AC Circuit Reviewer (Single Phase), Exams of Circuit Theory

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AC Circuits

Part 1

Single Phase (1 ) Syste~

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• \ ,.. i· ' ; .: MULTIVECTOR REVIEW AND TRAINING CitNTER . )

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ALTERNATING CURRENT (AC) CJRq.JJTS PAI<T 1

Single' Phase ( 1~}System (^) ,,

:!, If

.. I

'l/'.[' I

• ;

An alternating quantity either current or voltage is that which acts in alternate dirccti\m and whose magnitude undergoes a definite cycle of change in a definite interval of time.

Sine Wave/Sinusoidal Wave

Positive p¢ak ---- --------------------- --f

Negative peak

Peak to pe.ak -+-,----~~-----r-- value

---- --"------------- ____ :~ __l

·1·.·· .alternation 1-.-·~-

. o111111 I cycle _ · ' (360°).

·. Cycle· a complete change in value and direction of~m alternating quantity. A cycle of alternating voltage and current compktes in 360 electi'ical degrees. There are .two (2) altemations per cycle.

Frequency <0 .. the nu;nber of cycles per second expres:sed in heriz (HZ).

where:

f

'f..,.. freqtlency, h.ertz (HZ) or cps P --no. o( poles N -speed, rpm

Period (T} -- the time it takes to complete one cycle.

E

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Wave ~ leni!fh, (A.)--. the lengthI of one co.mplete wave or c)'cle or the distance traveled by the wave in one cycle.

Note:

velocity A.=-----=- frequency f

' '•

!. For e1Cctror'i1agnetic waves, the velocity (v) in air or vacuum is 186,000 niils or 3 x I o'" cm/s, which

is the speed of light. · \

2. For sound waves, the velocity (v) in air is I, 130 fi!s. ·

ACCircuits Page I ofiO

MULTI VECTOR REVIEW AND TRAINING ~NTER

. l'(l)

----/--------------·

i ____ "1' ______;__ --- iI^ ____^ il^ _____^ _ From the above figure.

where:

l'(t)"' Y 111 sin (!)t volts

l'(t)= instantaneous value ofvoltage, volts V"'"" niaximum value of voltage, volts (I)"' angular frequency, rad/sec t "' time, seconds : (tlt = radian · Ul = 2rrf

Similarly, if the above voltage wave is a current wave, then the equation is I i(t) "".1 111 sin OJ! Amp

where: i(t) = instantai1eous value of current, amperes · 1, 1 .-.= maximum value of curreht, amperes

' J

Note: mt whose unit is radia:n can be replaced by 0 expressed in degrees.

• I I'J(t)

~........._--- (^)! -S7--~~-.(~-I ~----. -· vit)

i ,· :

From the above figure,

where:

v 1 (t) = Ym sin cot volts vit) = Vm sin ( mt +G) volts v 3 (t) = V 111 sin (cot - 0) volts

+ e = leading or ahead

· - 0 = lagging or behind 0 = phase angle or angle of displacement

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MULTIVECTOR R,EVIEW A~D TRAINING cENTE{~

Instantaneous Value- the value of alternating quantity at any instant. I Maximum Value- the mnximum value nttnincd'by an alternating quantity during positive or negative half cycle. This is

also called the peak value, or amplitudcf·.~,f alternating quantity.. :

I

Average Value ihe nvcragc ofall the instaillanel)US values of half (either positive or nqgative) cycle of alternating

. quantity.^ I^ '^ I^ •^ :

For sini1soidal voltage wave, V•>< ~' 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

T V.ve== -^1 - f v(t)dt

T o

and for RMS of EffeCtive value

T Yrms = +J v 2 (t) dt 0

Forin. factor

Effective valu·e Average value

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

Maximum value Crest or peak factor = ------- Effective value

For sine wave. Crest or peak factor= 1.

. T

l•ve~ -+-:f i(t)dt ~ 0

I ·JT P(t):dt

T o

AC Circuits Page 3 of 10

.. ·

Pure Resistance. Power Tri.angle:

where:

V ~~ 1R

V · eflective or RMS voltage

''effective or RMS current R ~~ efTective or ac resistance

For sinusoidal Voltage supply,

1m ' I"' -::::-- ";' 0.707 (^1111)

'>/

: .. V 111 = I 111 H '. Phasor diagram:

I is in phase with V

I. Real or True or Active or Average Power

. P ~VI cos 8 watts VI p =~cos e watts 2 P = 12 Rwatts

'1 Reactive Power

I~:) ¥~-J_:_ 0 p ' I 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

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

Energy expended, W = Pt joules or watt-sec

Pure Inductance

v L

where: XL = inductive reactance in ohms XL= wL =" 2nfl~ n f= frequency, Hz i L := inductance. henry (I I)

For simisoidal voltage supply.

Phasor diagrnm:

1m I v

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

-+-r--------~Y ·Yrnlm. O Q =-- sm · vars 2

3. Apparent Power

S =VI volt-ampere (va) Vmlm

s·=-.""'

2

va

I lags V by 90'?

i.

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MULTIVECTORREVIEW AND TRAINING CENTER

.. For Pure L, since 0"' 90° P = VI cos q(jo "' 0 Q '"· VI sin <}Oo "' VI

s .·Vi

.'. s '()

Energy expetided.

Pure Capacitance

--~I.

joules.

t. I ~.. T c

~~-__j

V" IXc where:

1,/L

Xc "" citpacitive reactance in ohms I 1 · Xc=--=·-- (J)(.. 2nfC

C, ~~ capacitance,. farad (F)

For sinusoidal voltage supply,

'.

