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ELECTRICAL-REFERENCES

--&+.'!

REVISED 1990 EDITION

BY GEORGE V. HART

. .

A note from the author . UGLY'S ELECTRICAL REFERENCES is designed to be used as a quick on-the-job reference in the electrical industry. We have tried to include the most commonly required information in an easy-to-read format. Ugly's Electrical Reference is not intended to be a substitute for the National Electrical Code@. We salute the National Fire Protection Association for their dedication to protecting lives and property from fire and electrical hazards through sponsorship of the National Electrical Code.

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NATIONAL ELECTRICAL CODE' AND NEC? ARE REGISTERED TRADEMARKS OF THE NATIONAL FIRE PROTECTION ASSOCIATION, INC., OUNCY. M A

While the a u t h o r a n d publisher of UGLY'S E L E C T R I C A L R E F E R E N C E S have m a d e efforts t o insure that all information in this book is clear a n d accurate, neither a u t h o r nor publisher shall be held responsible for any inadvertent errors in content; n o r shall they be responsible for the interpretation o r application of material in this book.

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ISBN 0-9623229-1-1

,

-

UGLY'S ELECTRICAL REFERENCES

COPYRIGHT, 1978 BY GEORGE V. HART (AUTHOR) REVISED 1990 PRINTED IN U.S.A.

THIS BOOK MAY NOT BE REPRODUCED IN ANY FORM WITHOUT WRITTEN PERMISSION OF THE COPYRIGHT HOLDERS

GEORGE V. HART AND

SAMMIE HART

united printing arts

3509 Oak Forest Drive

.

Houston, Texas 77018. (713) 6884115

TABLE OF CONTENTS TITLE OHM'S LAW SERIES CIRCUITS PARALLEL CIRCUITS COMBINATION CIRCUITS ELECTRICAL FORMULAS TO FIND: AMPERES (I) HORSEPOWER (HP) WATTS (P) KILO-WATTS (KW) KILO-VOLT-AMPERES (KVA) CAPACITANCE (C), AND CAPACITORS INDUCTION (L) IMPEDANCE (Z) REACTANCE (INDUCTIVE-XL, AND CAPACITIVE-XC) RESISTOR COLOR CODE U.S. WEIGHTS AND MEASURES METRIC SYSTEM CONVERSION TABLES METALS AND SPECIFIC RESISTANCE (K) CENTIGRADE AND FAHRENHEIT THERMOMETER SCALES USEFUL MATH, FORMULAS THE CIRCLE FRACTIONS EQUATIONS SQUARE ROOT TRIGONOMETRY CONDUIT BENDING TAP, DRILL BIT, AND HOLE SAW TABLES MOTORS: RUNNING OVERLOAD UNITS BRANCH CIRCUIT PROTECTIVE DEVICES DIRECT CURRENT SINGLE-PHASE (A.C.) TWO-PHASE (A.C.) THREE-PHASE A.C. MOTORS TRANSFORMERS: CALCULATIONS VOLTAGE DROP CALCULATIONS SINGLE-PHASE CONNECTIONS BUCK AND BOOST CONNECTIONS FULL LOAD CURRENTS THREE-PHASE CONNECTIONS TWO-PHASE CONNECTIONS TWO-PHASE AND THREE-PHASE CONNECTIONS

PAGE 1-2 3 -4 5-7 8 - 12 13 14 - 19 20 - 21 22 23 - 24 25

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TABLE OF CONTENTS (Continued)

MISCELLANEOUS WIRING DIAGRAMS PROPERTIES OF CONDUCTORS ALLOWABLE AMPACITIES OF CONDUCTORS INSULATION CHARTS MAXIMUM NUMBER OF CONDUCTORS I N CONDLJIT - - -MAXIMUM NUMBER OF FIXTURE WIRES IN CONDUIT TABLES. METAL BOXES COVER REQUIREMENTS TO 600 VOLTS VOLUME REQUIRED PER CONDUCTOR CLEAR WORKING SPACE IN FRONT OF ELECTRICAL EQUIPMENT MINIMUM CLEARANCE OF LIVE PARTS GROUNDING ELECTRICAL SYMBOLS HAND SIGNALS FOR CRANES AND CHERRY PICKERS USEFUL KNOTS AMERICAN RED CROSS FIRST AID

OHM'S LAW

T H E R A T E OF T H E FLOW OF T H E CURRENT I S E Q U A L T O E L E C T R O M O T I V E FORCE D I V I D E D B Y R E S I S T A N C E . E L E C T R O M O T I V E FORCE = V O L T S = " E M CURRENT = AMPERES = :I" R E S I S T A N C E = OHMS = R"

VOLTS AMPERES

=

-

OHMS

SERIES CIRCUIT

PARALLEL C I R C U I T

A SERIES CIRCUIT I S A CIRCUIT T H A T H A S ONLY ONE P A T H THROUGH W H I C H T H E E L E C T R O N S MAY FLOW. NOTE: " T " S T A N D S FOR T O T A L .

A PARALLEL CIRCUIT I S A C I R C U I T T H A T H A S MORE T H A N ONE P A T H THROUGH W H I C H T H E E L E C T R O N S MAY F L O W .

ET = E l + E2 + E3

ET =

I

E l =

E2 =

E3

1

NOTE:

-

FOR A P A R A L L E L C I R C U I T H A V I N G ONLY TWO R E S I S T O R S , F O L L O W I N G FORMULA MAY BE U S E D . RT

=

R1 R2 R 1 + R2 X

-1-

THE

OHM'S LAW

A.

VOLTS

WHEN

AND O VOLTS OHMS

AMPERES =

0.

=

-

WHEN

= R

-

12' 60

WATTS

AMPERES =

AN0

=

2 AMPERES

VOLTS

A R E KNOWN:

WATTS VOLTS

I = -P = - 1440 = E 120 WHEN O

m AND

AMPERES = EXAMPLE:

I

-

I = P E

OR

A 120 V O L T C I R C U I T H A S A 1440 WATT L O A D . D E T E R M I N E THE CURRENT.

EXAMPLE:

C.

-

I = E R

OR

F I N D T H E C U R R E N T OF A 120 V O L T C I R C U I T W I T H A R E S I S T A N C E OF 6 0 OHMS.

EXAMPLE:

I

m A R E KNOWN:

WATTS

12 AMPERES A R E KNOWN:

+g

OR

1

=&

A C I R C U I T CONSUMES 625 W A T T S THROUGH A 12.75 OHM RESISTOR. DETERMINE THE CURRENT.

=E

=

=

C

fi

=

7 AMPERES

12.75

A.

ONE E L E C T R I C A L HORSEPOWER = 746 W A T T S E L E C T R I C MOTORS ARE R A T E D I N HORSEPOWER.

0.

ONE K I L O W A T T = 1000 W A T T S GENERATORS ARE RATED I N K I L O W A T T S .

-

SERIES CIRCUITS RULE 1:

T H E T O T A L CURRENT I N A S E R I E S C I R C U I T I S E Q U A L TO THE CURRENT I N ANY OTHER PART OF T H E C I R C U I T .

RULE 2 :

T H E T O T A L VOLTAGE I N A S E R I E S C I R C U I T I S EQUAL TO THE SUM OF T H E V O L T A G E S ACROSS A L L P A R T S OF T H E C I R C U I T .

T O T A L CURRENT = 1 ( 1 ) = 1 ( 2 ) = I ( 3 ) .

T O T A L VOLTAGE = E ( l ) + E ( 2 ) + E ( 3 ) , RULE 3 :

AND E T C .

