chapter 2: fundamentals of electric circuit

CHAPTER 2

FUNDAMENTALS OF

ELECTRIC CIRCUITS

EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING

INDEPENDENT SOURCES

• The voltage/current sources that have the capability of

generating a prescribed voltage or current independent of

any other element within the circuit.

• These sources may output a constant voltage/current, or

they may output voltage/current that varies with time.


PRINCIPAL ELEMENTS OF ELECTRICAL

CIRCUITS

1) Ideal Voltage Sources

An ideal voltage source is a two-terminal element thatmaintains the same voltage across its terminals regardlessof the current flowing through it.

• Vt = constant, no matter what the load current is.


Vt

IL

Vo

L

t

+

-

Vo

2) Ideal Current Sources

An ideal current source is a two-terminal element thatmaintains the same current regardless of the voltageacross its terminals.

• IS = constant, no matter what the load voltage is.


IO

VO

IS


DEPENDENT (CONTROLLED) SOURCES

• Dependent sources are whose output (current or voltage)

is a function of some other voltage or current in a circuit.

• The symbols typically used to represent dependent

sources are in the shape of a diamond.

BRANCH, NODE, LOOP, MESH

• Branch : any portion of a circuit with two terminals

connected to it.

• A branch may consist of one or more circuit elements.


• Node : the point of connection between two or more

branches.

• A node usually indicated by a dot in a circuit.

• Loop : any closed path through the circuit in which no

node is encountered more than once.



• Mesh : a loop that does not contain other loops.

Electric Current

• Electric current is defined as the time rate of change of

charge passing through a predetermined area.

𝑖 =∆𝑞

∆𝑡𝑜𝑟 𝑖 =

𝑑𝑞

𝑑𝑡

• The units of current are called Amperes, where

1 Ampere (A) = 1 Coulomb/second (C/s).

• In order for current to flow, there must exist a closed

circuit.


CURRENT AND KIRCHHOFF’S CURRENT

LAW

• In the circuit of this figure, the current i flowing from the

battery to the light bulb is equal to the current flowing from

the light bulb to the battery.

no current (and therefore no charge) is “lost” around

the closed circuit. This principle is known as

Kirchhoff’s current law (KCL).



Kirchhoff’s Current Law (KCL)

• One of the fundamental laws of circuit analysis.

• Establish in 1874 by G.R. Kirchhoff.

• “The sum of the currents at a node must equal zero.”

𝑛=1

𝑁

𝑖𝑛 = 0 𝑜𝑟

(𝐸𝑛𝑡𝑒𝑟𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡𝑠) = (𝐿𝑒𝑎𝑣𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡𝑠)


• Example of Kirchhoff’s current law:

At node 1:

−𝑖 + 𝑖1 + 𝑖2 + 𝑖3 = 0

𝑖 = 𝑖1 + 𝑖2 + 𝑖3

• In this illustration, currents

entering a node are defined as

negative and currents leaving

the node as positive.

Voltage

• The total work per unit charge associated with the motion

of charge between two points.

• The units of voltage are called Volts, where

1 Volts (V) = 1 Joule (J)/Coulomb (C).


VOLTAGE AND KIRCHHOFF’S VOLTAGE

LAW


Kirchhoff’s Voltage Law (KVL)

• The second fundamental laws of circuit analysis introduced

by G.R. Kirchhoff.

• The principle underlying KVL is that no energy is lost or

created in an electric circuit.

• In circuit terms, the sum of all voltages associated with

source must equal the sum of the load voltages.

• “The net voltage around a closed circuit is zero.”

𝑛=1

𝑁

𝑣𝑛 = 0


• Example of Kirchhoff’s voltage law:

𝑣1 = 𝑣2 where

𝑣2 = 𝑣𝑎𝑏 = 𝑣𝑎 − 𝑣𝑏

• In general, elements that provide

energy are referred as sources

and elements that dissipate energy

as loads.

Power

• The electric power generated by an active element, or

that dissipated or stored by a passive element, is equal to

the product of the voltage across the element and the

current flowing through it.

• The units of power are called Watts (Joules/second).


ELECTRIC POWER AND SIGN

CONVENTION

𝑷 = 𝑽𝑰

Passive Sign Convention

• State that if current flows from a higher to a lower voltage

(plus to minus), the power is dissipated and will be a positive

quantity.

• Example:

• Power generated (supplied) always equals power dissipated.


Power dissipated = vi Power dissipated = - vi

Power generated = vi

• An ideal resistor is a device that exhibits linear resistance

properties according to Ohm’s law,

which states that the

voltage across a

resistance is directly

proportional to the current

flowing through it.


RESISTANCE AND OHM’S LAW

𝑽 = 𝑰𝑹

FUNDAMENTAL OF ELECTRICAL ENGINEERING, First Edition, by Giorgio Rizzoni, © 2009 McGraw-Hill Companies, Inc.


• The value of the resistance R is measured in units of

ohm’s (), where

1 = 1 V/A

• For a resistor R, the power dissipated can be expressed

by

𝑷 = 𝑽𝑰 = 𝑰𝟐𝑹 =𝑽𝟐

𝑹

Open and Short Circuits

• Open circuit : a circuit element whose resistance

approaches infinity.