1111 J=- v

Phasor diagram: I

ote~9oo ~ v

I leads V by 90°

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

Energy expended,

joules

v"c

We= ___ joules We,.·_V,/C^ JOU..^ Ies 2 4

R and L in Series ·

v

f-

~

Note: I is constant x~. = ooL = 2ntl, n

Phaso~ diagram:

Vr~C

71

V•IZ

~~ >

where:

:VR = IR

v = -,JvR2 +·v~,

V =I -,JR 2. +X I.^2

V = IZ volts

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

Impedance Triangle:

·~x,

0 R!

P =VI (p.f.) :~ 12 R '"' V 1 /!R watts

Q =VI (r.f.) ~ 12 Xt. = V 12 /X1. vars

S = VI ·~ 12 Z = V^2 /Z va

Also,

AC Circuits. Page 5 or 10

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R and C in Series ----::;:.

i

1' VR

v

~.

f'.,.

l J^

Yc w

Phasor diagram:

~">

Yc,;,IXc V=IZ

where:

, (~ 2 i V = 'YVR + Vc- V ,;, I "H.? + Xc 2 V= IZ volts

·z=,JR2+Xc p.r. =cos e = vRIV = R/Z

r.f. =sin 8 = Vc!V = Xc!Z

tan 0 = Vc!VR = Xc!R

Impedance Triangle:

Also,

6 R

z

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

S =VI=' 12 Z = V^2 /Z va

RLC in Series

--

r~- (^) R

v f-

where:

Note:

Also,

L V,

x~_ =· (J)L = 2nfl, n I I Xc''~oc~ {

V"' ~VR 1 + (VL -Vd V =I ~R 2 +(X,- Xci V = IZ volts

Z·= (^) "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

Transpose XL and Xc if X(·-. X 1 • ' J P =VI (p.f.) = 12 R = V 1 //R watts

Q =VI (r.f.) = 12 (X~_ -Xc)

Q = (V1.^2 /XL)- (V/!Xc) S -=VI·"' 12 Z:= V 2 /Z va

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Rand Lin Parallel

·v ·.. R. · I.

1

·. r~fR J~~~~

~f.....

Note: V is colistant.

X1. = mL c ·:2nfL n.

IR ·.c VIR: I, ~ v;x,

Phasor dingraii1:

where:

Also,

V

lr = v /Z amp.

Z = impedance in ohms

z

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

. ') ., '

P =VI (p.f.) = IR-R = V'/R watts

Q= VI (r.f.) "''L 2 x~."" V^2 /X1. vars S = VI-r= 1 1, 2 Z '"' V 2 /Z va

Rand C in J>aralld · ~h

Xc=~--^ I^ = --'I

. wC 2nfC

I_R = V/R lc •= V/Xc

I

Phasor diagram: I( -----------~ ' ./~. : \

\'here:

Also,

P =VI (p.f.) c= 11 /R · V 2 /R watts

Q =VI (r.f.) = 1/ X(

Q = V 2 /Xc vars:

S"" VIr cc I/Z = V^2 /Z va

RLC in Parallel · ~ Ir

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

X( .. l_ _ we 2nfC

IR = V/R: l1. .. V/X 1 :It VtXc lr = ~l (IL - lc)' vx-; 2 x~ 2 :i:-rZ26<-~~x~)z-

IT = V/Z amperes

' I

AC Circuits Page 7 of 10

MULTitECTOR REVIEW AND TRAINING CENTER_

where:

Note:

RX 1 Xc z = -;::=;=::;:=::;::==:::;-

"x,2xc2·~- R^2 (Xc-Xd

p.f. = Ct)S 0- IR/1 1 ~ Z/R r.f=sin()c·(lr.~-lc)/Ir. I(Xc X,)^1 XrXc tan 0 =· {1 1 - lc)IIR 0 ' R (X, - X 1 )i XcX (^1)

Transpose 11 mid Ic if 11 • 1<. ·

) Impedance' in^ Com1llexI Form•

I. For PureR

z ~ R +iOn

z ,c I R L::oo n

, For Pure L z =, o 1 iX, n z-= I x; ·I L90^6 n

3. For Pure C z =0 ~jXcn z = I x'c I L -90° n
4. For Series RL Z = R + jX~,n

z"" ~R 2 ' x?L.+O

z = I z I L+8 n

where: 0° < 8 < 90°

5 .. For Series RC

Z = ~-" i_XcQ__ ·

· · Z = R^2 +X/ L- 8

z =: Iz I L-On where: oo f': 0 < 90°

6. ·For Series RLC : Z = R + j (Xr.- Xc) Q

Z = ~R 2 + (Xr. -Xc} L. ± 0

Z· lziL±Ori Note: r 0, if Xr. > Xc -0, ifXc > Xr.

In general, Z =. R ±.jX

where + jX =inductive reactance

• jX = capacitive reactance

*For the complex expression of an impedance,

its angle ranges from oo to 90° only either

positive or negative. Mathematically, 0° :50$ 90°.

lmpedam~cs in Scl'ics

1 I lmpedan£eS in Pnrnllcl · 1~

• --- ·! ~ ---- /1 Zn For two (2).impedances in parallel. _ z,z (^2) z, ~-~,-/- --1 --; Note : In the above equations, iinpedances must be in complex limn.

Admittance (Y) -- is the rccipr6cal of impedance/.

Y .._, --^ I mho or sicmen z

Conductance({;) -- is the reciprocal I or resistance R.

G = J mho or sicmen · R

Susceptance (H)- is the recipr6cal of reactance X.

B •= - I^ i:^ mI 10 or.^ sremen. x. '

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

In complex form,

where

Admittances in Series

y .=.p ±jB

' .ti ~

+ j~ = ~apaci~ive susceptance B.-

• (^) .J B,I= mductlve susce11tance (^13 ) 1i ![ ,'{ --^.^ I^ + ---I I ...... -t Yl Y, Y2 Y1 Yn

For two (2) admrtt~nces in series.

· Y,Y, Yir ~ ---· --- Y, + y Admittances in Parallel

Y 1 =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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Power in Complex Form

I. Voltage Conjugate Method

where:

Note:

S c I x Conjugate of V with

• resp~ct to the horizonial axis S · .JV '·-P±~ .. s "' ~ + o· /.. J () va . S - :app·arcnt power. va P -, real or true or average; power. watts Q ·^7 reactive power. vars I c- .current, amperes V ., voltage. ~'(>Its :

V " conjugate of the voltage, 'olts

· 0 =· power factor angle

+j = Ois cap<:~citive (p f. is le~ding) -j =Q is inductive (p.f. is lagging)

2. Current Conjugate Mothod. J S "' V x Conjugate of I \~.

where: l-

Note:

. rppect to the horizontal axis S ,. VI "'"· P'TlQ __

S ·~ @+()' / ' 0 va

• I = cor~jugatc of the current. amperes

-j = Q is capacitive (p.C is leading)

lj ~ Q is inductive (p.f. is lagging)

H.esona nee.. _1_ circuit is said to' be in resonance when the npJ{Iicd voltage V and the resttlting current I arc in phase. Thus at resonance, the equivalent complex impedance of the circuit consists of only resi~tance .R. Since V and I arc in phase. the power factor of a resonant circuit is unity. · ; I

Series Resonance (For RLC in. Series) At resonance;

Parallel Resonnnce/Antiresonance (For RLC in ParaiJ(.';J)

I. v~.~~ Vc

2. Xr. ·=Xc
3. L=-·-·~1- w}C
4. C=~L (l)o- ·

I

S. w_" 7" ~LC.

·.·where W 0 =angular frequency at

. resonance, racl/sec

(J. (^) f,.;-~-I

1. 2rcvLC where ~ "' resonant frequency; hertz (HZ)

2. Supply ~oltage, V c=VR

3. Z == R ~ a pure resistance .·. Z is minimum.

4. Total reactance, Xr "'' 0

I 0. I = V /Z = V /R

. .' .. r is maximum.