AND E T C .

T H E T O T A L R E S I S T A N C E OF A S E R I E S C I R C U I T I S E Q U A L TO T H E SUM OF THE R E S I S T A N C E S O F A L L T H E P A R T S O F THE CIRCUIT. TOTAL RESISTANCE = R ( 1 ) + R ( 2 ) + R ( 3 ) ,

AND E T C .

FORMULAS FROM OHM'S LAW AMPERES = RESISTANCE RESISTANCE =

VOLTS

AMPERES V O L T S = AMPERES X EXAMPLE:

RESISTANCE

OR

E

=

I

X

R

F I N D T O T A L V O L T A G E . T O T A L CURRENT. AND T O T A L RESISTANCE.

E ( 2 ) = 1 0 VOLTS I ( 2 ) = 0 . 4 AMP R ( 2 ) = 2 5 OHMS

E ( l ) = 8 VOLTS I

E ( 3 ) = 6 VOLTS I ( 3 ) = 0 . 4 AMP R ( 3 ) = 1 5 OHMS

I ( 1 ) = 0 . 4 AMP

R ( 1 ) = 2 0 OHMS

E(T) = ? I(T) = ? R(T) = ? C O N T I N U E D N E X T PAGE

PARALLEL CIRCUITS

T H E T O T A L CURRENT I N A P A R A L L E L C I R C U I T I S E Q U A L TO THE SUM OF T H E C U R R E N T S I N A L L T H E B R A N C H E S OF T H E C I R C U I T .

:1-

T O T A L CURRENT =

I(1)

+

1(2) +

I ( 3 ) . AND E T C .

T H E T O T A L V O L T A G E ACROSS ANY BRANCH I N P A R A L L E L I S E Q U A L TO T H E V O L T A G E ACROSS ANY OTHER B R A N C H AND I S A L S O EQUAL TO THE TOTAL VOLTAGE.

:2-

T O T A L VOLTAGE = E ( l ) = E ( 2 ) = E ( 3 ) . R U L E 3:

T H E T O T A L R E S I S T A N C E I N A P A R A L L E L C I R C U I T I S FOUND BY A P P L Y I N G O H M ' S LAW TO T H E T O T A L V A L U E S OF T H E C I R C U I T . TOTAL RESISTANCE =

TOTAL VOLTAGE T O T A L AMPERES

EXAMPLE:

F I N D THE T O T A L CURRENT. RESISTANCE.

I(T) = = I(T) =

I(1) + I(2) + 1(3) 2 + 1.5 + 1 4 . 5 AMP

R(T) =

E(T) I(T)

NOTE:

AND ETC

=

1 2 0 VOLTS 4 . 5 AMP

=

OR

.

RT =

ET IT

T O T A L V O L T A G E . AND T O T A L

E(T) = =

E(T) = 26.66

E ( l ) = E(2) = E(3) 120 = 120 = 120 1 2 0 VOLTS

OHMS R E S I S T A N C E

I N A P A R A L L E L C I R C U I T THE T O T A L R E S I S T A N C E I S A L W A Y S L E S S T H A N T H E R E S I S T A N C E OF ANY B R A N C H . I F T H E B R A N C H E S OF A P A R A L L E L C I R C U I T H A V E T H E SAME R E S I S T A N C E . T H E N EACH W I L L DRAW T H E SAME C U R R E N T . I F T H E B R A N C H E S OF A P A R A L L E L C I R C U I T H A V E D I F F E R E N T R E S I S T A N C E S . T H E N EACH W I L L DRAW A D I F F E R E N T C U R R E N T . I N E I T H E R S E R I E S OR P A R A L L E L C I R C U I T S . T H E L A R G E R T H E R E S I S T A N C E . T H E S M A L L E R T H E CURRENT DRAWN.

I

PARALLEL CIRCUITS

TO D E T E R M I N E T H E T O T A L R E S I S T A N C E I N A P A R A L L E L C I R C U I T WHEN THE T O T A L C U R R E N T . AND T O T A L VOLTAGE ARE UNKNOWN. 1

-

TOTAL R E S I S T A N C E EXAMPLE:

1 -

1

+ -

R(1)

+

R(2)

-

AND E T C .

R(3)

F I N O THE TOTAL RESISTANCE.

R(l) =

R(2) =

R(3)

6 0 OHMS

8 0 OHMS

1 2 0 OHMS

1 -

4

=

+

R(T)

-

3

+

2

=

USE LOWEST DENOMINATOR

I

COMMON (240)

IC

240

I\=/$_

R(T)'

-

CROSS M U L T I P L Y '240

9 X R(T)

=

1 X 240

OR

9RT

=

240

D I V I D E B O T H S I D E S OF THE E Q U A T I O N B Y " 9 " R(T) = NOTE:

2 6 . 6 6 OHMS R E S I S T A N C E THE T O T A L R E S I S T A N C E OF A NUMBER OF EQUAL R E S I S T O R S I N P A R A L L E L I S EQUAL TO THE R E S I S T A N C E OF ONE R E S I S T O R D I V I O E D B Y T H E NUMBER OF R E S I S T O R S . T O T A L RESISTANCE

= R E S I S T A N C E OF ONE R E S I S T O R

NUMBER OF R E S I S T O R S I N C I R C U I T C O N T I N U E D N E X T PAGE

I

PARALLEL CIRCUITS

FORMULA:

-R

=

R(T)

N

EXAMPLE:

F I N D THE TOTAL R E S I S T A N C E T H E R E ARE THREE R E S I S T O R S I N P A R A L L E L . E A C H H A S A V A L U E OF 1 2 0 OHMS R E S I S T A N C E . A C C O R D I N G TO T H E F O R M U L A . I F WE D I V I D E T H E R E S I S T A N C E OF ANY ONE OF T H E R E S I S T O R S BY T H R E E WE W I L L O B T A I N THE T O T A L R E S I S T A N C E OF T H E C I R C U I T .

R(T)

=

-R

OR

R(T)

=

120

N

3

TOTAL RESISTANCE R(T)

NOTE:

=

=

4 0 OHMS

?

TO F I N D T H E T O T A L R E S I S T A N C E OF ONLY TWO R E S I S T O R S I N P A R A L L E L . M U L T I P L Y T H E R E S I S T A N C E S . AND THEN D I V I D E THE PRODUCT B Y T H E SUM OF THE R E S I S T O R S .

FORMULA:

TOTAL RESISTANCE

R(l) R(1)

=

+

R(2) R(2)

EXAMPLE:

I

R(1) = 4 0 OHMS

I

.,, , ,

X

R(2)

R(1)

+

R(2)

40

R(T) R(T) = ?

R(1) -

= 120

X

80

=

26.66

OHMS

w

COMBINATION CIRCUITS I N C O M B I N A T I O N C I R C U I T S WE COMBINE S E R I E S C I R C U I T S W I T H P A R A L L E L C I R C U I T S . C O M B I N A T I O N C I R C U I T S MAKE I T P O S S I B L E TO O B T A I N THE D I F F E R E N T VOLTAGES OF S E R I E S C I R C U I T S . A N 0 D I F F E R E N T CURRENTS OF PARALLEL C I R C U I T S . EXAMPLE:

1.

