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

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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.

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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.

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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.

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

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

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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)

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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:

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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.

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

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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)

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

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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

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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:

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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.

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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:

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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:

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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

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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:

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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

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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

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

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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.

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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.

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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.

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

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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

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

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

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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

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

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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

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