AC Circuits 1

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AC Circuits Part 1 Single Phase (1 ) Syste~

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"THE ESTABLISHED LEADER IN EE REVIEW"

MULTIVECTOR · Review and Training Center

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MULTIVECTOR REVIEW AND TRAINING CitNTER ) .

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ALTERNATING CURRENT (AC) CJRq.JJTS PAI< ~' 0.636 Vm For sintlsoidal current wave, l.,e "' 0.636 rm Effcctivc Value · the vi1lue of al;ernating quantity wh'lch when applied to a given circuit for a given· time produces the same expenditure of energy as when d.c. is applied to th~ same circuit for the same itHerval of till\e. The effective . ,~,; value is also called· a~ the "root-mean-square" (rms) value. For sinusoidal voltage :wave, V''"'' ., o:707 V,., For sinusoidal current wave; r,;;;.; ~- 0.707 1m For other waves. avhage ~alue of voltage and current is given by

1 V.ve== - -

T

fo

l•ve~

T

v(t)dt

. T

-+-:f

i(t)dt

~

0

and for RMS of EffeCtive value

+J T

Yrms =

2

v (t)

dt

0

Forin. factor

_I_

T

·JT

P(t):dt

o

Effective valu·e Average value

:. For sine wave, Form· factor= I. I I

Crest or peak factor

Maximum value = -------

Effective value

For sine wave. Crest or peak factor= 1.414

AC Circuits Page 3 of 10

Pure Resistance .

Power Tri.angle:

I~:)

¥~-J_:_ 0 p ' I

V

~~

1R

0 =-power factor (p.f).anglc or phase angle p.f. ·· power factor · · cos 0 · PIS r.f. ' rcactivct-d.·or ' sin 0 ,. Q/S· tan 0 '" Q/P

where:

V · eflective or RMS voltage ''effective or RMS current R ~~ efTective or ac resistance

For Pure R. since 0 '' 0° p. f. ""cos 0° "'· I or unity r.f. ·~'~'sin 0° = 0 '

For sinusoidal Voltage supply,

P"'·S;

o~~o

Energy expended, W = Pt joules or watt-sec

1m ' I"' -::::-- ";' 0.707 1111

'>/2 Pure Inductance

: .. V 111

I111 H

=

'.

Phasor diagram:

v

L

where: I is in phase with V

XL = inductive reactance in ohms XL= wL =" 2nfl~ n

I. Real or True or Active or Average Power

f= frequency, Hz i

L := inductance. henry (I I)

. P ~VI cos 8 watts VI p =~cos

e watts

For simisoidal voltage supply.

2 2

P= 1 '1

I

Rwatts

1m

v2

Reactive Power Phasor diagrnm:

. Q =VI sin 8 volt-ampere reactives (~ars)

o

-+-r--------~Y

·Yrnlm . O Q = - - sm · vars 2 3. Apparent Power

I lags V by 90'? S =VI

Vmlm

s·=-.""'

volt-ampere (va) va

2

i.

1\C Circuits Page 4 of 10

MULTIVECTORREVIEW AND TRAINING CENTER .. For Pure L, since 0"' 90° P = VI cos q(jo "' 0 Q '"· VI sin

v = -,JvR2 +·v~,2 V =I -,JR 2. +X I. 2 V = IZ volts

capacitance,. farad (F)

For sinusoidal voltage supply,

where:

Z=

im~edance

in ohms + x~,2' p.f. =cos 0 =! VR!Y= R./Z r .. f. =sin 0 = V 1 !Y~ X1 /Z tan 8 = VtfV 1:( =X 1/R .

1111

z =-,JR.

J=-

'.

v2

Phasor diagram: Impedance Triangle:

I

ote~9oo ~

·~x,

v

0

I leads V by 90°

R

!