I

PARALLEL-SERIES CIRCUIT:

SOLVE FOR A L L M I S S I N G V A L U E S . E(3)=? ' I(3)=? '

R(3) =10

OHMS

E(4)=? I(4)=? R(4) =50

OHMS

TO SOLVE: 1. F I N D THE T O T A L R E S I S T A N C E OF EACH GROUP. S I M P L E S E R I E S C I R C U I T S , SO

BOTH GROUPS ARE I

R(1) + R(2) = R(A) 2 0 + 4 0 = 6 0 OHMS. T O T A L R E S I S T A N C E OF GROUP " A " Rlo3 1 + R ( 4 ) = R ( B ) + 5 0 = 6 0 OHMS, TOTAL R E S I S T A N C E OF GROUP " 8 " 2.

RE-DRAW THE C I R C U I T . C O M B I N I N G R E S I S T O R S ( R ( 1 ) + R ( 2 ) ) AND ( R ( 3 ) + R ( 4 ) ) SO THAT EACH GROUP W I L L HAVE ONLY ONE RESISTOR.

CONTINUED NEXT PAGE

COMBINATION CIRCUITS NOTE:

WE NOW HAVE A S I M P L E P A R A L L E L C I R C U I T , SO E(T) = 1 2 0 v =

E(A)

120

= =

E(B)

WE NOW HAVE A P A R A L L E L C I R C U I T W I T H ONLY TWO R E S I S T O R S , AND THEY ARE OF EQUAL V A L U E . WE HAVE A C H O I C E OF THREE D I F F E R E N T FORMULAS THAT CAN BE USED TO SOLVE FOR THE T O T A L R E S I S T A N C E OF C I R C U I T .

( 2 ) W H E N THE R E S I S T O R S OF A P A R A L L E L C I R C U I T ARE OF EQUAL V A L U E . R(T)

=

! = 60 N

l\=OL -

OR

R(T)/ 3.

=

3 0 OHMS

2

1 X R(T) =

1 X 30

OR

R ( T ) = 3 0 OHMS

'30

WE NOW KNOW THE V A L U E S OF E ( T ) . R ( T ) , E ( A ) . R ( A ) . E ( B ) . R ( B ) , R ( 1 ) . R ( 2 ) , R ( 3 ) , AND R ( 4 ) . NEXT WE W I L L SOLVE FOR I ( T ) , I ( A ) , I ( B ) , I ( l ) , 1 ( 2 ) , I ( 3 ) . AND I ( 4 ) .

E(B) =

I(B)

OR

120 60

-

I(3)

=

I(4)

OR

R( 0 ) I(B)

=

CONTINUED NEXT PAGE

-

2 2 =

2

=

2

I(B)

=

I(3) 1(4)

= =

2

x

.

I

COMBINATION CIRCUITS

4.

WE KNOW T H A T R E S I S T O R S # I a n d # 2 OF GROUP " A " ARE I N S E R I E S . WE KNOW TOO T H A T R E S I S T O R S # 3 a n d # 4 OF GROUP " 0 " ARE I N SERIES. WE HAVE D E T E R M I N E D THAT THE T O T A L R E S I S T A N C E OF GROUP " A - IS 2 A M P , AND T H E T O T A L RESISTANCE O F GROUP IS 2 AMP: BY U S I N G THE S E R I E S FORMULA WE CAN S O L V E FOR THE CURRENT V A L U E OF EACH R E S I S T O R .

5.

WE WERE G I V E N THE R E S I S T A N C E VALUES OF A L L R E S I S T O R S . R(1) = 2 0 OHMS, R ( 2 ) = 4 0 OHMS. R ( 3 ) = 1 0 OHMS. R(4) = 5 0 OHMS.

AND

-

BY U S I N G OHM'S LAW WE CAN D E T E R M I N E THE VOLTAGE DROP ACROSS EACH R E S I S T O R .

E(1) E(1)

= = =

R(1) X I ( 1 ) 2 0 X 2 4 0 VOLTS

E(2)

=

R(2) X 1 ( 2 ) 4 0 X 2 8 0 VOLTS

= E(2) EXAMPLE:

= 2.

E(3)

= = =

R(3) X 1(3) 10 X 2 2 0 VOLTS

E(4)

= =

E(4)

=

R(4) X 1 ( 4 ) 5 0 x 2 1 0 0 VOLTS

E(3)

SERIES PARALLEL CIRCUIT:

SOLVE FOR A L L M I S S I N G V A L U E S

E(2) = ? I(2) = ? R ( 2 ) = 20

4 C - r

E(1) = ? I(1) = ? R(1) =

u E(3) = ? 1(3) = ? R ( 3 ) = 30 GROUP " A "

-10-

E ( T ) = 110 V. I(T) = ? R(T) = ?

I . r

-

COMBINATION CIRCUITS

TO SOLVE: I

1. WE CAN S E E T H A T R E S I S T O R S # 2 AND # 3 ARE I N P A R A L L E L . AND C O M B I N E D T H E Y ARE GROUP " A " . WHEN THERE ARE ONLY TWO R E S I S T O R S . WE U S E T H E F O L L O W I N G F O R M U L A .

2.

WE CAN NOW RE-DRAW OUR C I R C U I T A S A S I M P L E S E R I E S C I R C U I T

I

R ( l ) = 1 0 OHMS

R ( A ) = 1 2 OHMS

I

=

E ( T ) = 1 1 0 VOLTS I(T) = R(T) =

GROUP " A "

3.

??

I N A SERIES CIRCUIT R(T) = R ( l )

+

R(A)

OR

R(T) = 10 + 12

OR

2 2 OHMS

BY U S I N G O H M ' S LAW

I N A SERIES CIRCUIT I(T) = I(1) = I(A) AND I ( A ) =

OR

I ( T ) = 5 AMP,

I(1) =

5 X 1 0 = 5 0 VOLTS

B Y U S I N G O H M ' S LAW

E(l) =

I ( 1 ) X R(1)

=

-

E(1) = E(A)

OR

E(T)

110

-

5 0 = 60 VOLTS = E ( A )

I N A PARALLEL CIRCUIT E(A) E(2)

:

=

E(2) = E(3) OR E ( A ) = 6 0 VOLTS. 6 0 V O L T S , AND E ( 3 ) = 6 0 V O L T S .

COMBINATION CIRCUITS

BY U S I N G O H M ' S LAW

PROBLEM:

S O L V E FOR T O T A L R E S I S T A N C E RE-DRAW C I R C U I T A S MANY T I M E S A S N E C E S S A R Y CORRECT ANSWER I S 1 0 0 OHMS

R-3

R-2

R-1

R-4

GROUP A

R-5

R-6

R-9

,.A,.

1

R-7

Ann

R-8

R-T

= ?

G I V E N VALUES: R-1

=

1 5 OHMS

R-2

=

3 5 OHMS

R-7

=

R-3

=

5 0 OHMS

R-8

= 3 0 0 OHMS

R-4

=

4 0 OHMS

R-9

=

R-5

=

3 0 OHMS

1 0 OHMS

6 0 OHMS

I

(

(

NOTE:

E

1000

PF

746

1000

KVA X

1000

E X P F X 2

KW X

746

IX X E F F X

1000

)

POWER FACTOR = P F =

E X

HP X Z E F F X PF X

2

I E X

HP X

1000 1.73

APPARENT POWER

-

KVA

KW

1.73

1.73

1.73

PF X

-

746

XEFF X

1000

I X PF X

E X

1.73

1000

PF X

746

PF X KVA X

E X

KW X

XEFF X

IX

E X

E X

THREE PHASE

POWER USED (WATTS)

PF X 2

E X I X P F X Z

E X

T w o PHASE FOUR WIRE

A L T E R N A T I N G CURRENT

D I R E C T CURRENT FORMULAS DO NOT USE ( P F . 2 . OR 1 . 7 3 S I N G L E PHASE FORMULAS 0 0 NOT USE ( 2 OR 1 . 7 3 ) TWO PHASE-FOUR WIRE FORMULAS DO NOT USE ( 1 . 7 3 ) THREE PHASE FORMULAS DO NOT USE ( 2 )

OUTPUT INPUT

746

I X Z E F F X PF

1000

E X I X P F

KVA X

E X

KW X 1 0 0 0

S I N G L E PHASE

E X

I

I

PERCENT E F F I C I E N C Y = XEFF =

746

IX ZEFF

E X 1

E:;

E X

1 I

1

HoRsEpowER

KILOVOLTAMPERES KVA*

KILOWATTS

I I I I

AMPERES WHEN "KVA" I S KNOWN

AMPERES WHEN "KW"

I T w I N o

ELECTRICAL FORMULAS FOR CALCULATING AMPERES, HORSEPOWER, KILOWAlTS, A N D KVA

I

I

I

TO FIND AMPERES DIRECT CURRENT: A.