• Short circuit : a circuit element with resistance

approaching zero.


Series Circuit

• Two or more circuit elements are said to be in series if the

current from one element exclusively flows into the next

elements.

• All series elements have the same current.

Series Resistors

• Equivalent series resistance:


𝑹𝑬𝑸 =

𝒏=𝟏

𝑵

𝑹𝒏 = 𝑹𝟏 + 𝑹𝟐 +⋯+ 𝑹𝑵

• Example 2.1:

For the circuit shown,

a) Find the equivalent resistance seen by the source.

b) Find the current I .

c) Calculate the voltage drop in each resistor.

d) Calculate the power dissipated by each resistor.

e) Find the power output of the source.

Given: V = 24 V, R1 = 1 , R2 = 3 , and R3 = 4 .



b) 𝐼 =𝑉

𝑅𝐸𝑄=

24 𝑉

8 Ω= 3 𝐴

c) 𝑉1 = 𝐼𝑅1 = 3 𝐴 1 Ω = 3 𝑉

𝑉2 = 𝐼𝑅2 = 3 𝐴 3 Ω = 9 𝑉

𝑉3 = 𝐼𝑅3 = 3 𝐴 4 Ω = 12 𝑉

Solution:

a) 𝑅𝐸𝑄 = 𝑅1 + 𝑅2 + 𝑅3

= 1 + 3 + 4

= 8

REQ

Note: 𝑉1 + 𝑉2 + 𝑉3 = 3 + 9 + 12 = 24 𝑉

The total voltage drop is equal to the voltage output of the source.


Note: 𝑃1 + 𝑃2 + 𝑃3 = 9 + 27 + 36 = 72 𝑉

The total power dissipated by the resistors is the same as the power output by the source.

e) 𝑃 = 𝐼𝑉 = 3 𝐴 24 𝑉 = 72𝑊

d) 𝑃1 = 𝐼2𝑅1

= 3 𝐴 2 1 Ω

= 9𝑊

𝑃2 = 𝐼2𝑅2

= 3 𝐴 2 3 Ω

= 27 𝑊

𝑃3= 𝐼2𝑅3

= 3 𝐴 2 4 Ω

= 36 𝑊

Parallel Circuit

• Two or more circuit elements are said to be in parallel if

the elements share the same terminals.

• All parallel elements have the same voltage.

Parallel Resistors

• Equivalent parallel resistance:

or

where


𝟏

𝑹𝑬𝑸=

𝟏

𝑹𝟏+

𝟏

𝑹𝟐+⋯+

𝟏

𝑹𝑵

𝑹𝑬𝑸 =𝟏

𝟏 𝑹𝟏+ 𝟏 𝑹𝟐

+⋯+ 𝟏 𝑹𝑵

𝑉1 = 𝑉2 = 𝑉3 = 𝑉4

Various Parallel Resistors Networks



• Example 2.2:

For the circuit shown,

a) Find the equivalent resistance seen by the source.

b) Find the total current I .

c) Calculate the currents in each resistor.

d) Calculate the power dissipated by each resistor.

e) Find the power output of the source.

Given: V = 24 V, R1 = 1 , R2 = 3 , and R3 = 4 .


Solution:

a) 1

𝑅𝐸𝑄=

1

𝑅1+

1

𝑅2+

1

𝑅3=

1

1+

1

3+

1

4= 1.583 Ω

∴ 𝑅𝐸𝑄 =1

1.583= 0.632 Ω

b) 𝐼 =𝑉

𝑅𝐸𝑄=

24 𝑉

0.632 Ω= 37.975 𝐴 ≈ 38 𝐴

REQ

I


c) 𝐼1 =𝑉

𝑅1=

24 𝑉

1 Ω= 24 𝐴

REQ

I

𝐼2 =𝑉

𝑅2=

24 𝑉

3 Ω= 8 𝐴

𝐼3 =𝑉

𝑅3=

24 𝑉

4 Ω= 6 𝐴

Note: 𝐼1 + 𝐼2 + 𝐼3 = 24 + 8 + 6 = 38 𝐴

The sum of the individual current is equal to the current output of the

source.


Note: 𝑃1 + 𝑃2 + 𝑃3 = 576 + 192 + 144 = 912 𝑉

The total power dissipated by the resistors is the same as the power output by the source.

e) 𝑃 = 𝐼𝑉 = 38 𝐴 24 𝑉 = 912 𝑊

d) 𝑃1 = 𝐼12𝑅1

= 24 𝐴 2 1 Ω

= 576 𝑊

𝑃2 = 𝐼22𝑅2

= 8 𝐴 2 3 Ω

= 192𝑊

𝑃3= 𝐼32𝑅3

= 6 𝐴 2 4 Ω

= 144 𝑊

Series and Parallel Resistor Combinations

• Example 2.3:

The Wheatstone Bridge consists of two series circuits that are connected

in parallel with each other.

1) Find the value of the voltage

Vab = Vad - Vbd in terms of the four

resistances and the source voltage Vs.

2) If R1 = R2 = R3 = 1 k, Vs = 12 V,

and Vab = 12 mV, what is the value

of Rx.



Solution:

1) 𝑉𝑎𝑑 =𝑅2

𝑅1+𝑅2𝑉𝑠 and 𝑉𝑏𝑑 =

𝑅𝑥

𝑅3+𝑅𝑥𝑉𝑠

Thus,

𝑉𝑎𝑏 = 𝑉𝑎𝑑 − 𝑉𝑏𝑑 = 𝑅2

𝑅1+𝑅2−

𝑅𝑥

𝑅3+𝑅𝑥𝑉𝑠

2) 0.012 =1000

1000+1000−

𝑅𝑥

1000+𝑅𝑥12

𝑅𝑥 = 996 Ω

R1

R2

R3

RX

Vs

a Vab b

c

d

R1

R2

R3

RX

Voltage Divider Rule (VDR)

• VDR is useful in determining the voltage drop across a

resistance within a series circuit.

where VX = the voltage drop across the measured resistor,

RX = the resistance value of the measured resistor,

REQ = the circuit total resistance,

VS = the circuit applied voltage


𝑽𝑿 =𝑹𝑿

𝑹𝑬𝑸𝑽𝑺

+ V1 -

+ V3 -

+

V2

-S

• Example 2.4:

Determine the voltage across the R2 and the R3.

Solution:


𝑉2 =𝑅2𝑅𝐸𝑄

𝑉𝑆 =20

10 + 20 + 3060 = 20 V

𝑉3 =𝑅3𝑅𝐸𝑄

𝑉𝑆 =30

10 + 20 + 3060 = 30 V


Current Divider Rule (CDR)

• CDR is useful in determining the current flow through one

branch of a parallel circuit.

where IX = the current flow through any parallel branches,

RX = the resistance of the branch through which the

current is to be determined,

REQ = the total resistance of the parallel branch,

IS = the circuit applied current

𝑰𝑿 =𝑹𝑬𝑸

𝑹𝑿𝑰𝑺

• Example 2.5:

Find each of the branch currents in the figure shown below.


Solution:

𝑅1 𝑅2 𝑅3

1

𝑅𝐸𝑄=

1

𝑅1+

1

𝑅2+

1

𝑅3=

1

3 𝑘Ω+

1

8 𝑘Ω+

1

24 𝑘Ω=

1

2 𝑘Ω

𝑅𝐸𝑄 = 2 𝑘Ω

Thus,

𝐼𝑅1 =𝑅𝐸𝑄𝑅1

∙ 𝐼 =2 𝑘Ω

3 𝑘Ω12 𝑚𝐴 = 8 𝑚𝐴


∙ 𝐼 =2 𝑘Ω

8 𝑘Ω12 𝑚𝐴 = 3 𝑚𝐴


∙ 𝐼 =2 𝑘Ω

24 𝑘Ω12 𝑚𝐴 = 1 𝑚𝐴

𝐼𝑅1 + 𝐼𝑅2 + 𝐼𝑅3 = 8 + 3 + 1 = 12 𝑚𝐴 ( KCL )


• For the particular case of two parallel resistors,

and, the current passing through

R1 and R2 are

𝐼1 =𝑅𝐸𝑄

𝑅1∙ 𝐼 =

𝑅1𝑅2𝑅1+𝑅2

𝑅1∙ 𝐼

𝐼2 =𝑅𝐸𝑄

𝑅2∙ 𝐼 =

𝑅1𝑅2𝑅1+𝑅2

𝑅2∙ 𝐼


𝑹𝑬𝑸 = 𝑹𝟏 𝑹𝟐 =𝑹𝟏𝑹𝟐

𝑹𝟏 + 𝑹𝟐

I

I1 I2

𝑰𝟏 =𝑹𝟐

𝑹𝟏 + 𝑹𝟐∙ 𝑰

𝑰𝟐 =𝑹𝟏

𝑹𝟏 + 𝑹𝟐∙ 𝑰

Note: It only works for two parallel resistors.


MEASURING DEVICES

Ohmmeter

• The ohmmeter is a device that, when

connected across a circuit element,

can measure the resistance of the

element.

• The resistance of an element can be

measured only when the element is

disconnected from any other circuit.


Ammeter

• The ammeter is a device that, when connected in series

with a circuit element, can measure the current flowing

through the element.

1. The ammeter must be placed in series with the

element whose current is to be measured (e.g.,

resistor R2).

2. The ammeter should not restrict the flow of current

(i.e., cause a voltage drop), or else it will not be

measuring the true current flowing in the circuit. An

ideal ammeter has zero internal resistance.


Measurement of current


Voltmeter

• The voltmeter is a device that can measure the voltage

across a circuit element.

1. The voltmeter must be placed in parallel with the

element whose voltage it is measuring.

2. The voltmeter should draw no current away from the

element whose voltage it is measuring, or else it will not

be measuring the true voltage across that element.

Thus, an ideal voltmeter has infinite internal resistance.

Measurement of voltage


chapter 2: fundamentals of electric circuit

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