P =VI (p.f.) :~ 12R '"' V1/!R watts Q =VI (r.f.) ~ 12 Xt. = V12 /X1. vars S = VI ·~ 12Z = V 2/Z va

. . For Pure C, since 0 = .90° P~"O;Q=Vl,S=VI

S=Q·

Also,

Energy expended,

v"c

We= _ _ 2

joules

V,/C

We,_.·_ 4

.. I es

JOU

AC Circuits . Page 5 or 10

MULTIVECTOR REVIEW AND TRAINING dENTER R and C in Series

RLC in Series

----::;:.

--7

i

r~-

1' VR

~.

v

f'.,.

l

v

Yc

L V,

f-

w

J

R

,_

Phasor diagram:

x~_

~"> Yc,;,IXc ,

2

i

1

Xci

where:

V= IZ volts where:

=

Vc!VR

Note: Transpose XL and Xc if X(·-. X 1 • '

R

J

2

= Xc!R

P =VI (p.f.) = 1 R = V 1//R watts Q =VI (r.f.) = 12 (X~_ -Xc) Q = (V1. 2/XL)- (V/!Xc) 2 S -=VI·"' 1 Z:= V 2/Z va

Impedance Triangle:

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R"' +(XL- Xc)"' p.f. =cos 0 = VRIV = R/Z ' r.f. =sin 0 = (VL- Vc)IV -~ (X 1 - Xc)IZ : tan 0 =' (V 1 --Vc)IVR ,, (X 1 Xt )/R Z·=

V ,;, I "H.? + Xc 2

tan 0

{2

V"' ~VR + (VL -Vd 2 V =I ~R +(X,V = IZ volts

V = 'YVR + Vc-

·z=,JR2+Xc2 p.r. =cos e = vRIV = R/Z r.f. =sin 8 = Vc!V = Xc!Z

n

I I Xc''~oc~

V=IZ (~

=· (J)L = 2nfl,

Also,

z P"' VI (p.f.) ~ 12R = VR 2/R watts Q =VI (rJ.) = ! 2 Xc = V~_ 2 /Xc vars S =VI=' 12Z = V 2/Z va

Also,

1\C Circuits Page 6 or I 0

MlJL TIVECTOR REVtEW AND TRAINING CElNTER I

Rand Lin Parallel

Phasor diagram: I(

·.

1 ·v

-----------~

r~fR J~~~~ R. · I.

·..

'

../~ : \

~f .....

Note:

V is colistant. X1. = mL c ·:2nfL n . IR ·.c VIR: I, ~ v;x, Phasor dingraii1: \\'here:

>V

P =VI (p.f.) c= 11/R · V2/R watts Q =VI (r.f.) = 1/ X( Q = V2/Xc vars: S"" VIr cc I/Z = V 2/Z va

lr = v /Z amp. where:

Z = impedance in ohms Also,

z p.f. =cos 0 = IR!Ir = Z/R r.f. =sin 8 =IL/h = ZIX1. tan 8 = 11/IR = RIXL

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RLC in Parallel · ~

Ir

'

P =VI (p.f.) = IR-R = V'/R watts 2 Q= VI (r.f.) "''L x~."" V 2/X1. vars 2 2 S VI-r= 11, Z '"' V /Z va

=

Also,

Xc"" (I)L .., 2nfL

n

X( .. _ l __ _

Rand C in J>aralld ·

we

~h

2nfC

IR = V/R: l1. .. V/X 1 :It lr = ~l (IL - lc)'

VtXc

vx-;2 x~2 :i:-rZ26 Xc -0, ifXc > Xr.

In general, Z =. R ±.jX

.=.p ±jB

' .ti ~

+ j~

=

B

- J = . ,I 1i

=

where

I

x. '

Inductive susceptance. 13 1 = I/X 1 Capaciti-.:e susceptance. Be·= I 1X,

5 .. For Series RC

Note:

mho or sicmen ·

R

where: 0° < 8 < 90°

Z

mho or sicmen

z

6

~apaci~ive susceptance B.mductlve susce11tance 13 1

![

Admittances in Series

,'{ .I

Yl

I

- - + --Y2 Y1

Y,

I ...... -t

Yn

For two (2) admrtt~nces in series.