WHEN HORSEPOWER I S KNOWN: HORSEPOWER X

AMPERES =

I =

OR

746

HP X

E X

VOLTS X EFFICIENCY

746 %EFF

WHAT CURRENT W I L L A T R A V E L - T R A I L E R T O I L E T DRAW WHEN E Q U I P P E D W I T H A 1 2 V O L T , 1 / 8 HP MOTOR, H A V I N G A 9 6 % EFFICIENCY RATING? I = - = HP - = X 746 E X % E F F B.

746 X

1/8

-.. 9 3 ' 2 5 - 8 . 0 9 AMP 11.52

-

1 2 X 0 . 9 6

WHEN K I L O W A T T S ARE KNOWN: AMPERES =

KILOWATTS X

1000

VOLTS

OR

I =

KW X

1000

E

A 7 5 KW. 2 4 0 V O L T . D I R E C T CURRENT GENERATOR I S U S E D TO POWER A V A R I A B L E - S P E E D CONVEYOR B E L T AT A ROCK C R U S H I N G PLANT. D E T E R M I N E THE CURRENT.

I =

KW

1000

=

E

I

7 5 = 3 1 2 . 5 AMPERES 240

SINGLE PHASE: A.

WHEN WATTS. V O L T S . AMPERES =

AND POWER-FACTOR

ARE KNOWN:

WATTS V O L T S X POWER-FACTOR

D E T E R M I N E T H E CURRENT WHEN A C I R C U I T H A S A 1 5 0 0 WATT L O A D A POWER-FACTOR OF 8 6 % , AND O P E R A T E S FROM A S I N G L E - P H A S E 2 3 0 V O L T SOURCE.

I =

I5O0 230 X 0.86

=

1500 197.8

= 7.58

AMP

-

TO FIND AMPERES

--

SINGLE PHASE:

0.

WHEN HORSEPOWER I S KNOWN: AMPERES =

*

HORSEPOWER X 7 4 6 V O L T S X E F F I C I E N C Y X POWER-FACTOR

D E T E R M I N E THE AMP-LOAD OF A S I N G L E - P H A S E . 1 / 2 H P . 1 1 5 VOLT MOTOR. THE MOTOR HAS AN E F F I C I E N C Y R A T I N G OF 9 2 % . AND A POWER-FACTOR OF 8 0 % .

I = C.

4 . 4 AMP

WHEN K I L O W A T T S ARE KNOWN: AMPERES =

KILOWATTS X 1 0 0 0 V O L T S X POWER-FACTOR

I

OR

=

KW X 1 0 0 0 E X PF

A 2 3 0 VOLT S I N G L E PHASE C I R C U I T ' H A S A 1 2 KW POWER L O A D , AND OPERATES A T 8 4 % POWER-FACTOR. D E T E R M I N E THE CURRENT.

D.

WHEN K I L O V O L T - A M P E R E I S KNOWN: AMPERES =

KILOVOLT-AMPERE X 1 0 0 0 VOLTS

OR

I =

KVA X 1 0 0 0 E

A 1 1 5 V O L T , 2 K V A , S I N G L E PHASE GENERATOR O P E R A T I N G A T F U L L LOAD W I L L D E L I V E R 1 7 . 4 AMPERES. (PROVE) I =

REMEMBER:

2 X 1000 115

=

2000 = 115

17.4

AMP

BY D E F I N I T I O N AMPERES I S THE RATE OF THE FLOW OF THE CURRENT.

TO FIND AMPERES

TWO-PHASE, FOUR WIRE: NOTE:

A.

FOR THREE W I R E , TWO-PHASE C I R C U I T S . T H E CURRENT I N T H E COMMON CONDUCTOR I S 1 . 4 1 GREATER T H A N I N E I T H E R OF T H E OTHER TWO CONDUCTORS.

WHEN WATTS,

V O L T S . AND POWER-FACTOR

ARE KNOWN:

WATTS

AMPERES =

VOLTS X

P

POWER-FACTOR X

2

E X

PF X

2

D E T E R M I N E T H E CURRENT WHEN A C I R C U I T H A S A 1 5 0 0 WATT L O A D , A POWER-FACTOR OF 8 6 % . A N 0 O P E R A T E S FROM A TWO P H A S E . 2 3 0 V O L T SOURCE. P

I = I =

0.

-

1500

E X PF X

2

230 X 0.86 X 2

-

1500 395.6

3 . 7 9 AMP

I

WHEN HORSEPOWER I S KNOWN: AMPERES =

HORSEPOWER X 7 4 6 VOLTS X

E F F I C I E N C Y X POWER-FACTOR X

2

OR

D E T E R M I N E THE A M P - L O A D OF A TWO-PHASE. 1 / 2 H P . 2 3 0 V O L T MOTOR. THE MOTOR H A S AN E F F I C I E N C Y R A T I N G OF 9 2 % . AND A POWER-FACTOR OF 8 0 % . HP X

I=-=-

746

1/2 X

E X % E F F X P F X 2

=

-3 7=3 339

746

2 3 0 X 0 . 9 2 X 0 . 8 0 X 2

1.1 AMP

NOTE :

CoNsUMEo APPARENT POWER

=

& KVA

= POWER-FACTOR

(PF)

-

u

TO FIND AMPERES

T W O - P H A S E , F O U R WIRE: I

C.

WHEN K I L O W A T T S ARE KNOWN: KILOWATTS X

AMPERES =

VOLTS X

I =

1000

POWER-FACTOR X

2

KW X 1 0 0 0 E X P F X 2

A 2 3 0 V O L T . TWO-PHASE C I R C U I T , H A S A 1 2 K W POWER L O A D , AND DETERMINE THE CURRENT. O P E R A T E S A T 8 4 % POWER-FACTOR.

1

=

I

0.

3 1 AMP

WHEN K I L O V O L T - A M P E R E I S KNOWN: AMPERES =

KILOVOLT-AMPERE VOLTS X

m

X

1000

2

OR KVA X

I = 1

1000

E X 2

A 2 3 0 V O L T . 4 K V A , TWO-PHASE GENERATOR O P E R A T I N G A T F U L L (PROVE) LOAD W I L L D E L I V E R 8 . 7 AMPERES.

I =

4 X

1000

230 X 2

=

4000 = 460

8 . 7 AMP

TO FIND AMPERES THREE-PHASE: A.

WHEN WATTS, VOLTS.

AND POWER-FACTOR

ARE KNOWN:

WATTS

AMPERES =

V O L T S X POWER-FACTOR

X

1.73

D E T E R M I N E T H E CURRENT WHEN A C I R C U I T H A S A 1 5 0 0 WATT L O A D . A POWER-FACTOR OF 8 6 % . AND O P E R A T E S FROM A T H R E E - P H A S E . 2 3 0 V O L T SOURCE.

=

8.

4 . 4 AMP

WHEN HORSEPOWER I S KNOWN: HORSEPOWER X 7 4 6

AMPERES =

V O L T S X E F F I C I E N C Y X POWER-FACTOR X 1 . 7 3

OR

D E T E R M I N E T H E A M P - L O A D OF A T H R E E - P H A S E , 1 / 2 H P , 2 3 0 V O L T MOTOR. T H E MOTOR H A S AN E F F I C I E N C Y R A T I N G OF 9 2 % , AND A POWER-FACTOR OF 8 0 % .

=

-3 7=3

293

1.27

AMP

m

TO FIND AMPERES

m

THREE-PHASE: C.

WHEN K I L O W A T T S ARE KNOWN: KILOWATTS X

AMPERES =

VOLTS X

1000

POWER-FACTOR X

a

1.73

0 1

I =

KW X E X

1000

PF X

1.73

I

A 2 3 0 V O L T , T H R E E - P H A S E C I R C U I T , H A S A 1 2 KW POWER L O A D , AND O P E R A T E S A T 8 4 % POWER-FACTOR. D E T E R M I N E T H E CURRENT.

-

I =

KW X

1000

12.000

-

E X P F X 1 . 7 3 I 0.

=

12,000 -

=

2 3 0 X 0 . 8 4 X 1 . 7 3

334.24

36AMP

WHEN K I L O V O L T - A M P E R E I S KNOWN: AMPERES =

KILOVOLT-AMPERE X E X

I .

1000

1.73

=

KVA X E X

1000 1.73

A 2 3 0 V O L T , 4 K V A , THREE PHASE GENERATOR O P E R A T I N G A T F U L L LOAD W I L L D E L I V E R 1 0 AMPERES. (PROVE)

I = m

KVA X E X

I NOTE:

1000 = 1.73

4 X

1000

230 X

1.73

-

4000 397.9

= TO B E T T E R UNDERSTAND THE P R E C E D I N G FORMULAS:

1. TWO-PHASE CURRENT X 2 = S I N G L E - P H A S E CURRENT. 2 . T H R E E - P H A S E CURRENT X 1 . 7 3 = S I N G L E P H A S E CURRENT. 3 . T H E CURRENT I N THE COMMON CONDUCTOR OF A TWO-PHASE ( T H R E E W I R E ) C I R C U I T I S 1 4 1 % GREATER T H A N E I T H E R OF T H E OTHER TWO CONDUCTORS OF T H A T C I R C U I T .

TO FIND HORSEPOWER

DIRECT CURRENT: VOLTS X

HORSEPOWER =

AMPERES X

EFFICIENCY

746

A 1 2 V O L T MOTOR DRAWS A CURRENT OF 8 . 0 9 AMPERES, AND H A S AN E F F I C I E N C Y R A T I N G OF 9 6 % . D E T E R M I N E THE HORSEPOWER. I

HP =

E X I X % E F F

1 2 X 8 . 0 9 X

746

=

0.1249

0.96

746

- -9 3 . 1 9 746

=

SINGLE-PHASE: HP =

VOLTS X

AMPERES X

EFFICIENCY X

POWER-FACTOR

746

A S I N G L E - P H A S E . 1 1 5 V O L T ( A C ) MOTOR H A S AN E F F I C I E N C Y R A T I N G OF 9 2 % , A N 0 A POWER-FACTOR OF 8 0 % . D E T E R M I N E THE HORSEPOWER I F THE A M P - L O A D I S 4 . 4 AMPERES.

-

TWO-PHASE:

HP =

V O L T S X AMPERES X E F F I C I E N C Y X POWER-FACTOR

X

2

746

D E T E R M I N E T H E HORSEPOWER OF A TWO-PHASE. 2 3 0 V O L T ( A C ) MOTOR. THE MOTOR H A S AN E F F I C I E N C Y R A T I N G OF 9 2 % . A POWER-FACTOR OF 8 0 % . AND AN AMP-LOAD OF 1.1 AMPERES.

I

TO FIND HORSEPOWER THREE-PHASE: . I

HP =

I

V O L T S X AMPERES X E F F I C I E N C Y X POWER-FACTOR

X

1.73

746

A T H R E E - P H A S E . 4 6 0 V O L T MOTOR DRAWS A C U R R E N T OF 5 2 A M P E R E S . THE MOTOR H A S AN E F F I C I E N C Y R A T I N G OF 9 4 % . A N D A POWER FACTOR OF 8 0 % . D E T E R M I N E T H E HORSEPOWER.

TO FIND WATTS

THE E L E C T R I C A L POWER I N ANY P A R T OF A C I R C U I T I S E O U A L TO T H E C U R R E N T IN T H A T P A R T MULTIPLIED B Y T H E VOLTAGE A C R O S S T H A T P A R T OF THE C I R C U I T . A WATT I S T H E POWER U S E D WHEN ONE V O L T C A U S E S ONE AMPERE TO FLOW I N A CIRCUIT. ONE HORSEPOWER I S T H E AMOUNT OF ENERGY R E Q U I R E D TO L I F T 3 3 . 0 0 0 POUNDS. ONE F O O T . I N ONE M I N U T E . THE E L E C T R I C A L E Q U I V A L E N T OF ONE HORSEPOWER I S 7 4 5 . 6 W A T T S . ONE WATT I S THE AMOUNT OF ENERGY R E Q U I R E D TO L I F T 4 4 . 2 6 P O U N D S , ONE FOOT. I N ONE M I N U T E . WATTS I S POWER, AND POWER I S T H E AMOUNT OF WORK DONE I N A G I V E N T I M E . 1.

W H E N VOLTS A N D AMPERES ARE K N O W N : A.

POWER ( W A T T S ) = V O L T S

X

AMPERES

A 1 2 0 V O L T A - C C I R C U I T DRAWS A CURRENT OF 5 A M P E R E S : D E T E R M I N E THE POWER C O N S U M P T I O N . P

=

I

E X

=

120 X 5

=

6 0 0 WATTS

WE CAN NOW D E T E R M I N E T H E R E S I S T A N C E OF T H I S C I R C U I T . (1.)

POWER

=

P = 600 -

25 (2.)

NOTE:

POWER

=

=

RESISTANCE

X

R

X

OR

R

OR

(I)' R

=

=

(120)'

R

=

2 4 OHMS

OR

=

600

R

X

25

2 4 OHMS

OR

RESISTANCE

R X 600

(AMPERES)Z

R

P

=

=

R

14,400

600

REFER TO FORMULAS OF THE O H M ' S LAW C H A R T ON P A G E 1.

-22-

I

DIRECT CURRENT:

KILOWATTS

V O L T S X AMPERES 1000

=

A 1 2 0 V O L T ( D C ) MOTOR DRAWS A CURRENT OF 4 0 AMPERES DETERMINE THE K I L O W A T T S .

SINGLE-PHASE: KILOWATTS

=

VOLTS X

AMPERES X POWER-FACTOR 1000

A S I N G L E - P H A S E . 1 1 5 V O L T ( A C ) MOTOR DRAWS A C U R K E N T OF 2 0 A M P E R E S . AND H A S A POWER-FACTOR R A T I N G O F 8 6 % . D E T E R M I N E THE KILOWATTS.

=

1.978

=

TWO-PHASE: KILOWATTS

=

VOLTS X

x

AMPERES X POWER-FACTOR 1000

2

A T W O - P H A S E , 2 3 0 V O L T ( A C ) MOTOR W I T H A P O W E R - F A C T O R O F 9 2 % . DRAWS A CURRENT OF 5 5 AMPERES. DETERMINE THE K I L O W A T T S . KW =

E X I

X P F X 2 1000

2 3 0 X 55 X 1000

0.92X

2

TO FIND KILOWAlTS THREE-PHASE: KILOWATTS

=

V O L T S X AMPERES X POWER-FACTOR X 1.73

-

1000

A T H R E E - P H A S E , 4 6 0 V O L T MOTOR DRAWS A CURRENT OF 5 2 A M P E R E S . AND H A S A POWER-FACTOR R A T E D A T 8 0 % . D E T E R M I N E T H E K I L O W A T T S . II

K I R C H H O F F ' S LAWS F I R S T LAW

(CURRENT1

-

T H E SUM OF T H E C U R R E N T S A R R I V I N G AT ANY P O I N T I N A C I R C U I T MUST E Q U A L T H E SUM OF T H E CURRENTS L E A V I N G T H A T P O I N T . SECOND LAW

I

(VOLTAGE)

THE T O T A L V O L T A G E A P P L I E D TO ANY C L O S E 0 C I R C U I T P A T H I S ALWAYS EQUAL TO THE SUM OF THE V O L T A G E DROPS I N T H A T P A T H .

OR THE A L G E B R A I C SUM OF A L L THE V O L T A G E S ENCOUNTERED I N ANY LOOP EQUALS ZERO.

I

TO FIND KILOVOLT-AMPERES

SINGLE-PHASE: KILOVOLT-AMPERES

=

V O L T S X AMPERES 1000

A S I N G L E - P H A S E , 2 4 0 V O L T GENERATOR D E L I V E R S 4 1 . 6 6 A M P E R E S A T FULL LOAD. DETERMINE THE K I L O V O L T - A M P E R E S R A T I N G . KVA

=

-=

E X I

240 X 41.66

1000

_

-

1000

10,000 - = 1000

TWO-PHASE: KILOVOLT-AMPERES

I

=

AMPERES 1000

A T W O - P H A S E . 2 3 0 V O L T GENERATOR D E L I V E R S 5 5 AMPERES DETERMINE THE KILOVOLT-AMPERES R A T I N G . KVA

=

-

E X IX 2 1000

=

I

230 X 55 X 2 1000

-

-

25,300 1000

=

25.3

THREE-PHASE: KILOVOLT-AMPERES

= 1000

A T H R E E - P H A S E , 4 6 0 V O L T GENERATOR D E L I V E R S 5 2 AMPERES DETERMINE THE KILOVOLT-AMPERES R A T I N G .

NOTE:

KVA

=

A P P A R E N T POWER = POWER B E F O R E U S E D . SUCH A S THE R A T I N G OF A TRANSFORMER.

TO FIND CAPACITANCE (C1:

C

=

a E

OR C A P A C I T A N C E

=

COULOMBS VOLTS

C A P A C I T A N C E I S THE PROPERTY OF A C I R C U I T OR BODY T H A T P E R M I T S I T TO STORE A N E L E C T R I C A L CHARGE E Q U A L TO THE ACCUMULATED CHARGE D I V I D E D B Y T H E V O L T A G E . EXPRESSED I N FARADS. A.

I

TO D E T E R M I N E T H E T O T A L C A P A C I T Y OF C A P A C I T O R S , AND / OR CONDENSERS CONNECTED I N S E R I E S .

D E T E R M I N E THE T O T A L C A P A C I T Y OF FOUR E A C H , 1 2 M I C R O F A R A D C A P A C I T O R S CONNECTED I N S E R I E S .

C(T) 8.

=

3 MICROFARADS

TO D E T E R M I N E T H E T O T A L C A P A C I T Y OF C A P A C I T O R S . AND / OR CONDENSERS CONNECTED I N P A R A L L E L .

-

DETERMINE THE T O T A L C A P A C I T Y OF FOUR EACH. 1 2 M I C R O F A R A D C A P A C I T O R S CONNECTED I N P A R A L L E L . C(T) C(T)

+

=

=

4 8 MICROFARADS

A FARAD I S THE U N I T OF C A P A C I T A N C E OF A CONDENSER T H A T R E T A I N S ONE COULOMB OF CHARGE W I T H ONE V O L T D I F F E R E N C E OF POTENTIAL. 1 FARAD

=

1 , 0 0 0 , 0 0 0 MICROFARADS

-26-

I

-

6-DOT COLOR CODE FOR MICA AND MOLDED PAPER CAPACITORS

D I R E C T I O N OF '

O

"

~

T

K

OR C L A S S

TI

+-

2 NI D D I G ~ \

MULTIPLIER TOLERANCE

L/

t---

TYPE

JAN. MICA

COLOR

BLACK BROWN RED ORANGE YELLOW GREEN BLUE VIOLET GRAY WHITE E I A . MICA GOLD MOLDED P A P E R S I L V E R BOD'I

1ST ZND DIGIT DIGIT 0 1 2 3 4

5 6 7 8 9

0 1 2

3 4 5 6 7 8 9

MULTIPLIER

1 10 100 1.000 10.000 100,000 1,000,000 10,000,000 100.000.000 1.000.000.000 .1

.O1

lOLERANCE CHARACTERISTIC OR C L A S S ( % )

** + *i

1 2

f 3 4 5 6

i 7 i 6

+

9

+lo *20

A P P L I E S TO TEMPERATURE COEFFICIENT OR M E T H O D S OF T E S T I N G

-

M A X I M U M PERMISSIBLE CAPACITOR KVAR FOR USE WITH OPEN-TYPE THREE-PHASE SIXTY-CYCLE INDUCTION MOTORS 1 8 0 0 RPM 1 2 0 0 RPM 3 6 0 0 RPM MOTOR R A T I N G MAXIMUM R E D U C T I O N MAXIMUM REDUCTION MAXIMUM REDUCTION I N LINE HP CAPACITOR I N LINE I N L I N E CAPACITOR CAPACITOR CURRENT RATING RATING CURRENT CURRENT RATING KVAR % % KVAR % KVAR 10 15 20 25 30

3 4 5 6 7

10 9 9 9 8

3 4 5 6 7

11 10 10 10 9

40 50 60 75

9 12 14 17

8 8 8 8

9 11 14 16

9 9 8 8

11 13 15 18

22 27 32.5 40

8 8 8 8

21 26 30 37.5

8 8 8 8

25 30 35 42.5

100 125 150 200

14 13 12 11 11 10 10 10 10 9 9 9 9

6 0 0 RPM

7 2 0 RPM

9 0 0 RPM

3.5 5 6.5 7.5 9

10 15 20 25 30

5 6.5 7.5 9 10

21 18 16 15 14

6.5 8 9 11 12

27 23 21 20 18

7.5 9.5 12 14 16

31 27 25 23 22

40 50 60 75

12 15 18 21

13 12 11 10

15 19 22 26

16 15 15 14

20 24 27 32.5

20 19 19 18

27 32.5 37.5 47.5

10 10 10 10

32.5 40 47.5 60

13 13 12 12

40 47.5 52.5 65

17 16 15 14

100 125 150 200

NOTE. I F C A P A C I T O R S OF A LOWER R A T I N G T H A N T H E V A L U E S G I V E N I N T H E T A B L E ARE U S E D , T H E P E R C E N T A G E R E D U C T I O N I N L I N E CURRENT G I V E N I N THE T A B L E SHOULD BE REDUCED P R O P O R T I O N A L L Y . REPRINTED WITH PERMISSION FROM NFPA 70 1990 NATIONAL ELECTRICAL CODEv COPYRIGHT 1989 NATIONAL FIRE PROTECTION ASSOCIATION OUINCY MA 02269 THIS REPRINTED MATERIAL IS NOT THE COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIRETY

-

POWER-FACTOR CORRECTION T A B L E V A L U E S X KW L O A O = K V A O F C A P A C I T O R S N E E D E D T O C O R R E C T F R O M E X I S T I N G T O D E S I R E D POWER F A C T O R . EXISTING POWER FACTOR %

100%

95%

90%

85%

80%

75%

50 52 54 55 56 58 60 62 64 65 66 68 70 72 74 75 76 78 80 82 84 85 86 88 90 92 94 95

1.732 1.643 1.558 1.518 1.479 1.404 1.333 1.265 1.201 1.168 1.139 1.078 1.020 0.964 0.909 0.882 0.855 0.802 0.750 0.698 0.646 0.620 0.594 0.540 0.485 0.426 0.363 0.329

1.403 1.314 1.229 1.189 1.150 1.075 1.004 0.936 0.872 0.839 0.810 0.749 0.691 0.635 0.580 0.553 0.526 0.473 0.421 0.369 0.317 0.291 0.265 0.211 0.156 0.097 0.034

1.247 1.158 1.073 1.033 0.994 0.919 0.848 0.780 0.716 0.683 0.654 0.593 0.535 0.479 0.424 0.397 0.370 0.317 0.265 0.213 0.161 0.135 0.109 0.055

1.112 1.023 0.938 0.898 0.859 0.784 0.713 0.645 0.581 0.548 0.519 0.458 0.400 0.344 0.289 0.262 0.235 0.182 0.130 0.078

0.982 0.893 0.808 0.768 0.729 0.654 0.583 0.515 0.451 0.418 0.389 0.328 0.270 0.214 0.159 0.132 0.105 0.052

0.850 0.761 0.676 0.636 0.597 0.522 0.451 0.383 0.319 0.286 0.257 0.196 0.138 0.082 0.027

C O R R E C T E D POWER F A C T O R

T Y P I C A L PROBLEM: W I T H A L O A O O F 5 0 0 KW A T 70% POWER F A C T O R . I T I S D E S I R E D T O F I N D THE K V A O F C A P A C I T O R S R E Q U I R E O TO CORRECT THE POWER F A C T O R T O 85%. SOLUTION: FROM T H E T A B L E S E L E C T T H E M U L T I P L Y I N G F A C T O R 0 . 4 0 0 C O R R E S P O N D I N G T O T H E E X I S T I N G 70%. A N D T H E C O R R E C T E D 8 5 % POWER 0.400 X 500 = 200 KVA O F C A P A C I T O R S R E Q U I R E O . FACTOR. REPRINTED WITH PERMISSION FROM NFPA 70 1990. NATIONAL ELECTRICAL CODE' .COPYRIGHT 1989. NATIONAL FIRE PROTECTION ASSOCIATION. OUINCY, MA 02269 THIS REPRINTED MATERIAL IS NOT THE COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCEDSUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIRETY

TO FIND

INDUCTION (L): I N O U C T I O N I S T H E P R O D U C T I O N OF M A G N E T I Z A T I O N OF E L E C T R I F I C A T I O N I N A BODY B Y THE P R O X I M I T Y OF A M A G N E T I C F I E L D OR E L E C T R I C CHARGE. OR OF THE E L E C T R I C CURRENT I N A CONDUCTOR BY THE V A R I A T I O N OF THE M A G N E T I C F I E L D I N I T S V I C I N I T Y . EXPRESSED I N HENRYS. A.

TO F I N O THE T O T A L I N D U C T I O N OF C O I L S CONNECTED I N S E R I E S .

D E T E R M I N E T H E T O T A L I N D U C T I O N OF FOUR C O I L S CONNECTEO I N SERIES. EACH C O I L H A S AN I N D U C T A N C E V A L U E OF FOUR HENRYS. L(T)

B.

+

=

Li1)

=

16 HENRYS

+

Lj2)

-

-

+

TO F I N O THE T O T A L I N D U C T I O N OF C O I L S CONNECTEO I N PARALLEL.

D E T E R M I N E THE T O T A L I N O U C T I O N OF FOUR C O I L S CONNECTED I N PARALLEL. EACH C O I L H A S AN I N D U C T A N C E V A L U E OF FOUR HENRYS.

1

-

L(T)

OR

L(T) X

4

=

1 X

4

OR L ( T )

=

4

4

L(T) =

1 HENRY

AN I N O U C T I O N C O I L I S A O E V I C E . C O N S I S T I N G OF TWO C O N C E N T R I C C O I L S AND AN I N T E R R U P T E R . T H A T CHANGES A LOW STEADY VOLTAGE I N T O A H I G H I N T E R M I T T E N T A L T E R N A T I N G VOLTAGE BY E L E C T R O M A G N E T I C I N D U C T I O N . MOST O F T E N USED A S A SPARK C O I L .

I

TO FIND

-

IMPEDANCE (Z): I M P E D A N C E I S T H E T O T A L O P P O S I T I O N TO AN A L T E R N A T I N G CURRENT PRESENTED B Y A C I R C U I T . EXPRESSED I N OHMS. A.

1

WHEN V O L T S AND AMPERES ARE KNOWN: IMPEDANCE

VOLTS AMPERES

=

Z

OR

=

E I

D E T E R M I N E T H E I M P E D A N C E OF A 1 2 0 V O L T A - C C I R C U I T T H A T DRAWS A CURRENT OF FOUR AMPERES.

3 0 OHMS B.

WHEN R E S I S T A N C E AND REACTANCE ARE KNOWN:

1

D E T E R M I N E THE I M P E D A N C E OF AN A - C C I R C U I T WHEN THE R E S I S T A N C E I S 6 OHMS. AND THE REACTANCE I S 8 OHMS.

I

-

=

C.

1 0 OHMS

WHEN R E S I S T A N C E , I N D U C T I V E REACTANCE, REACTANCE ARE KNOWN:

AND C A P A C I T I V E

D E T E R M I N E THE I M P E D A N C E OF AN A-C C I R C U I T W H I C H H A S A R E S I S T A N C E OF 6 OHMS. AN I N D U C T I V E REACTANCE OF 1 8 OHMS. AND A C A P A C I T I V E REACTANCE OF 1 0 OHMS.

z

= = =

7/R'

+

yS2 d=

+

( X(L) (

18

-

=

10)'

X(C) =

7/100

)'

d =

m 1 0 OHMS

TO FIND

REACTANCE (XI: REACTANCE I N A C I R C U I T I S THE O P P O S I T I O N TO AN A L T E R N A T I N G CURRENT CAUSED B Y I N D U C T A N C E AND C A P A C I T A N C E . E Q U A L TO THE D I F F E R E N C E BETWEEN C A P A C I T I V E AND I N D U C T I V E R E A C T A N C E . EXPRESSED I N OHMS. A.

JX.J

I N D U C T I V E REACTANCE

I N D U C T I V E REACTANCE I S THAT ELEMENT OF REACTANCE I N A C I R C U I T C A U S E D BY S E L F - I N D U C T A N C E .

X(L) =

2

X

3.1416

X

FREQUENCY X

INDUCTANCE

D E T E R M I N E T H E REACTANCE OF A FOUN-HENRY 6 0 CYCLE, A-C C I R C U I T . X(L)= =

8.