II. p.f. =I or unity

12. S=P=VI.
    13. Q=O

,.

1\t resonance, I. lr. = lc

2. X~_=; Xc
   3. L=--,-^ I n.>o-c

4. c =---,-^ I

ro 0 "L

I

5. COo = r:-;::;- 1 . 'LC

I ';

6. r, = __, 2rc..JLC
7. Supply volt~\ge, V =VR
8. Z =R ~ a pure resistance :. Z ib maximUI!.).
9. Total reac·tahce, X- 1 ~ 0 l
10. 11 = IR = V(Z= V/R :. 11 is minimur_!}. II. p. f. ~- I or unity
11. S~cP=VI
12. Q ~o

.; ·'j' i~ tl

1\C Circuits Page 9 of',IO

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Parallel Resonance, Two-Bratich Circuit

At resonance.

_, ill

R 12 - LIC

Rc2- L/C

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

RL^2 - LIC

Rc 2 -L/C

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

Q = 2 7t energy dissipaied per cycle

1. For Series RL,

Q=~=~= 2nfL

R R R

2. For Series RC, Xc I. I o=--=--=--

. R .. wCR 2rrfCR

3. For Series RLC. a{ resonance

Oo= [,._~-^ f,

f2-f, BW where:

' J

(I)(}~· (til(!)-;

w,·c 2nf, ··-·---~J... ; (I)~. 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.

I Effective Value of a Nonsinusoidal Current or Voltage Wave

I=

,.-------'.---------·--,..----

I^1 1111 22 1m/

I (^) de^2 + ~-1·^ --,^ ·t---^ I 2 2 I 2

' (^) ~~~~~·' t --··- ,

---~:;-- .. f ~ 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 i.

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

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

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 e · · _AC CIRCUITS. ~~~

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

' '

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

f. 100 V, 50 Hz B. 100 V, 100Hz C. 50-/2 V, 50 Hz D. 50-/2 V, 100Hz

'.^ I

REE - April 2007 · · ~"'-

4. What is the complex expression for a given alternating current, i = 250 sin(wt- 2 deg)?

A.227+j106'. B.160-j75 C.227-j106. D.160+j

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?

A.O B.10A C.5A D.7.07A

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?

. A. OA B. 1.5A ·C. 3.0A D. 0.532A

.;

10. The average value ofthe function i = 50 sin 'wt + 30 sin 3wt is equal to .io;f-:1

A. 31.8 A B. 25 A 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 B. 200 A C. 200/rr 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 p

.. AC CIRCUITS ·. · ~)

13. The form factor of a half-wave rectified alternating current is A.1.11 s. 1.57 · e. 1.73 D. 1.

.14. Three alternating.^ currents are given by i^1 =^^141 sin^ (rot^ +^ 45°^ );^ i^2 =^30 sin^ (^ <tlt^ +goo);^ i^3 =^20 cos^ (^ o1t^ -120°) · Find the equation of resultant current. A 167.4 sin (rot +45.66°) B. 74.6 sin wt C. 143.8 sin (wt+51.4°) D. 64.7 sin (rot- 30°)·

REE - Sept. 2006 · ,

15. When the sole purpose of ac is to produce heat, the selection of conducto~ is based on ___ value of the current. · A. Average. B. Instantaneous C. RMS D. Peak . 16. The maximum value of a sine wave AC voltage which ~uce he~t in a resistor at the same average rate as· 115 V of direct current is ___ _

A. 81.3 V B. 115 V C. 162.6 V l D. 230 V

i

17. A sinusoidal voltage source has a peak value of 150 volts. What equivalent DC voltage source would

produce the same heating effect in a 1 ohm resistor? i

A.15V B.212V: C.95V , D.106V

· 18. The effective value of v( t) = 100 ·t A, sin (l)t is known to be 1Q3.1. The amplitude A of the sine term is

A. 25 B.4.85 C. 35.48 D. 100

19. An alternating curreht and a direct current flows simultaneously in the sa1'.~e conductor. If the effective

value of the AC is 8 A and DC is 12 A, wha't will an AC ammeter read when 1 connected in. the circuit? A. 14.42 A B. 12 A C. 11.66 A i~ D. 16.49 A l:l ,I

20. An impedance draws a current i = 10 cos ((J)t- 30°) A from a v = 220 sin OJt vd!ts.^ '^ What is' the power? A. 550W ·B.1100W C.190.5W , D.1320W

REE - October 2000

21. A series circuit has an applied voltage of V = 220 sin( OJt + 30°) and draws. a current h 10 sin( wt - 30°). , ' I What is the average power and powedactor of the circuit? A. 1,905 W, 86.6% lagging B. 1,905 W, 86.6% lagging G. 2,200 W, 100% D. 1,100 W, 50% lagging ~
22. The effective voltage across a circuit element is· (20 + j10) V arid the effective current through the element is 4 - j3 A. Calculate the true and the reactive power taken by the element.

A. 50 watts & 1OOVars lagging B.,20 watts & 100 Vars leading

C. 110 watts & 20 Vars lagging D. 110 watts & 20 Vars leading

23. A 11C)-volt AC line feeds two circuits in parallel. The currents are (2.2 - j6.5) A and (1.8 - j3.5) A. The power consumed by the two circuits is approximately---'---- A. 1,185 watts B. 440 watts C. 755 watts D. 433 watts

, 24. Find .the average power in a resistance R = 10 ohms if the current in series form is i = 10 sin wt + 5 sin :. 3wt + 2 sin 5wt amperes.

A. 65.4 watts B. 645 watts C. 546 watts D. 5.46 watts

AC Circuits 1 Page 2 of 6

MUL T. IV~CTOR REVIEW AND TRAINING CENTER f.?tp ·. ·. ·. AC CIRCUITS i, • ~

...

. REE - May 2009 · ·.. '..

