ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Lecture 12 First Order Transient Response

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Chapter 4 Transients 1.Solve first-order RC or RL circuits. 2. Understand the concepts of transient response and steady-state response.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 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, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transients The 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 integro-differential equations.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Discharge of a Capacitance through a Resistance KCL at the top node of the circuit: iCiC iRiR

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Discharge of a Capacitance through a Resistance We need a function v C (t) that has the same form as it’s derivative. Substituting this in for v c (t)

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Discharge of a Capacitance through a Resistance Solving for s: Substituting into v c (t): Initial Condition: Full Solution:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Discharge of a Capacitance through a Resistance To find the unknown constant K, we need to use the boundary conditions at t=0. At t=0 the capacitor is initially charged to a voltage V i and then discharges through the resistor.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Discharge of a Capacitance through a Resistance

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Discharge of a Capacitance through a Resistance The time interval t= RC is called the time constant of the circuit RC circuits can be used for timing applications (e.g. garage door light)

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Larry Light-Bulb Experiment i R (t)

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Charging a Capacitance from a DC Source through a Resistance

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Charging a Capacitance from a DC Source through a Resistance KCL at the node that joins the resistor and the capacitor Current into the capacitor: Current through the resistor:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Charging a Capacitance from a DC Source through a Resistance Rearranging: This is a linear first-order differential equation with contant coefficients.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Charging a Capacitance from a DC Source through a Resistance The boundary conditions are given by the fact that the voltage across the capacitance cannot change instantaneously:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Charging a Capacitance from a DC Source through a Resistance Try the solution: Substituting into the differential equation: Gives:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Charging a Capacitance from a DC Source through a Resistance For equality, the coefficient of e st must be zero: Which gives K 1 =V S

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Charging a Capacitance from a DC Source through a Resistance Substituting in for K 1 and s: Evaluating at t=0 and remembering that v C (0+)=0 Substituting in for K 2 gives:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Charging a Capacitance from a DC Source through a Resistance

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. If the initial slope is extended from t=0, when does it intersect the final value V S ? Charging a Capacitance from a DC Source through a Resistance A line with this slope would intersect V S after a time 

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Larry Light-Bulb Experiment

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. DC Steady State In steady state, the voltage is constant, so the current through the capacitor is zero, so it behaves as an open circuit.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. DC Steady State In steady state, the current is constant, so the voltage across and inductor is zero, so it behaves as a short circuit.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. DC Steady State The 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, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. DC Steady State

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. DC Steady State Find the steady state current i a and voltage v a

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. DC Steady State Open circuit the capacitor

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. DC Steady State Short circuit the inductor

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. DC Steady State Find the steady state currents i 1, i 2, i 3

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. DC Steady State Open circuit the capacitor

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. DC Steady State Short circuit the inductor 5  i 1 =20V/10  =2A 10  Current divider gives i 2 =i 3 =1A 1A