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Teoria de Circuitos Elétricos 1: Introdução a Circuitos Resistivos, Notas de aula de Circuitos Elétricos

Aulas de circuitos eletrônicos com base no livro Dorf

Tipologia: Notas de aula

2019

Compartilhado em 18/09/2019

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TEORI A DE CIRCUITOSELETRÔNICOS 1
Aula 1: Resistive Circuits
Prof. Marcelino Andrade
Faculdade UnB Gama
10 de agosto de 2017
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Baixe Teoria de Circuitos Elétricos 1: Introdução a Circuitos Resistivos e outras Notas de aula em PDF para Circuitos Elétricos, somente na Docsity!

Aula 1: Resistive Circuits

Prof. Marcelino Andrade

Faculdade UnB Gama

10 de agosto de 2017

Contents

Indroduction (3.1) Kirchhoff’s Laws (3.2) Series Resistors and Voltage Division (3.3) Parallel Resistors and Current Division (3.4) Series Voltage Sources and Parallel Current Sources (3.5) Circuit Analysis (3.6) Analyzing Resistive Circuits Using MATLAB

Introduction to Electric Circuits 9th Edition by James A. Svoboda, Richard C. Dorf

Indroduction (3.1)

Resistive Circuits

We say that circuit drawings A and B represent the same circuit when the following three conditions are met.

(A)

(B)

♣ There is a one-to-one correspondence between the nodes of drawing A and the nodes of drawing B. ♣ There is a one-to-one correspondence between the elements of drawing A and the elements of drawing B. ♣ Corresponding elements are connected to corresponding nodes.

Kirchhoff’s Laws (3.2)

Kirchhoff’s Laws

In 1847, Gustav Robert Kirchhoff, a professor at the University of Berlin, formulated two important laws that provide the foundation for analysis of electric circuits.

Kirchhoff’s current law (KCL): The algebraic sum of the currents into a node at any instant is zero. ♣ Kirchhoff’s voltage law (KVL): The algebraic sum of the voltages around any loop in a circuit is identically zero for all time.

Kirchhoff’s laws are a consequence of conservation of charge and conservation of energy.

Kirchhoff’s Laws (3.2)

Kirchhoff’s Laws

EXERCISE 3.2-1 - Determine the values of i 3 , i 4 , i 6 , v 2 , v 4 and v 6.

Answer:i 3 = − 3 A, i 4 = 3 A, i 6 = 4 A, v 3 = − 3 V, v 4 = − 6 V, v 6 = 6 V

Series Resistors and Voltage Division (3.3)

Series Resistors and Voltage Division

Let us consider a single-loop circuit ...

Single-loop circuit with a voltage source vs.

In general, we may represent the voltage divider principle by the equation:

vn = Rnvs R 1 + R 2 + ... + RN

where vn is the voltage across the nth resistor of N resistors connected in series.

Parallel Resistors and Current Division (3.4)

Equivalent Circuit

♣ Equivalent circuit for a series resistors

Rs = R 1 + R 2 + ... + RN =

∑^ N

n= 1

Rn (3)

♣ Equivalent circuit for a parallel resistors

Gp =

Rp

R 1

RN

∑^ N

n= 1

Rn

Series Voltage Sources and Parallel Current Sources (3.5)

Series Voltage Sources

Voltage sources connected in series are equivalent to a single voltage source.

Series Voltage Sources and Parallel Current Sources (3.5)

Series and Parallel Sources

E X A M P L E 3.5-1a - Show similar circuits

♣ Determine the value of the current i 1 and voltage v 2.

E X A M P L E 3.5-1c - Show similar circuits

♣ Determine the value of the current i 1 and voltage v 2.

Circuit Analysis (3.6)

Analyzing Resistive Circuits

( a ) Set of series and parallel resistors.

( b ) Equivalent circuit.

Rs = R 1 + R 2 + R 3 (5)

Rp =

Gp

Rs = R 1 + R 2 + R 3 (7)

vo =

Rp Rs + Rp

vs (8)

The analysis of a circuit by replacing a set of resistors with an equivalent resistance, thus reducing the network to a form easily analyzed.

Analyzing Resistive Circuits Using MATLAB

Analyzing Resistive Circuits Using MATLAB

EXERCISE 3.7-1 - Determine the values of the resistor voltages and currents.

Answer:i 2 = 0. 4 A, i 4 = − 0. 05 A, i 5 = 0. 05 A, i 6 = 0. 05 A, v 2 = 16 V, v 4 = − 4 V, v 5 = 2. 4 V and v 6 = 1. 6 V