2S. The current thru an inductor with inductance of 1 mH is given as i(t) =' d.01 0 sin 10

6 t A. What is the .voltage across this inductor? ·.

A. tOO sin 106 tv. B. 10cos 106 tv. C. 10 sin 106 tv ~\ D. 100 cos 10

6 tv

REE - Sept. 2007 ·. ·... 1 ·· · ·. · ·. '

26. The capacitive reactance Xc of a clrcult varies inversely as the capacitance C of the circuit: If the

capacitance of a certain circuit is decrease.d by 2S%, by what percentage will Xc change?

A. 2S% increase

I • 2

B. 20% increase C. 66-% increase

· D. 33-% increase

REE - April 2005

27. A circuit has resistance of 20 0 and a reactance of 30 0. What is the powedactor of the circuit? ·

. A. 0 ' B. O.SS C. 0.832 , D. 0.

REE - Sept. 2006. i.

28. A two-element series circuit with R =1S ohms, L = 20 mH has an impedance of 30 ohms and an unknown angle. What is the frequency in .. HZ?

A. 2.0 ·. B. 21 C. 2,067 i D. 207

. REE -Sept. 2008 .·. ·:. , i ..

29.. rhe current in a series circuit of R = 10 'ohms and L = 60 mH lags the ap~lied. I voltage by 80 degrees. · What is the impedance in ohms?. : · · ··

A 10 + jO.S671 n B. 10 + j56.71 n c. 10 + jS.671 n D. 10 + jS67.1 n

REE - April 2007

30. What are the tWo elements in a series circuit connection having a current and voltage of · i ~ 13.42 sin( SOOt - 53.4 · )A and v =· 1SO sin( SOOt + 1 o· )v? · : A. R =8.8 nand L =3S mH ' B. R =6.3 nand L =2S m~- c. R =7.S nand L =3o·mH D. R= s nand L = 20 mH ·

REE...;. Sept. 2009 · ·. I • ·.::

31. A two-element series :circuit has voltage V•= 240 + jO v and current I =48!- j36 A. Find the current in

amperes w~ich results when the resist.ance .is reduced to so.% of its former ralue...

A. 46.20 + j69.20 B. -46.20 + J69.20 C. 46.20- J69.20 I' D. -46.20- j69.20 · •;

REE - Sept. 2001... '·

32. In laboratory. experiment, the impedance of a coil was obtained at .60 Hz afnd 30 Hz. These are 7S. ohms and S7.44 ohms, respectively. What is the inductance of the coil? j

A. 1SO mH B. 182.S mH C. 42.S mH , · D. 2.1 mH

REE - May 2009 ;

33. A 200 mH inductor and an 80 ohm resistpr are connected in parallel across :a 100 v rms, 60 HZ source. By what angle does the total current lead or Jag behind the voltage? 1

A. 46.70° leading B. 43.40° leading C. 43.40° lagging. D. 46.70° lagging

REE - Sept. 2004 ' ..

34. A capacitor in series w/ a 200 0 resistor draws a current of 0.3 ampere from 120 volts, 60Hz source.

What is the value of capacitor in microfarad? A. 8.7 · B. 9.7. C. 6.7 D. 7.

AC Circuits 1 Page 3 of 6

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MULTIV~CTOR~'VtEWAND TRAlNINY ~ENTER e

.. · AC CIRCUITS , ~ ,!^. ,. iI

· 35. A series circuit consisting of a 66.2 !Jf. capacitor and a variable resistor. For what two values of;.

resistance wili the power taken by the circuit be 172.8 watts, if the impressed 60-cycle emf is 120

·W~? ,

A. 83.33 & 3.33 ohms B. 53.33 &:3o ohms c. 5.33 & 3.0 ohms ; D. 83.33 & 5.33 ohms

. REt - April 2004 · 36. Reactances are connected in series: 'Xc 1 = 100, Xc 2 = 40, XL1 =30, Xi 2 =70. What is the net

reactance?

A. 80 Xc

REE - Sept. 2005

B. 40 Xc

37. Which is not 180 degrees apart?

A. XL and lc. B. Xc and I~

c. 60 XL

D. Xc and XL

i REE - April 2007. · :. · 38. A series RLC circ'uit has elements R = 50 0, L = 8 mH and C = 2.22 microfarads. What is the equivalent impedance of the circuit it'the frequency is 796 HZ? · A.50+j500 B.50+j130.0 C.50-j500 : D.50-j

REE- May 2008.

39. A coil of inductance 0.1 H and resistance of 4 ohms is connected in series with a condenser of 70.4.

microfarad capacit~nce across a 60 Hz, 240 V line. How much current will flow on the circuit?

A. 55 A. B. 60 A C. '57.5 A D. 52.5 A

40. An impedance of 100 n resistance and an unknown inductance is connected across the capacitor. The · resulting impedance is a pure resistance of 500 Q if ro = 105 rad /sec. Calculate the values of inductor and capacitor. ·

A. 1 J,Jf & 2 mH B. 5 J,Jf & 1 mH C. 7 J,Jf & 3 mH D. 0.04 J,Jf & 2 mH

;. REE- May 2009. '.41. Find the value of an impedance which absorbs a complex power of 2300 VA with a 30 degrees angle at 115 V rms. · · A. 5 + j7 n B. 7 + j5 o c. 5 + j3 n D. 3 + j5 n

REE - April 2007'

42. Find the value of an impedance which requires -1 ,540 vars at 230 v rms and 10 A rms. A. 17.40- j15.00 B. 16.82- j.15.69 C. 17.09- j15.3.9 D. 17.62- j14.

REE - Sept. 2006

· 43. A 10 ohm resistor is connected in parallel to an impedance Z = 3 + j4 ohms. Find the circuit power

factor. ' ·. A. 0.600 lagging ..J B. 0.86.6 lagging C. 0.609 lagging D. 0.809 lagging

44. The maximum values of alternating voltage and current are 40o·v and 20 A, respectively, in a cir<;!Jit connected to. 50 Hz supply and these quantities are sinusoidal.· The instantaneous values of voltqge and current are 283 V and 10 A respectively at t = 0 both ihcrea.sing positively. What is the po#er: factor of the circuit? · · A. 0.707 B. 0.

'• '

C. 0.

J J D. 0.

AC Circuits 1 J!lage4 of 6

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MULTIVECTOR REvil!:W AND TRAININd \cENTER e · ··. AC CIRCUITS · ~ ~

REE - October 1996

45. The resistor of 6 nand unknown impedance coil in series draw 12 A from a 120 V, 60Hz line. If the

real power taken frbm the line is 1152 watts, what is the coil inductance? · ·

A15.9mH B.10mH. C.20mH D.1.59mH

REE -·May.

