture Explain the operation of a RC circuit in dc circuits As the capacitor releases energy when there is: a transition.
Source-Free RLC Circuit
Assist.Prof. Sibel ÇİMEN Electronics and Communication Engineering University of Kocaeli.
Parallel RLC Network. Objective of Lecture Derive the equations that relate the voltages across a resistor, an inductor, and a capacitor in parallel as:
Amplifier Circuit This amplifier circuit DC analysis.
Chapter 8 Second-Order Circuits
Chapter 7. First and second order transient circuits
Series RLC Network. Objective of Lecture Derive the equations that relate the voltages across a resistor, an inductor, and a capacitor in series as: the.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Lecture 12 First Order Transient.
Lecture 10 - Step Response of Series and Parallel RLC Circuits
EENG 2610: Circuit Analysis Class 12: First-Order Circuits
Fluid flow analogy. Power and energy in an inductor.
SECOND ORDER CIRCUIT. Forced Response of Parallel RLC Circuit (Step response Parallel RLC Circuit) Second order circuit When switch is open, a STEP current.
Electricity Define Electricity: Electrons: Short Circuit: Current: Battery: Voltage:
Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Chapter 9 The RLC Circuit.
Lecture 13 final part. Series RLC in alternating current The voltage in a capacitor lags behind the current by a phase angle of 90 degrees The voltage.
Step Response Series RLC Network.
1.4. The Source-Free Parallel RLC Circuits
Alexander-Sadiku Fundamentals of Electric Circuits
Assist.Prof. Aysun Taşyapı ÇELEBİ Electronics and Communication Engineering University of Kocaeli.
Lecture 19 Review: First order circuit step response Steady-state response and DC gain Step response examples Related educational modules: –Section
SECOND ORDER CIRCUIT. Revision of 1 st order circuit Second order circuit Natural response (source-free) Forced response SECOND ORDER CIRCUIT.
1 SECOND ORDER PSpice Example. 2 Example 1 (i)Find v(t) for t > 0. (ii) Find i(t) for t >0. v i 24 V R = 6  1 H 0.25 F  t = 0 The switch has been.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Lecture 13 RC/RL Circuits, Time.
1 EKT101 Electric Circuit Theory Chapter 5 First-Order and Second Circuits.
Exponential Growth/Decay
1 SECOND ORDER Examples. 2 What Circuit do we require and why? 1. Circuit at t = 0 - This circuit is required to find the initial values of i(0 - ) and/or.
Self-Inductance and Circuits Inductors in circuits RL circuits.
To obtain Coefficient A1 and A2
Alternating Current Circuits. AC Sources  : angular frequency of AC voltage  V max : the maximum output voltage of AC source.
Chapter 5 First-Order and Second Circuits 1. First-Order and Second Circuits Chapter 5 5.1Natural response of RL and RC Circuit 5.2Force response of RL.
Source-Free Series RLC Circuits.
Capacitance (II) Capacitors in circuits Electrostatic potential energy.
Response of First Order RL and RC
CHAPTER 5 DC TRANSIENT ANALYSIS.
A Requirement for Superposition. Objective of Lecture Introduce the property of linearity Chapter 4.2.
Electricity Definitions
Circuits and Systems Theory
Inductance and Capacitance Response of First Order RL and RC
EKT 101 Electric Circuit Theory
EKT101 Electric Circuit Theory
EKT101 Electric Circuit Theory
1.4. The Source-Free Parallel RLC Circuits
General Solution for the Steady-State Characteristics of the Series Resonant Converter Type k CCM Mode index k and subharmonic number 
Mechatronics Engineering
Lecture 19 Review: Steady-state response and DC gain
Topics to be Discussed Steady State and Transient Response.
Lecture 13 - Step Response of Series and Parallel RLC Circuits
Source-Free RLC Circuit
Step Response Parallel RLC Network.
An alternate solution technique
* 07/16/96 What is Second Order?
Chapter 8 Second Order Circuits
Quiz 10 min.
Chapter 8 – Second-Order Circuits
First Order Circuit Capacitors and inductors RC and RL circuits.
Chapter 8 – Second-Order Circuits
Presentation transcript:

An example on how to solve for A1 and A2 Series RLC Network An example on how to solve for A1 and A2

Series RLC Circuit Note that these two circuits are the same. A single pole/double throw switch (a switch that is composed of one wire that is moving between two points in the circuit) in the upper circuit causes the 5 V supply to be removed from the circuit at t = 0 s. The step function in the lower circuit causes the output of the voltage supply to change from 5 V to 0 V at t = 0 s.

Step Function The unit step function, u(t), in this circuit has the following definition:

Determining the type of solution Note that this calculation can be done before we have even identified which voltage or current will be should be used when writing the solution to the 2nd order equation.

Boundary Conditions In d.c. steady state, the capacitor will act like an open circuit (∞W)and the inductor will act as a short circuit (0 W). The initial and final conditions for the voltages and currents are:

You find that that A1 = 5 V (3 sig. figs. ) and A2 = -7 You find that that A1 = 5 V (3 sig. figs.) and A2 = -7.97 x 10 -6 V = -7.97 mV