ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Lecture 13 RC/RL Circuits, Time.

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ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Lecture 13 RC/RL Circuits, Time Dependent Op Amp Circuits

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Circuits The steps involved in solving simple circuits containing dc sources, resistances, and one energy-storage element (inductance or capacitance) are:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 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 K 1 + K 2 e st.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. 4. Substitute the solution into the differential equation to determine the values of K 1 and s. (Alternatively, we can determine K 1 by solving the circuit in steady state) 5. Use the initial conditions to determine the value of K Write the final solution.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis Find i(t) and the voltage v(t) Apply KVL around the loop: i(t)= 0 for t < 0 since the switch is open prior to t = 0

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

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

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

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

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

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis Transient starts by opening switch Prior to t = 0 inductor acts as a short circuit so that: v(t) = 0for t < 0 i(t) = V S /R 1 for t < 0

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis After t = 0 current circulates through L and R, dissipating energy in the resistance R

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

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

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis with a Current Source After the switch is opened, i R (0 + ) = 2A, I L (0 + ) = 0 Find v(t), i R (t), i L (t)

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis with a Current Source After the switch is closed, i R (0 + ) = 2A, I L (0 + ) = 0

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis with a Current Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis with a Current Source i L (0 + ) = 0  K 1 = -K 2

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis with a Current Source In steady state the inductor becomes a short circuit  i L (∞) = K = 2A = 2A

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis with a Current Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis Prior to t = 0 i(0) = 100V/100  = 1A Find i(t), v(t)

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis Prior to t = 0 i(0) = 100V/100  = 1A After t = 0:

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

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

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

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RC and RL Circuits with General Sources First order differential equation with constant coefficients Forcing function

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RC and RL Circuits with General Sources The general solution consists of two parts.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 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, Fourth Edition, by Allan R. Hambley, ©2008 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, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Step-by-Step Solution Circuits containing a resistance, a source, and an inductance (or a capacitance) 1. Write the circuit equation and reduce it to a first-order differential equation.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. 2. Find a particular solution. The details of this step depend on the form of the forcing function. 3. Obtain the complete solution by adding the particular solution to the complementary solution x c =Ke -t/  which contains the arbitrary constant K. 4. Use initial conditions to find the value of K.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source Take the derivative:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source The complementary solution is given by: The complete solution is given by the sum of the particular solution and the complementary solution:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source Initial conditions: 2sin(0 + ) = 0 v C (0 + ) = 1V v R (0 + ) + v C (0+) = 0  v R (0 + ) = -1V i(0 + ) = v R /R = -1V/5000  = -200  A = 200cos(0)+200sin(0)+Ke 0 = K  K= -400  A

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source What happens if we replace the source with 2cos(200t) and the capacitor initially uncharged v c (0)=0?

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source What happens if we replace the source with 2cos(200t) and the capacitor initially uncharged v c (0)=0?

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source Take the derivative:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source The complementary solution is given by: The complete solution is given by the sum of the particular solution and the complementary solution:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source Initial conditions: 2cos(0 + ) = 2 v C (0 + ) = 0V v R (0 + ) + v C (0+) = 2  v R (0 + ) = 2 i(0 + ) = v R /R = 2V/5000  = 400  A = 200cos(0)-200sin(0)+Ke 0 = K  K= 200  A

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with an Exponential Source What happens if we replace the source with 10e -t and the capacitor is initially charged to v c (0)=5?

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with an Exponential Source Take the derivative:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with an Exponential Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with an Exponential Source The complementary solution is given by: The complete solution is given by the sum of the particular solution and the complementary solution:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Initial conditions: 10e 0+ = 10 v C (0 + ) = 5V v R (0 + ) + v C (0+) = 10  v R (0 + ) = 5 i(0 + ) = v R /R = 5V/1M  = 5  A = 20e 0 + Ke 0 = 20 + K  K= -15  A Transient Analysis of an RC Circuit with an Exponential Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with an Exponential Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Integrators and Differentiators Integrators produce output voltages that are proportional to the running time integral of the input voltages. In a running time integral, the upper limit of integration is t.

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

If R = 10 k , C = 0.1  F  RC = 0.1 ms

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

Differentiator Circuit

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