46. A coil and resistor each have a 160 v voltage across them when connected in series to 240 v, 60 HZ·

source. If they draw a 1 A current from the source, what is the resistance of the coil?

A 25 ohms · B. 10 ohms C.·15 ohms D. 20 ohms

:~1. A feeder. supplies two .l~ads, one. at .50 amperes at .50% 9p wer factor, the. other 150 amperes at unity

' power factor.. The total current supplied by the feeder is ap roximately..

..! A180A •. B. 200A C. 175A ' · D. I 150A •

48. A single phase, 7.46 kW motor is supplied from a 400 V, 50 Hz AC mains. If its efficiency is 85% and

the power factor is0.8 lagging, find the r&active component of the input current.

A 16.46A. B. 21.96A C. 27.43A D. 21 A

REE - October 1996

49. Acapacitor is rated 100 KVAR, 380 V, 50 Hz. What will its rating be at 60 Hz, 220V?

A 50 KVAR R 40 KVAR C. 90.9. KVAR D. 57.7 KVAR

50. A fluorescent lamp and iis inductive ballast 1 draw a 1.0 A current at 50% lagging power factor from a

120 V, 60 Hz source~ What is the overall power factor when a Z6.5 IJf capacitor is connected across

the fixture? : ·

A 0.832 lagging B.0.832 leading ·I C. 0.5 leading : D. 0.5 lagging

51. A 3 HP, 120 V, 60Hz' induction motor operating at 80% efficiency and 0.,896 lagging power factor is to

be used temporarily with 240 V, 60 Hz source. What resistance in series with the motor will be required

for the motor to .have 120 V across its·terminals at full load?

A 6,68 0 B. 4.77 0 C. 13 .. 76 0 D. 9.54 0

REE - Sept. 2009...

52. A single-phase motor fs taking 30 A, from a 220 v, 60 Hz supply, the power factor being 80% lagging.

What value of capacitor in microfarad connected across the circuit will be necessary to raise the power

factor to unity?. · I • · •

A.214 B.220 C.217 D.

REE ....: October 2000

53. A series circuit consists of a 20-ohm resistance,. a 150 mH inductance and an unknown capacitance.

The circuit is supplied with a voltage v ~ 100 sin 377t. Find the value of capa

1

citance at resonance. ·

A. 42 IJF B. 47 IJF ' ' C. 34.65' IJF' ' 'I'I D. 72.57 IJF

54. A 50 0 resistance; 30 n inductive reactance and 25 n capacitive reactance are connected in',series

across a 100 V, 60Hz supply. What will be its resonant frequency?

A 65.726 Hz B. 53 Hz C. 25Hz. D. 54.77 Hz

REE - April 2005.

55. A series RLC circuit has R = 10 n, L = 40 ·x 10-e Hand C = 60 x 10-^12 F:· What is the resonant of the

circuit in MHZ?

A 20.17. B. 3.24 C. 4.49 I D. 1.

AC drcuits 1 Page 5 of 6

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MULTIVECTOR REVI~W ~ND TAAININJ CENTER~

· AC CIRCUITS ' :! ~

REE -April 2001 t

56. A 5 mH pure inductance is connected in parallel with one microfarad capC!Citor. What frequency will the

circuit be in antiresonance? ' '

A 250Hz B. 2250Hz C. 60Hz. D. 100Hz

REE - Sept. 2010

57. A series RLC circuit resonates at 1,000 rad/sec. If C = 30 microfarads and it is known that the

impedance at resonance is 2.4 ohms, compute the value of Q of the circuit. ·

·. A 13.9.. B. 10.4 · C. 20.~. D. 41.

58. A coil having a Q-factor.of 5 is connected in series with n ideal capacitor across an ac source of 60 V.

Calculate the voltage across the capacitor at resonanc.

A 150 V. B. 300 V C. 12 V D. 65 V

,_ '

REE - April 2004

, 59. A tuned circuit is resonant at 4 MHz. Its Q is 100. What is its bandwidth?

. A 4 kHz B. 400 kHz C. 400 Hz. , D. 40kHz

REE - Sept. 2007.

60. A circuit has load impedance Z = 50 +j80 0. Determine the parallel impedanc;:e required to correct the

power factor to unity....

A 32 JJF B. 36 JJF C. 24 JJF D. 28 JJF

61. A circuit consisting of aGapacitor in serie.s with a resistance of 10 ohms is connected in parallel with a coil having a reactance; and resistance qf 17.32 ohms and 10 ohms, respectively. What is the

reactance of the capacitor that will draw minimum current from a 230 V, 60 Hz supply?

A 17.32 n · B. 10.32 n c. 5.78 n D. 22.1a n

REE - April 2007..

62. An impedance Z3 is in series to the.two impedances z, = 10· + j10 n and Z2 = 20- j30 n that are

connected paralleL What is the value of the impedance Z 3 in ohms that will'.produce resonant circuit?

Aj13.1. B.-j5.4 C.j5.4 D.-j13.

AC Circuits 1 Page 6 of 6

MUL TIVECTOR REVIEW AND TRAINING CENTER

AC CIRCUITS --PART I

SINGLE PHASE SYSTEM..- SUPPLEMENTARY PROBLEMS

e

. ...- '

1. A voltage wave is given by v = 100 sin314t. How long does it take this wave form to complete one·

,_ fourth cycle? ·

' A. 20 ms B. 10 ms C. 5 ms D. 1 ms

1 2. When a 15 V square wave is connected across a 50-volt AC voltmeter, it will read __ _

A..21.21 V :· B.10.61 V C.15V ' D. 9.55V

3. Calculate the effective value of v(t) = 100 sin400t + 50 sin800St 10 cos 1200t V.

A. 79.5V.. · B. 57.9V C.112~-~. D.121.52V

4. A circuit draws a current of (3 - j8) A from a source of (1 00 + j37) V. Find the true power of the

circuit.

A. 4 W ~.B. 596 W C. 300 W D. 296 W

•'

5. Across a 230 V, 60 Hz·power supply is a 15 ohm non-inductive resistor. What is the equation of the

voltage and resulting current?