·

+ jX =inductive reactance - jX = capacitive reactance

Yir

~

Y,Y,

- - - · --Y, + y2

Admittances in Parallel *For the complex expression of an impedance, its angle ranges from oo to 90° only either positive or negative. Mathematically, 0° :50$ 90°.

Y1 =Y 1 +Y2+Y 3 +

Note: · In the above equations. admittances must be expressed in complex rorm. i\C Circuits Page li or I 0

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MULTIVECTOR REVIEW AND TRAINING CENTER . .

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Power in Complex Form

I. Voltage Conjugate Method S c I x Conjugate of V with • resp~ct to the horizonial axis S · .JV '·-P±~ . . s "' ~ + J () va where: . S - :app·arcnt power. va P -, real or true or average; power. watts Q · reactive power. vars I c- .current, amperes V ., voltage. ~'(>Its : V " conjugate of the voltage, \'olts ·0 =· power factor angle Note: +j = Ois capo-c

4.

C=~L · (l)o-

4.

S.

I w_" 7" ~LC .

5. COo

Vc

w}C

·.·where W0 =angular frequency at . resonance, racl/sec (J.

I. lr.

lc

=

X~_=;

Xc

I c =---,ro 0 "L I = r:-;::;- 1

. \'LC

6.

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r, = _ __,_ 2rc..JLC

I

f,.;-~-

2rcvLC where ~ "' resonant frequency; hertz (HZ) 7. Supply ~oltage, V c=VR 8. Z == R ~ a pure resistance .·. Z is minimum. 9. Total reactance, Xr "'' 0 I 0. I = V /Z = V /R ,. . .' .. r is maximum. II. p.f. I or unity 12. S=P=VI. 1

.

=

7. Supply volt~\ge, V =VR 8. Z =R ~ a pure resistance :. Z ib maximUI!.). 9. Total reac·tahce, X-1 ~ 0 l

IR = V(Z= V/R :. 11 is minimur_!}. II. p. f. ~- I or unity 12. S~cP=VI 10. 11

=

13. Q

~o

13. Q=O .;

·'j'

i~ tl

1\C Circuits Page 9 of',IO

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MULTIVECTOR REVIEW AND TRAlNING CENTER

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Parallel Resonance, Two-Bratich Circuit

J

\(til(!)-;

w,·c 2nf, ; (I)~

.

··-·---~J...

2711':

. . lr ·. \ r, t;. ~ 4. For Parallel RI.C.

R

Q = - - - .,. o>.,CR "

(I).,L

This is the rec,procal of .. f(lr series RI.C.

At resonance.

