Module 1 - Sinusoids

8/18/2019 Module 1 - Sinusoids

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Sinusoids

EE 102 Circuits 2

Module 1


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Introduction

Hover Dam

Hover Dam Alternators

Commercial power supplies arealternating voltage sources that

follows the sine function known as

sinusoidal function or simply

sinusoids. Circuit currents resulting from

sinusoidal voltages are also

alternating in nature called

alternating current (AC).


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Alternating Current and Voltages

An alternating voltage or electromotive force (EMF) can berepresented by the following sine function

t T E e ft E e

t E e

m

m

m

)!"(#sin#sin

sin

π π

ω

=

=

=

where

e $ instantaneous value of the alternating voltage% volts (&)

Em $ ma'imum or peak value of the alternating voltage% volts (&)

$ angular velocity of the alternating voltage% rad!sec

f $ freuency of the alternating voltage% hert* (+*)

, $ period of the alternating voltage% sec

t $ time in sec.

t

e

E m

T


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Instantaneous Value of a Sinusoid

-efers to the voltage (or current) values at any point in time ina sine wave or sinusoid.

nstantaneous values of voltage and current are usually

symboli*ed by lower case v and i respectively.

t

v

v1

v2

-v3

t 1 t 2

t 3


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/ An alternating voltage is given by the sinusoidal function%

v = 325 sin277t. Find

a) ,he angular velocity% freuency% and period of thefunction.

b) ,he time when the voltage is half its peak value.


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The Phase Angle

-efers to the amount of shift (in degrees or radians) of the sinewave from the vertical a'is.

t

v

shift φt

v

shift φ


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The Phase Angle

A sinusoid shifted to the right of the vertical a'is can bee'pressed by the sine function

)sin( φ ω −= t V v m

t

v

φ

v =V m sinωt

v =V m sin(ωt-φ)

A sinusoid shifted to the right is said to lag from its original

position by the amount of shifting 0 degrees


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The Phase Angle

A sinusoid shifted to the left of the vertical a'is can bee'pressed by the sine function

)sin( φ ω += t V v m

t

v

φ

v =V m sin(ωt-φ)

v =V m sin(ωt+φ)

A sinusoid shifted to the left is said to lead from its original

position by the amount of shifting 0 degrees


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The Phase Angle

,wo or more sinusoids of the same freuency and phase angleare said to be in-phase with each other.

1inusoids that are in2phase reaches its minimum and

ma'imum values at the same time.

t

v

v1 =V 1m sin(ωt-φ)

v =V 2m sin(ωt-φ)


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The Phase Angle

1ince sin(t345o) $ cos t% the sinusoid v = V m cosωt is asinusoid with a phase angle of 90o leadin (shifted 45o to left).

t

v

90o

V m

v =V m sin(ωt+90o )= V m !osωt

1imilarly% the sinusoid v = -V m cosωt % is a sinusoid with a

phase angle of 90o lain (shifted 45o to right).

t

v

90o

V m

v =V m sin(ωt-90o )= -V m !osωt


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The Phase Angle

)

#

cos()45cos(sin

)#

sin()45sin(cos

π ω ω ω

π ω ω ω

−=−=

+=+=

t t t

t t t

o

o

)45sin(sin

)"65cos(cos

o

o

t t

t t

±=−

±=−

ω ω

ω ω


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The Phase Angle

1ines and cosines of the sum and difference of two angles

"a"a"a

"a"a"a

"a"a"a

"a"a"a

sinsincoscos)cos(

sincossinsin)sin(

sinsincoscos)cos(

sincoscossin)sin(

+=−

−=−

−=+

+=+


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/ E'press the sinusoid v = V m !os(ωt+#5o ) in terms of the sine

function.

7 1olution8

9etting : $ t3;


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/ Find the phase difference of the sinusoidal functions

i1=15 !os(10t+50o ) and i2=20 !os(10t-100

o ).

7 1olution8

i 1 is a function shifted 45o+50o= 1#0o to the left of the vertical

a'is while i 2 is a function shifted 45o-100o= -10o to the left of

the vertical a'is (or actually 310o to the right ) . ,herefore% the

phase difference between the two functions is%

Ans.140o ! 10o = 150o

T$is means i 1 is leadin i 2 "% 150o.

t

v

i2 =20 !os(10t-100o )i1 =15 !os(10t+50

o )

140o

10o


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/ Seat&or'

=etermine the freuency and the phase angle difference

between the two voltages v1 = 4 cos(5t + 30

o

) volts andv2 = -2 sin(5t + 20

o ) volts.


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Effectie Value of Sinusoids

Also called the root2mean2suare or -M1. t is the value of the sinusoid that has the same heating effect

as if it were a =C.

>iven the sinusoid v = V m sinωt % the effective value V of this

sinusoid is given by the formula%

mm V

V V ?5?".5

#==


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)eriation of Effectie Values

i

Rv = V m sinωt

t & '

t V

'

vi m

m ω ω

sinsin

===

where I m = V m /R.

,hus% current i is in2phase withsource voltage v.

t

v

i =I m sin ωt

v =V m sin ωt ,he instantaneous power p

dissipated by the resistor at any

given time is%

t & V

t & t V vi

mm

mm

ω

ω ω

#sin

sinsin

=

×==


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lotting the power yields%

t

p p = V m I m sin2 ωt

ω

π

ω

π #

,he incremental energy d

dissipated by the resistance

during incremental time dt %

is d = dt. Calculating fortotal energy consumed

from time t$5 to t$B!

yields%

mm

mmmm

& V

dt t & V dt t & V dt

ω

π

ω ω ω π

ω

π

ω

π

#

sinsin5

#

5

#

5

=

=== ∫ ∫ ∫


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mm & V ) ω

π

#=

V& *

& V *

& V & V

) *

mm

mmmm

=

×=

===

##

#

#

ω

π ω

π

ω

π

,he power dissipated over one pulse of the power graph is

the total energy dissipated divided by the total time this

energy was consumed%

where is the effective

value of the sinusoidal voltage and

is the effective value of the

sinusoidal current.

#

mV V =

#

m &

& =


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/ Meralco supplies a sinusoidal voltage with an effective value

(-M1) of #5 &ac to home users.

a) Dhat is the peak2to2peak value of this sinusoidal voltage

b) f a flat iron with a resistance of < ohms is connected to

this line what is the resulting -M1 current to the iron

c) Dhat is the power consumption of the flat iron in b)

d) Dhat is the energy dissipated in kilowatt2hr for two hours

of use of the flat iron in b)


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Than' *ou

Module 1 - Sinusoids

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