A. e = 398.4 sin60t, i = 21.6 sin60t

C. e = 230 sin377t, i =15.3 sin377t

B. e =325.5 sin377t, i =21.6 sin377t D. e = 230 sin1 ~Wt, i = 15.3 sin120t ·

6. A resistor and a coil are conne~ted in series with a voltage source. If the voltage across the coil is

10sin(866t + 70°) V and the current flowing thru the resistor is 2cos(866t - 80°) A; what is the

resistance.of the coil? ..

A. 4.92 0 B. 2.5 0 C .. 50 : D. 4.33 0

7. Ten impedances' · connected in parallel draws the following individual curr~nt;

5.L0°, 5.L5°, 5.L1 0°, 5L15°, 5.L20°, 5.L25°, 5£30°, 5L35°, 5L40°, 5.L45°. What is the effective value

of the total current? (shortcut solution will be discussed on refresher)

A. 48.444 A B. 34.255 A C. 25.345 A D. 84.389 A

8 .. Using Problem 7, what is the equivalent impedance that could replace the iiT)pedances if the source

voltage is 100 sin 150t V? '

A. 1.325L- 30° h B. 6.026.L- 2.5° 0 C. 32.51.L50° 0 ,I D. 1.46.L- 22.5° 0 !

. 9. Using Problem 7, what is the equivalent power factor of the circuit?

A. 0.924 B. 0.866 C. 0.707 D. 0.

10. Using Problem 7, .what element should be connected across the circuit so that the current would be

in phase with the source? . 'I.

A. 54 mH B. 25.4 mH C. 13 mH . :D.I 31 mH

11. An impedance has a lagging power:factorand takes 250 W when connect~d across ·100 V, 60Hz

source. What power will it absorb when connected across 100 V, 50 Hz source?

A. 325W B. 352W , C. 322W D. 235W

12. A t~lephone circuit makes power available at a pair of terminals. The open c!rcuit voltage across the

terminals is 1 volt and the impedance looking into the terminals is 500 ~ j500 0. What is the

maximum power that can be drawn from ttie circuit?

A. 0.002 W B. 0.0005 W C. 0.001 W D. 0.0014 W

AC Circuits 1 Page 1 of 7

MULTIVECTOR REVIEW AND TRAINING CENTER

. AC CIRC,UITS -PART I :.

13. The ohmic resistance of a large magnetic contactor is measured to be 20 0. A 230 Vis impressed

on the contactor and the current is taken as 3.2 A. Neglecting core loss, determine the inductance

of the contactor..

A. 261 mH. B. 183 mH C. 315 mH · D. 251 mH

14. A coil has a resistance of .6 oh~s and an. induct~nce of 0.02 H. When a hon-inductive resistor is·

connected in series with the coil, the current drawn when connected to a 220 V DC source is equal

, to the current c;lrawn by the coil alone across a 220 V, 60 Hz· source. Determine the resistance of

· the non-inductive resistor. ·

A. 3.63 n. B: 6.39 n · c. 3.69 n ~ D. 3.96 o

15.Aseries RL circuit,has L. ~ 0.02 Hand an impedance of 17.85 n. When~ sinusoidal voltage is

applied, the current le~gs the voltage by 63.5°; What is the v lue of the angular frequency?.

A. 400 rad/sec. · R 500 rad/sec C. 600 rad/sec l D. 800 rad/sec

REE - April 1997

16. A current of 2.5 A ·flows through a 'series circuit consisting of a 100 n re!:jistor and an unknown

capacitor across a source 0f 460 V, 50 Hz. What is the value of the capacitive reactance?

A. Xc = 91.86 0. (^) ' B.. Xc (^) ..= 39.19 (^0) ' C.. Xc = 154.45 0 D. Xc = 184.0 0 ·I I

. REE- Apri11996 ·. i ·. ·

· 17. A circuit consists of a 4 0 resistor and 300 :I-IF capacitor in series. It is cohnected across a 60 Hz

voltage source with a SOOV peak voltage. What is the phasor form of the cur~ent?

A. 1=19.7L57.5° B, 1=8.84L73.2° ·.· C. 1=36.5L65.7° :D.I=10.5L65.7°

..

18. A small single phase·, 240 V induction motor is tested in parallel with 160 n resistor. The motor

takes 2 amperes and the total current Is 3 amperes. What is the power of the: whole circuit?

A. 800 W ·.s: 360 W. ·· C. 220 W : D. 580 W !

19. .A capacitor is placed in parallel with two inductive loads, one of 20 A at 30o, 1 lagging and another of

40 A at 60° lagging. What current in amperes should flow in the capacito.~ so that the circuit will

· have a unity power factpr? · · 1.

A. 35.8 A B. 44.6 A C. 28.8 A · !I D. 50.2 A

j! '·

20. A single-phase AC generator is supplying power to a load of 3,200 W at 230 V and a power factor

of 60%. If the total loss is 10% of the load, what would be the savi.~gs in wat~s if the pow~r factor is

raised to 80%? · · ·. ,

A. .64 B. 240 C. 427 : i D. HO

I

21. When a 240 V, 50 Hz supply is applied to a parallel combination of 15 0 resistor and inductor, the

total current is 22.1 amperes. What must be the value of the frequency f~r a total current of 34

amperes?

A. 25Hz B. 60Hz ' C. 75Hz 'D. 100Hz

22. A coil of 10 0 resistance and 0.1 H inductance is connected in parallel with a capacitor of unknown

capacitance. If the total impedance of the combination is 100 0, determine the value of the

capacitance.

A. 50 IJF B. 100 IJF C. 150 IJF D. 200 IJF

AC Circuits 1 Page 2 of 7

. MULTIVECTOR REV.IEW AND TRAINING CENTER

AC CIRCUITS -PART I

REE - March 1998

23. A coil of 50 0 resistance and 150 mH, inductance is connected in parallel with a 50 ~F capacitor.

. The source voltage is 100 sin( w t + 30°) V. What is the equation of the line current?

A 1.91 sin( w t + 52.5°) ". B. 1.25 sin( w t + 75.5°)

C. 1.25 sin( cot- 62°) ~, D. 1.32 sin( (I) t- 75.5°)

24. An impedance equal to 4.4L60° 0 is connected across a 220 v source. What should be the value''

of the second impedance in parallel with the first, if the total pbwer delivered to the circuit is to be

. 16.5 kW and the overall power factor is to be·unity? ·

A 2.21L:30.1 ° 0. ·B. 3.33L:- 40.9° Q ' c. 5.63L30° 0 ' I D. 6.5'3L- 45° 0

2s: An induCtive reactance. of.B ohms is connected i {a;allel With a capacitive reactance of 18 ohrris.