I

Effective Value of a Nonsinusoidal Current or Voltage Wave

,.-------'.---------·--,..---I=

__,_

R12

ill

I

1

I de 2 + ~-1· 2

1m/

1111 22 --, ·t--2

I

'

~~~~~·

,

--··-

---~:;--­

LIC

-

2

' t

I

.. f ~

Rc2- L/C

I

where: I = effective· w11uc of thc current V =effective value of the voltagc Ide= de component of the current Vdc =de COlllJ?Onent of the voltage , 1111 = maximtlln value of the ac cornpiHlcfll of the current, subscript indicates thc degree ofti)e harmonic. (i.e .. I for· the fundan~ental or predor\1inant sinusoidal:fomponent. 2 for senind harmonic,~~ for third harmonic and so on) -~ vm = rnaximu\11 value of the ac component ofthe volt;ige, subscript indicates · . the degree pfthe harrnonic

RL 2 - LIC

t;= - - - ?.rr{Lf.

Rc 2 -L/C

· Quality Factor· Q

The qu;\lity f~1ctoi· of cofls, capacitors and circuits is defined by m~ximum

Q

stored energy

= 27t energy dissipaied per cycle

1. For Series RL,

Q=~=~= 2nfL R 2. For Series RC,

R

R

i .

Xc

I

.

I

Harmoni~s- components of th~ current or voltage in which

o=--=--=-. R ..

wCR

2rrfCR

3. For Series RLC. a{ resonance

Oo=

the frequencies are lll~ll~iples or the fundamental, Note: ' r ·~ fundament\tl fi·equency 2f= 2"" hanno'.nic fi·equcncy 3f= 3'" harmopic frequency

_[_,._~ _f,_

f2-f,

-

BW

where: fl ~nd

12 are the frequencies corresponding · to the half-power polnts.

BW =bandwidth }Vhich is the distance between half-power points. measured in hertz (HZ)

1\C Circuits Page 10 o( 10

MUL TIVECTOR REVIEW AND TRAINING CENTER _AC CIRCUITS

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PART I- SINGLE PHASE SYSTEM EXERCISES:

1. In an experiment, a sinusoidal wave form is observedto complete 8 cycles in 25 msec. Determine the frequency of the wave form. A. 320Hz ~ B. 40Hz C. 200Hz D. 64Hz REE - Sept. 2007 . . 2. Wavelength is the distance traveled by an electronic wave .during the time of one cycle. Given a ' wavelength of 12 meters, what is the frequency? A. 250 KHZ B. 25 KHZ C. 250 MHZ D. 25 MHZ '

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3. If emf in a circuit is given by e = 100 sin 628t, the maximum value of voltage and frequency is .

f\.

B. 100 V, 100Hz

100 V, 50 Hz '

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D. 50-/2 V, 100Hz

C. 50-/2 V, 50 Hz I

REE - April 2007 · · 4. What is the complex expression for a given alternating current, i A.227+j106' . B.160-j75 C.227-j106

~"'-

=250 sin(wt- 2

deg)? D.160+j75

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5. A sinusoidal voltage wave has an RMS value of 70.71 V and a frequency of 60 Hz. Determine the · value of the voltage 0.0014 second after the wave crosses the wt axis. A. 70.71 V B. 100 V C. 50 V D. 141.42 V REE - Sept. 201 0 1 6. In a single-phase circuit VA = 84.85 + j84.85 v and V 8 = 96.59- j25.88 v With respect to a reference · node 0. Calculate Vba· ' A.-11.74-j110.73v B.11.74+j110.73v C.11.74-j110.73v ; D.-11.74+j110.73v REE- April 2003 7. What is the rms value of a square wave with an amplitude of 10 A and frequency of 1 Hz? B.10A C.5A D.7.07A A.O REE - October 2000 · 8. A sinusoidal current wave has a maximum yalue of 20 A. What is the average value of one-half cycle? A 5 B. 12.7 C. 14.14 D. 0 REE- April1997. . , 9. A wire carries a current, i = 3 cos314t amperes. What is the average current over 6 seconds? B. 1.5A ·C. 3.0A D. 0.532A . A. OA

10. The average value ofthe function i A. 31.8 A B. 25 A

=50 sin 'wt + 30 sin 3wt is equal to __

.;

.io;f-:1_ _

C. 38.2 A

:; D. 51.43 A ii

r"

11. The rms value of a .half-wave rectified current is 100 A Its value for full ~ave rectification would be _ _ _ _ amperes. A.141.4A C. 200/rr A B. 200 A t D. 40/rr A I

12. A half wave rectified sine wave has an average value of 100 amp.'What is. the effective value? A.157A B.444A C.70.71A I D.100.A AC Circuits 1 Page 1 of 6

MULTlVECTOR REVIEW.· AND TRAINING CENTER . .

AC CIRCUITS

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13. The form factor of a half-wave rectified alternating current is A.1.11 s. 1.57 · e. 1.73

D. 1.0

currents are given by i1 = 141 sin (rot + 45° ); .14. Three alternating .

i2 = 30 sin (