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Mindanao State University-Iligan Institute of Technology

College of Engineering

Electric Circuit Theory 1

LECTURE 01

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1. Concepts in Engineering Circuit Analysis

1.1 Circuit, Circuit Elements and Experimental Laws

1.1.1 Introduction1.1.2 Current, Voltage and Power

1.1.3 Circuits and Circuit Elements

1.1.4 Ohms Law

1.1.5 Kirchoffs laws

1.1.6 R, L, C and Source Combination1.1.7 Voltage and Current Division

1.2 Useful Techniques in Circuit Analysis

TOPIC OUTLINE

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2. The Transient Circuit

2.1 Source-Free RL and RC Circuits

2.2 Application of Unit-Step Forcing

Functions

2.3 The RLC Circuit

3. Sinusoidal Steady-State Analysis

3.1 The Phasor Concept

3.2 The Sinusoidal Steady-State Response

3.3 Average Power and RMS Values

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Grading System:

Quizzes: 25%

First Prelim Exam: 25%

Second Prelim Exam: 25%Final Exam: 25%

Passing: 60%

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Part 1: Concepts in Circuit Analysis

1.1 Circuit, Circuit Elements and Experimental Laws

1.1.1 Introduction:

Systems of Units: International System of Units (SI)

Basic Units:

length meter (m) mass kilogram (kg)

time second (sec) current ampere (A)

Derived units:

force newton (N) resistance ohm (;)

work, energy joule (J) inductance henry (H)

voltage volt (V) capacitance farad (F)

power watt (W) charge coulomb (coul)

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Some Useful Prefixes:

Factor prefix Abbreviation Example

1012 tera - T 5.88 teramiles (Tmi)

109

giga - G 3 gigahertz (Ghz)

106 mega - M 5 megohms (M;)

103 kilo - K 15 kilowatts (kw)

10-3 milli - m 2.5 milliampere (mA)

10-6 micro - Q 5 microhenry (QH)

10-9 nano - n 2 nanofarad (nF)

10-12 pico - p 6 picofarad (pF)

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1.1.2 Current, Voltage and Power

i

A. Current (i) - charge in motion

- unit: ampere (A)

- unit named after: Andre Marie Ampere, a French physicist

i = dq/dt where q = charge in coulomb (coul)

Representation:

B. Voltage (v) - potential difference

- unit: volt (V)

- unit named after: Alessandro Giuseppe Antonio Anastasio Volta,

an Italian physiscist (Were fortunate indeed since hisfull name is not used as the unit of voltage)

+v-

Representation:

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+

2 V-

3 A-5 A

+4 V-

(a) p = (2) (3) = 6 Wabsorbed power

Examples:

(b) p = (4) (-5) = -20 W absorbed power(20W generated power)

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1.1.3 Circuits and Circuit Elements

Circuit an interconnection of simple electrical devices in which there is

at least one closed path for which current may flow.

Circuit Element mathematical model of two-terminal electrical device

Types:

A. Active Elements capable of delivering power to some external

device

Independent Sources

Independent Voltage Source (IVS) characterized by a terminal voltagewhich is completely independent of the current through it

+|

3V + 10 V -

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2i

2 A

+ |

3v

yIndependent Current Source (ICS)

characterized by a current which is completely independent of the voltage across it.

Dependent Sources

Dependent Voltage Source (DVS):

Dependent Current Source (DCS):

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B. Passive Elements capable of receiving power (may able to storeenergy)

Resistance (R) accounts for the conversion of electromagnetic energy into heat

Inductance (L) accounts for the storage of energy in a magnetic field

Capacitance (C) accounts for the storage of energy in an electric field

Mutual Inductance (M) accounts for the transfer of energy from one part of a device

to another part by a magnetic field

Network interconnection of two or more simple circuit elements

Circuit a network with at least one closed path

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1.1.4 Ohms Law

;

v = R i

i R

+ v -

Voltage across many types of conducting materials isdirectly proportional to the current flowing through the material.

where: R = resistance

Resistance - total opposition of direct current (DC)

- ratio of voltage to current- measured in ohm ( )

-unit named afterGeorge Simon Ohm, a German physicist

p = vi = i2R = v2/R

;

Conductance (G) - reciprocal of resistance

- ratio of current to voltage

- measured in mho or siemen (S)

Short Circuit (SC) - a resistance of zero ohm

- current through it may have a value

Open Circuit (OC) - infinite resistance

- current is zero (no current flow)

Resistors

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CAPACITORS

Unit: farad (F)

Note: q =Cv

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INDUCTORS

Unit: henry (H)

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1.1.5 Kirchoffs Laws

- named afterGustav Robert Kirchoff, German Professor

electrical conductors/leads connector of circuit elements

- has a zero resitance/perfectly conducting

node a point at which two or more elements have a common connection

PITFALL: Sometime networks are drawn so as to trap an unwary student

into believing that there are more nodes present than is actually are.

Path terms used for set of nodes and elements when there is no node

encountered more than once

Loop a closed path (node started is the same node ended)

Branch a single path in a network

- composed of one simple element and the node at each end

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Exercise: Hayt 7th ed, p31/#2.20

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A. Kirchoffs Current Law (KCL)

Algebraic sum of all currents entering any node is zero

IC ID

IBIA

CE = CL [On a node, sum of currents entering equals sum of currents leaving]

So, IA + IB = IC+ ID

B. Kirchoffs Voltage Law (KVL)

Algebraic sum of the voltages around any closed path is zero

+

V1

-

+

V3

-

+ V2 --V1 + V2 + V3 = 0

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Example: RLC Circuit

Applying KVL:

[Equilibrium Equation]

An equation relating the desired response function (vc) to the

source function V.

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Example: Determine an equilibrium equation for the response function Vo

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+ +

iiR iC

KVL at left mesh: -Vs + R1i + Ldi/dt + Vo = 0 (eq. 1)

KCL at node A: i = iR + iC = Vo/R2 + CdVo/dt (eq. 2)

SOLUTION:

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1.1.6 Resistance, Inductance, Capacitance and Source Combination

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Inductors in Series and Parallel

A. Series

B. Parallel

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Capacitors in Series and Parallel

A. Series

B. Parallel

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Example: LC Network

Example: LC Network in which no series or parallel combinations

of either the inductors or capacitors can be made

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1.1.7 Voltage and Current Division

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Series Inductors and Voltage Division:v1 = L1/Leq

v2 = L2/Leq

vN = LN/Leq

Parallel Capacitors and Current Division: i1 = C1/Ceq

i2 = C2/Ceq

iN =CN/Ceq

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Example 1:

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Solution 1:

Io

A

KVL at left mesh: -60 + 1kIo + 5kIo = 0; Io = 60/6k = 10 mA

By Current Division:

i1 = (500)(6mVo)/(2500) = 500(6m)(50)/2500 = 60 mA

i2 = (2000)(6mVo)/(2500) = 2000(6m)(50)/2500 = 240 mA

KCL atA: Io = Io + i3 ; i3 = 0

So, Vo = (5k)(Io) = (5k)(10m) = 50 V

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[Ref: Hayt, 7th edition]

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[Ref: Hayt, 7th edition]

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