Th1s comb1nat1on ·1s then connected m senes 7 !~~a~ vanable res1stance. For what value of

resistance will the power factor be 0.5? ·

A.8.3140 8.3.1380· C.i13.810 D.1.3810 ·

2,6: What size of c,ondenser must be placed across an inductance having a resistance of 10 ohms and

· reactance of 20 ohms to draw minimum current from the line when the combination is connected

across a 60-cycle line?(Assume a condenser of negligible resistance).

. A 20 1,1f B.106 J..Jf C. 10 J..Jf. D. 6.33 J..Jf , '.. I

27'. A circuit draws 25 A when connected across a source offrequency f1: Determine the current drawn

by the same circuit at resonance if t 1 is half the resonant frequency. ·. l

A 12.5 A B. 17.68 A C. 35.35 A l D. 50 A

28. A series RLC circuit is connected across a 120 V, 60 Hz source and draws a leading current of 5 A

. Determine the voltage across the capacitor at resonance if R =50 and L = 25 mH.

A 47.12 V. B. 164.5V C. 236.6 V • D. 422.6 V

29. The Q-factor of a coil is given by .,.

A the ratio of its maximum energy stored to' its energy dissipated per cycle

B. its power factor

C. the reciprocal of its power factor ' :

D. the ratio of its re.sistance to its indUctive reactance

~0, A coil takes apparent power and reac~ive power of 100 VA and 80 VAR, respectively. What is the Q

factor of the coil? ·

A1.33 B.10 C.8 D.

' ~i

31. A coil is to be wound with Q-factor of 8. A lamp rated 120 V, 480 W is con~ected in series with the

coil and connected across 230 V, 60 Hz source. What is the impedance ~f the coil if the voltage

across the lamp is maintained at 120 V. : · · ' ii

A. 35.2L82.9° 0 B. 27.5L:72.4° 0 C. 45.5.C82.9o n !: D. 40.5L72.4° n

32. An inductive coil having a resistance of 25 .ohms and inductance of 0.2 H is connected in parallel

with a 100 J..JF capacitor. Find the frequency at which the total· current taken is in phase with the

·.supply voltage. · · · · [ ·

. A 35.6 Hz B. 46.5 Hz C. 29.5 Hz 1 D. 52.9 Hz

AC Circuits 1 Page 3 of 7

MUL TIVECTOR. REVIEW AND TRAINING CENTER , AC CIRCUITS -PART I

33. The series circuit of R = 3 0 and X = 4' ~ and a parallel circuit of R' !3rid X' have the sar:ne

impedance and power factor. Calculate the yal~and X'.. '

A. 8.33 o. 6.25 n B. 2.2s o. 3.83 n c. 7.47 o. 7.51 n D. 5.62 o, 9.84 o

34. A parallel circuit with one branch of R = 5 0. and a single unknown element in the other branch has the following applied.voltage and the current e = 10cos(50t + 60')V and I= 5.38cos(50t- 8.23')A. The unknown element is , ·

A. L = 0.04 H B. L = 0.02 H C. C = 10 IJF ; D. C = 5 IJF

. 35. The voltage across the resistor, inductor and capacitor connected in seri~s is 60 V, 90 V, 10 V respectively. What is the voltage across this circuit? 'i A.160V B.140V C.100V (^) ~ ·;D.·,: 50V ,I REE - April 1997. j

36. Determine the power factor angle e in the series circuitwhich consist of R =^1 :25 0, L =0.2 H, across a power supply of 200 V, 30HZ. ' A. 36.4' B. 46.4' C. 52.4' ; I D. 56.4'

'

REE- March 1997

37. A load of 20 + j35· 0 is connected. across ·a 220 V source. Determine the power factor and the VARS .. A. 49.6%, 1042 VARS: C. 52.2%, 1023 VARS

B. 85.3%, 975 VARS

D. 42.3%, 1087 VARS

38. An incandescent lamp load, generally considered to be made up of resistor; take 4.8 kW from a 120-volt source. The instantaneous maximum value of power is ___ _ A 4,800 watts B. 9,600 watts C. 2,400 watts D. 480 watts
39. A factory takes 50 kW at 0.75 lagging power factor from a 480 V source. In order to raise the power factor to 0.9 lagging, a synchronous motor rated 10 HP with an efficiency of 87.5% is used. Find the power factor of the motor.

A. 0.416 lagging B. 0.825 lagging · C. 0.4761eading D. 0.825 leading

. REE - April 2004

40. The wavelength of international distress of frequency is 600 m. What is the equivalent frequency in

kHz? ·

A.200 B. 300 C.400 D. 500

41. The voltage across a given circuit is 75 ;+ j50 V. What is the power supplied to the circuit if the

current through it is (8 -j5) A?

A 850 W B. 550 W C.750 W D. 350 W

REE - Sept. 2002

42. A resistor R and capacitor C are connected in series across a 100 V, 60 cycle source. The reading of an ammeter connected in the circuit is 2 A and the reading of a voltmeter connected across the capacitor is 80 V. Calculate the values of R qnd C. A. 66 0 & 30 IJf B. 30 0 & 60 !Jf C. 30 0 & 66 !Jf D. 36 a· & 60 IJf

' J AC Circuits 1 Page 4 ol7

MUL TIVECTOR REVIEW AND TRAINING CENTER

A,C CIRCUITS -PART I I.

REE-^.^ September 2003 b^...

43. The following are in series: R = 1, 00 0, L =100 J..IH and C = 20,000 pF.' The voltage across the · circuit is 100 V. What is. the total impe ance expressed in ohms if the frequency is 3 MHZ? A.18820 B.10000 C.21320 'D.18850

REE - October 1994

44. A capacitor branch having a ratio of X to R of 5 is paralleled with an impedance consisting of 4 0

· resistance and 3 0 inductive reactaflce. The power factor of the resulting circuit is 0.8 leading. Find

the size of the capacitor in J..lf if the frequency is 60 Hz. ·

. A. 778.92 J..lf. B. 923.31 J..lf C. 552.31 J..lf D. 392.13 J..lf

45. An impedance of 3 - j3 0 is connec'ted in parallel with 5 + j2 0. The voltmeter connected across 3 0 resistance measures 45 V. Calculate the total current of the circuit. A. 22.4A B. 41.3A C.13.4A i D. 7.91 A t(
46. The open circuit voltage of an alternator is .127 V ·and its internal impedan'~e is 1 OL:1 oo 0. Find the voltage across a load· of 30L:- 30° 0. · A. 100L:- 9.7° V B. 97 L: -10° V C. 79L10° V D. 100L: --7.9° V

