PowerPoint PresentationChapter 4 TransientsSlide 3TransientsSlide 5Discharge of a Capacitance through a ResistanceSlide 7Slide 8The time interval τ = RC is called the time constant of the circuit.Slide 10Slide 11DC STEADY STATESlide 13Slide 14RL CIRCUITSSlide 16Slide 17Slide 18Slide 19RL Transient AnalysisSlide 21Slide 22Slide 23Slide 24Slide 25RC AND RL CIRCUITS WITH GENERAL SOURCESSlide 27Slide 28Step-by-Step SolutionSlide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37SECOND-ORDER CIRCUITSSlide 39Slide 40Slide 41Slide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Slide 51Slide 52Slide 53Slide 54ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.Chapter 4TransientsELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.Chapter 4Transients1. Solve first-order RC or RL circuits.2. Understand the concepts of transient response and steady-state response.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.3. Relate the transient response of first-order circuits to the time constant.4. Solve RLC circuits in dc steady-state conditions.5. Solve second-order circuits.6. Relate the step response of a second-order system to its natural frequency and damping ratio.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.TransientsThe time-varying currents and voltages resulting from the sudden application of sources, usually due to switching, are called transients. By writing circuit equations, we obtain integrodifferential equations.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.Discharge of a Capacitance through a Resistance   0RtvdttdvCCC  0 tvdttdvRCCC stCKetv 0ststKeRCKseELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.RCs1 RCtCKetv RCtiCeVtv iCVv 0ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.The time interval τ = RC iscalled the time constant ofthe circuit. tssCeVVtvELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.DC STEADY STATEThe steps in determining the forced response for RLC circuits with dc sources are:1. Replace capacitances with open circuits.2. Replace inductances with short circuits.3. Solve the remaining circuit.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.RL CIRCUITSThe steps involved in solving simple circuits containing dc sources, resistances, and one energy-storage element (inductance or capacitance) are:ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.1. Apply Kirchhoff’s current and voltage laws to write the circuit equation.2. If the equation contains integrals, differentiate each term in the equation to produce a pure differential equation.3. Assume a solution of the form K1 + K2est.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.4. Substitute the solution into the differential equation to determine the values of K1 and s . (Alternatively, we can determine K1 by solving the circuit in steady state as discussed in Section 4.2.)5. Use the initial conditions to determine the value of K2.6. Write the final solution.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.RL Transient Analysis LtReKti22Time constant isRLELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.RC AND RL CIRCUITS WITH GENERAL SOURCESThe general solution consists of two parts.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.The particular solution (also called the forced response) is any expression that satisfies the equation.In order to have a solution that satisfies the initial conditions, we must add the complementary solution to the particular solution.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.The homogeneous equation is obtained by setting the forcing function to zero. The complementary solution (also called the natural response) is obtained by solving the homogeneous equation.ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan