Unit System Electrical Elements

8/18/2019 Unit System Electrical Elements

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•SI Unit system•Basic Electrical Quantities and AssociatedUnits: Charge, Current, Voltage, and Power

•Current Direction and Voltage Polarity•Basic Circuit elements

•Independent Voltage and Current Sources

•Dependent Sources•Ohm’s Law


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• Through out this course a standard system of

units SI which is: Metric system will beadopted

– International System of Units (SI) adopted in1960 by the General Conference on Weights andMeasures.

– It consist of Six basic units:– Meter (m), Kilogram (kG), Second (s), Ampere

(A), degree Kelvin (0K), Candela (cd)


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• Standard SI prefixesStandard SI prefixesStandard SI prefixesStandard SI prefixes

SI base units


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Units of some electric circuits quantities

1 [W] = 1 [J/s](W)(W)(W)(W) Scottish engineer, James

WattWattWattWatt

Power

1 [V] = 1 [J/C](V)(V)(V)(V) Italian scientist, Alesandro

VoltaVoltaVoltaVolta

Voltage

1 [A] = 1 [C/s](A)(A)(A)(A) French scientist, Andre Marie

AmpereAmpereAmpereAmpere

Current

1 [J] = 1 [N–m](J)(J)(J)(J) British physicist, James P.

JouleJouleJouleJoule

Energy

1 [N] = 1 [kg–m/s2](N )(N )(N )(N ) English scientist, Sir Isaac

NewtonNewtonNewtonNewton

Force

(C )(C )(C )(C ) French scientist, Charles

Augustin de CoulombCoulombCoulombCoulomb

Charge

1 [lb] = 0.453593 [kg]lbkg kg kg kg Weight

1 [in] = 0.0254001 [minch(m)(m)(m)(m) Meter Length

Quantity Unit (SI)Quantity Unit (SI)Quantity Unit (SI)Quantity Unit (SI)


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Charge and Current• Charge: the nature of charge is based on the

concept of atomic theory.

• Normally for constant charge we use Q &q for time-varying charge (e=-1.602x10-19C)

• Current: is defined as the rate of change of

charge w.r.t time. I , i(t )

Q, q(t )


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• Conventional current direction:

is in the direction of positive charges flow

through a given conductor.• Current reference direction :

i1

(a)a b

7A

(b)


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Current flowing into a general element

So, the total charge entering the element since t 0 up totime t is given by

i

∫∫ =−==t

t

t q

t qTotal Coulombind it qt qdqq

00

)()()( 0)(

)(τ τ

∫ +==>t

t t qd it q

0

)()()( 0τ τ

dt

dq

t

qi

ot

=∆

∆=

→∆

lim


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>

≤≤−

≤≤−

≤≤−

≤≤

≤≤

<

=

s12t,0

s12ts10,C)12t(2

s10ts8,C4s8ts4,C)t6(2

s4ts2,C4

s2t0,Ct2

0t,0

)t(q

>

≤≤

≤≤

≤≤−

≤≤

≤≤

<

=

s12t,0

s12ts10,A2

s10ts8,0s8ts4,A2

s4ts2,0

s2t0,A2

0t,0

)t(i

Example 1. Given q(t), find i(t).


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>

≤≤−

≤≤−

≤≤−

≤≤

≤≤

<

=

s12t,0

s12ts10,C)12t(2

s10ts8,C4s8ts4,C)t6(2

s4ts2,C4

s2t0,Ct2

0t,0

)t(q

>

≤≤

≤≤

≤≤−

≤≤

≤≤

<

=

s12t,0

s12ts10,A2

s10ts8,0s8ts4,A2

s4ts2,0

s2t0,A2

0t,0

)t(i

Example 2. Given q(0)=0 and i(t), find q(t)


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Example 3

• Given

• Find the net charge passing through point A between –0.25 s and +0.25 s

• Answer: ∆q = 2.04 C

>

<

= − 0tAe5

0tA5

)t(i t4

A

i(t)

5

t-0.25s 0.25s


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Solution to Example 3

∫=−2

1

t

t

12 dt)t(i)t(q)t(q

( )

0 0.25

0.25 0

0 0.25

4

0.25 0

0.25

0 4

0.25

0

1

0

1

( ) ( )

5 5

15 | 5 (4 )

4

11.25 54

1.25 1.25 1

2.04

t

t

y

q i t dt i t dt

dt e dt

t e d t

e dy

e

−

−

−

−

−

−

−

∆ = +

= +

= + ×

= + ×

= + × −

=

∫ ∫

∫ ∫

∫

∫

A

i(t)

5

t-0.25s 0.25s

Set y=4t


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Common types of electrical currentsCommon types of electrical currentsCommon types of electrical currentsCommon types of electrical currents

(a) DC (b) AC sinusoidal (c) Exponential c(a) DC (b) AC sinusoidal (c) Exponential c(a) DC (b) AC sinusoidal (c) Exponential c(a) DC (b) AC sinusoidal (c) Exponential curr urr urr urr

ent decay (d) Damped sinusoidal currentent decay (d) Damped sinusoidal currentent decay (d) Damped sinusoidal currentent decay (d) Damped sinusoidal current

It is essential to determine bIt is essential to determine bIt is essential to determine bIt is essential to determine b

oth direction and quantity ofoth direction and quantity ofoth direction and quantity ofoth direction and quantity of

currentcurrentcurrentcurrent


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DC: i(t)=constantWe assign a direction of current flow for

the purpose of analysis only.

+: flows along the direction we specified

-: flows opposite to the direction wespecified

i(t) always opposite to the direction of

electron motion.

t

i 5A

-1A

Alternating current (AC or ac)Direction and amplitude change periodically with time.

A fundamental form of AC current is

i(t)=A sin ( ω ωω ω t + φ φφ φ )

Peak value Frequency

Phase

Common types of electrical currents cont..Common types of electrical currents cont..Common types of electrical currents cont..Common types of electrical currents cont..


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• Voltage, potential difference

– Is defined as the work done to move apositive unit charge through an element

Unit : Volt equivalent to 1[V ] = [1 J /1 C ]

So voltage is energy/unit charge

V =dW/dq

Voltage / potential difference

qV W •=


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DC Voltage polarity

Terminal A is 5 volt negative w.r.t terminal B

Terminal A is 5 volt positive w.r.t terminal B

It is essential to determine the polarity and thevalue of the voltage. For AC circuits the polarity

is only valid for a definite given time e.g. at t=0


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Example1

Examle2

(a)

v6+ -

(b)

23V+-

(a) (b)

-5V5V

+

-

+

-

A A

B B

Double script notation : ab bav v= −

( ) 5 [ ], ( ) 5 [ ]ab baa v V b v V = = −


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Example• A generates 20W of power, find

VA

• Find power absorbed by B

• If element C generates 3W ofpower, find Vc

• If element D absorbs 27W of

power, find ID

• If element E absorbs 2W ofpower, find IE

• Find the power absorbed byelement F

Answers:(1) VA = 5 V, (2) PB = 14 W, (3) VC = 1 V, (4) ID = 9 A, (5) IE = 2 A, (6) PF = 20 W.


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The passive elementsResistor R: is the resistance that resists the flow of the currentthrough a conductor. It is measured in Ohm ()

Capacitor C: It is an element for storing the electrical energy inform of static field. The capacitance of a capacitor is measured

in FARAD. (F)

Inductor L: It is an element of storing the electrical energy inform of magnetic field. The inductance of an inductor ismeasured in HENRY. (H)


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The Active Elements

They deliver energy to the passive elements of the circuit

VoltageSources

CurrentSources

IndependentVoltage Sources

Dependent

Voltage Sources

Independent

current Sources

Dependent

current Sources


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– Independent current source – It provides constant current value irregardless of

the voltage across it

The current is constant and completelyindependent of the voltages and currents of theother elements,

Symbol :


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– Dependent current source – It provides current value depending on

voltages or currents in other elements of

the same circuit or other surroundingcircuit.

They are of two types:

i1

Ki i1

a

b

V1 KG V1

a

b

+

-

1-Voltage-controlledcurrent source

2- current-controlledcurrent source


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– Dependent Voltage sources

– It provides voltage value depending on voltagesor currents in other elements of the same circuitor other surrounding circuit.

They are of two types:

1-Voltage controlled Voltage source 2- current-controlled Voltage source

V1 KvV1+-

a

b

+

-

Example: Find the value of VL


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• Delivering and absorbing power in thesources

+

-

(a) Delivering

12V

(b) Absorbing

12V+

-

2A

2A

AbsorbingDelivering

II

++ --


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Electrical Circuit: General Definition

Node: It is a point of connection of three or moreof the circuit elements, e.g. Points A, B, M, and D

Branch: It is part of the circuit elements betweentwo nodes, e.g.AD, AB, BM1,BM2, AD1,AD2

Loop: Any closed path by the circuit elements,e.g

. Loop I1 and Loop I2


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( ) 2sin(100 )

( ) 0.2sin(100 )

v t t

i t t

=

=

2 0.4( ) ( ) ( ) 0.4sin (100 ) (1 cos(200 ))2

p t v t i t t t = = = −

Instantaneous power

Eample: For the shown circuit find the instantenous power

Solution:


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Problem1• For the shown circuit if the voltage and currentare as shown:

• Find the time when the absorbed power is maximum

• Find the maximum power

• Find the total energy given to the element

elementv

i

+

_

Aet ivolt et vt at t t 500500 5.1;80000;0 −− ••=•=>

0;0;0 ==<

ivt at


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Solution

Which result in t . sms it is the time at maximum poer t . sms it is the time at maximum poer t . sms it is the time at maximum poer t . sms it is the time at maximum poer

he maximum poer is ien ashe maximum poer is ien ashe maximum poer is ien ashe maximum poer is ien as

t t t et et et ivP

10002500500 1200005.180000 −−− •=••••=•=

01200000002120000 100021000 =••−••= −− t t et et dt

dP

for the maximum power, diffrentiate the above equation w.r.t

time and equate to zero

To find the total energy absorbed mW emsP 96.64)102(120000)2( 223 =••= −−

dt et idt vPdt W t 10002

000

120000 −∞∞∞

•=•== ∫∫∫

J ee

t et W t t

t µ 240]

10

2

1000

2[120

0

6

1000100010002

=−•−•−=

∞−−

−


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Solution

_1 _ 1

_ 2 _ 2

( ) ( 1 ) 3(1 ) ( 1)

( 2 ) ( ) ( 2) (3 1)

t

ab so rbed A so urce

t

absorbed V source

P v t A e

P V i t e

−

−

= × − = − × −

= − × = − × −

_

1 _1 _ 2 _ 2 _

3(1 ) (3 ) 3(1 ) ( 1)

(3 1) (3 1) ( 2) (3 1)3(1 ) (3 1)

(3 3) (3 1)

3[1 ( 1)] (3 1)0

total absorbed

absorbed A source absorbed V source

t t t

t t t

t t

t t

t t t

PP P P P

e e e

e e ee e

e e

e e e

− − −

− − −

− −

− −

− − −

= + + +

= − × + − × −

+ − × − + − × −

= − × −

+ − × −

= − + − × −

=

1 1 1

2 2 2

( ) ( ) 3(1 ) (3 )

( ) ( ) (3 1) (3 1)

t t

t t

P v t i t e e

P v t i t e e

− −

− −

= × = − ×

= × = − × −

_ 0total absorbed P =

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Unit System Electrical Elements