REE .:... October 1998

47. The maximum instantaneows voltage and current output of an alternatdr are 300 V and 20 A, respectively. What is the power out'put in watts if the. voltage leads the curr~nt by 30°? A. 2598 B. 5196 C. 6000 I ; D. 3000
48. The following data are given for a series RL and a· series RC which ar:e connected in parallel: XL = 15 0, .Xc = 25 0, Rc = 15 0. For what value of RL will the circuit be in r 1 esonance? A.169ohms B.916ohms C.16.9ohms I D.91.6ohms
49. A series circuit composed of 0.2 H inductor and 74 J..lf capacitor is connected to a 60 V variable frequency source, At what frequency will the current be 4 A with lagging power factor: A. 47.767 Hz B. 74.68 Hz C. 60Hz D. 50 Hz

· 50. A 30 ohm resistor is connected in parallel with an inductor of inductive reactance XL· The combination is then connected in series with a capacitor of. reactance Xc. ·what is the value of XL and Xc if the total impedance is 1.92 ohms? A. 7.84, 7.34 8.47.4, 47.3 C. 44.8, 84.2 D. 84.7. 34.7

REE - April 1997

51. The phase shift between the voltage and current vectors is due to the following loads except one. A. magnet coils B. electric flat iron · C. power capacitors D. fluorescent lamp I ~ 52. The length of time between a point in one cycle to the same point of the hext cycle of an AC wave is the ___ _ A. frequency B. period C. magnitude D. polarity
52. ·An alternating current varying sinusoidally with frequency of 50 Hz has an RMS value of 20 A. At what time measured from the positive maximum value will the instantaneous current be 14.14 A? A. 1/600 sec , B. 1/200 sec C. 1/300 sec D. 1/400 sec

AC Circuits Page 5 of

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MUL TIVECT~ REVIE·w AND TRAINING C~NTER

· AC CIRCUITS -PART I il

!; •,

REE - April 2001

54. An alternating rectangular wave has a maximum value of 10 V and a fr;equency of 1

second. What is the average value of the wave? ~

A 5 V. '8. 10 V C. 0 i D. 7.07 V

cycle per

55. Determine the rms value of a semi-circular current wave which has a maximum value of a.

A 0.816 a B. 0.237 a C. 0.866 a ; D. 0.707 a

56. Find the reading of an AC voltmeter connected across th'e series source of

100 sin (wt- n /2) and 100 sin (tlt.

A 100 B. 130. 65 C. 170.71 D. 184.78

REE - Sept. 2006

57. A two-element series circuit hasV = 240 + jO volts and current 51.96- j30 A. What is the current in

· amperes which results when .the resistance is reduced 50% of its former value?

A 68.57- j59.39 B. 68.57 + j59.39 C. 59.39- j68.57 D. 59.39 + j68.57

REE - Sept. 2002

,sa. A 10 0. inductive resistor is conneCted in series with an unknown capacitance. At 60 Hz the

, impedance of the circuit is 10 + j1 t 72. At 30 Hz the impedance of the circuit is 10- j5. What is the

value of L in millihenrys?

A. 50 B,500 C. 100 D.250

REE - March 1998

59. A 50 llf and 100 llf capacitors are connected in series and across a 1OOsin (wt + 30°) voltage. Write

the equation of the current.

A. 1.26 sin(wt + 120°) B. 1.26 sin(wt + 90°) C. 5.65-sin(wt + 120°) D.^ 5.65 sin((l)t^ +^ 90°)

60. Two impedances ZA = 4 ~ j6 0 and Z 9 = 6 + j 12 0 are connected in parallel. The apparent power for

the impedance B is 1490 VA Determine the total apparent power.

A 4250 VA "B. 3290 VA C. 2652 VA D. 8031 VA

61. Two coils A and Bare connected in series across a 240 V, 50 Hz supply. The resistance of A is 50

and the inductance of B is 0.015 H. If the input from the supply is 3 kW and 2 kVAR, find the

inductance of A and resistance of B.

A 0.0132 H & 8.3 0 B. 0.215 H & 3.8 0 C. 0.026 H & 12 0 D. 0.031 H & 5.3 0

62. A current of 5 A flows through a non-inductive resista.nce in series with a choking coil when supplied

at 250 V, 50 Hz. If the voltage across the resistance. is 120 V and across the coil is 200 V, calculate

the power absorbed by the coil in watts.

A. 168.75 W B. 137.5 W. c.:51.37 W D. 75.31 W

63. A series RLC circuit consists of 20 _ohms resistance, 0.2 H inductance and an unknown

.: capacitance. What is the value of the capacitance if the circuit has a leading angle of 45° at 60 Hz?

A 35.181JF B. 47.91JF C. 27.81JF D. 30.71JF

64._ If e = 100 sin {. wt + 30°)-. 50 cos 3wt + 25 sin (5wt + 150°) and i = 20 sin (wt + (^). 40°) + (^10) : sin (3wt -+- 30°)

-5 sin(5(1)t - 50°). Calculate the power in watts.

A1177 B.919 C.1043 :D.1224

AC Circuits 1 Page 6 of 7

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MULTIVECTOR REVI'W AND TRAINING CENTER

AC CIRCUITS -PART I

REE - March 1998

65. A 25 .0 resistor connected in series with a coil of 50 .0 resistance and 150 mi-l inductance. What is

the power factor of the circuit?

A 85% '. B. 80% C. 90% D. 75%

66. A fluorescent lamp taking 80 W at 0. 7 pf lagging from a 230 V 50 Hz supply is to be corrected to

unity power factor. What element should be placed in for this purpose?

A series, 59.4 JJF. B. parallel, 4.95 J.IF C. series, 31.2 IJF D. parallel, 8.36 J.IF

REE- .October 1998

67. One leg of a radio tuned:;circuit has a capa<;:itance of 1. x 10- 9 F. It is tuned at 200 kHz. What is the

inductance of t~e other leg in Henry?

A 6.33 x 10- 4 B. 20 x 10- 3 C. 8.25,x 10· 5 .D. 120 x 10- 3

REE - April 2004

68: A series RLC circuit has a resistance of 15 0 and an inductive reactance of 1,500 .0. What is the Q-

factor of the circuit at reso'nance? ' '

A 120 B. 100 C. 140 D. 150

AC Circuits 1 Page 7 of 7