6.28 X F 1 5 0 7 OHMS

X

L

6.28

=

X

C O I L ON A

60

X

4

C A P A C I T I V E REACTANCE C A P A C I T I V E REACTANCE I S THAT E l C I R C U I T CAUSED B Y C A P A C I T A N C E . X(C) =

REACTANCE

1 2

X

3.1416

X

FREQUENCY X

CAPACITANCE

D E T E R M I N E THE REACTANCE OF A FOUR M I C R O F A R A D CONDENSER ON A 6 0 CYCLE, A-C C I R C U I T .

=

663

OHMS

0.0015072 A HENRY I S A U N I T OF I N D U C T A N C E . EQUAL TO THE I N D U C T A N C E OF A C I R C U I T I N W H I C H THE V A R I A T I O N OF A CURRENT A T THE RATE OF ONE AMPERE PER SECOND I N D U C E S AN E L E C T R O M O T I V E FORCE OF ONE V O L T .

-

RESISTOR COLOR CODE

r

1 1.51 O I G I T (PERCENT)

1ST OIGIT

2ND DIGIT

BLACK

0

0

1

BROWN

1

1

10

RE0

2

2

100

ORANGE

3

3

1,000

YELLOW

4

4

10,000

GREEN

5

5

100,000

BLUE

6

6

VIOLET GRAY

7 8

7 8

1.000,OOO 10.000.000 100.000.000

WHITE

9

9

COLOR

MULTIPLIER

1.000.000,000

GOLD

.1

SILVER

.O1

NO COLOR

TOLERANCE (PERCENT)

2 5

%

+ 10% + 20%

U.S. WEIGHTS AND MEASURES

LINEAR MEASURE

12 3 5.5 40 8

INCHES FEET YARDS RODS FURLONGS

= = = = =

1 1 1 1 1 1

INCH = FOOT = YARD = ROD, P O L E , OR PERCH = FURLONG = MILE =

2.540 3.048 9.144 5.029 2.018 1.609

CENTIMETERS DECIMETERS DECIMETERS METERS HECTOMETERS KILOMETERS

-

MILE MEASUREMENTS

1 1 1 1 1 1

STATUTE SCOTS IRISH RUSSIAN ITALIAN SPANISH

MILE MILE MILE VERST MILE MILE

5.280 5.952 6.720 3.504 4.401 15,084

=

= = = = =

FEET FEET FEET FEET FEET FEET

OTHER LINEAR MEASUREMENTS 1 1 1 1

HAND SPAN CHAIN KNOT

= = = =

=

4 INCHES 9 INCHES 2 2 YARDS 1 NAUTICAL M I L E 6 0 8 0 FEET

1 1 1 1

LINK FATHOM FURLONG CABLE

I

=

= = =

7.92 6 10 608

INCHES FEET CHAINS FEET

SQUARE M E A S U R E 144 9 30-114

40 4 640 1 36

SQUARE I N C H E S = = SQUARE F E E T SQUARE YAROS = = = RODS = = ROODS = ACRES SQUARE M I L E = = SECTIONS

1 1 1 1 1 1 1 1 1 1

SQUARE FOOT SQUARE YARD SQUARE ROO SQUARE P O L E SQUARE PERCH ROOD ACRE SQUARE M I L E SECTION TOWNSHIP

CUBIC O R SOLID MEASURE 1 1 1 -

1 1 1

= C U . FOOT = C U . YARD FOOT = C11. - GALLON (WATER) = GALLON ( U . S . ) = GALLON ( I M P E R I A L ) =

1728 27 7.48 8.34 231 277-1/4

CU. I N C H E S CU. FEET GALLONS LBS. C U . I N C H E S OF WATER CU. I N C H E S OF WATER

-

U.S. WEIGHTS AND MEASURES

-

LIQUID MEASURE 1 1 1 1 1

GILLS PINTS QUARTS GALLONS GALLONS

PINT QUART GALLON FIRKIN BARREL

= = = = =

4 2 4 9 42

1 QUART 1 PECK 1 BUSHEL

= = =

2 PINTS 8 QUARTS 4 PECKS

( A L E OR B E E R ) ( P E T R O L E U M OR CRUDE O I L )

DRY M E A S U R E

_ -

WEIGHT MEASUREMENT(MASS) A.

A V O I R D U P O I S WEIGHT: 1 1 1 1

B.

m

DRAMS OUNCES POUNOS POUNDS

= = = =

3.17 20 20 12

GRAINS GRAINS PENNYWEIGHTS OUNCES

=

112 20 2240

POUNDS LONG H U N D R E D W E I G H T S POUNDS

CARAT PENNYWEIGHT OUNCE POUND LONG HUNDREOWEIGHT 1 LONG TON

=

=

C.

A P O T H E C A R I E S WEIGHT: 1 1 1 1

D.

16

TROY WEIGHT: 1 1 1 1 1

-

16 100 2000

= OUNCE = POUND HUNDREDWEIGHT = TON =

SCRUPLE DRAM OUNCE POUND

= = =

=

20 GRAINS 3 SCRUPLE 8 DRAMS 1 2 OUNCES

= = = =

1.296 3.888 31.1035 313.2420

GRAMS GRAMS GRAMS GRAMS

K I T C H E N W E I G H T S AND MEASURES: 1 U.S. P I N T 1 S T A N D A R D CUP 1 TABLESPOON 1 TEASPOON

= =

= =

16 8 0.5 0.16

F L . OUNCES F L . OUNCES F L . O U N C E S ( 1 5 CU. F L . OUNCES ( 5 C U .

CMS.) CMS.)

-

Xr-?

METRlC SYSTEM

PREFIXES: I\. 0. C. D.

I

WEGA KILO HECTO DEKA

=

1,000,000 1,000 100 10

= = =

E. F. G. H.

DECI CENT1 MILL1 MICRO

= =

= =

0.1 0.01 0.001 0.000001

UNEAR MEASURE: i E U N I T 'c

1 f w - l

THE METER = 39.37 INCUFQ.

CENTIM DECIME

3

DEKAMETER HECTOMETER KILOMETER MYRIAMETER

= =

'

1

I

I

/ I 1 I

'

1 1 1 1

=

=

10 MILLIMETERS 10 CENTIMETERS = 10 DECIMETERS = =,3. 10 METERS = 10 DEKAMETERS = 10 HECTOMETERS = 10.000 METERS

011 I N . 3.9370113 I N S . 1.0936143 YDS.

*

10.936143 YDS. 109.36143 YDS. 0.62137 M I L E

SQUARE MEASURE: THE U N I T I S THE SQUARE METER = 1549.9969 SQ. INCHES:

1 1 1 1 1 1

SQ. SQ. SQ. SQ. SQ. SQ.

CENTIMETER DECIMETER METER DEKAMETER HECTOM€TER KILOMETER

= = =

= r

=

100 100 100 100 100 100

SQ. SQ. SQ. SQ. SQ. SQ.

MILLIMETERS CENTIMETERS DECIMETERS METERS DEKAMETERS HECTOMETERS

= = =

=

0.1550 SQ. 15.550 SQ. 10.7639 SQ. 119.60 SQ.

THE U N I T I S THE "ARE" = 100 SO. METERS: CENTIARE DECIARE ARE DEKARE HEKTARE (HECTO-ARE) 1 SQ. KILOMETER

1 1 1 1 1

= = = = =

10 10 10 10 10

MILLIARES CENTIARES DECIARES ARES DEKARES

= = = = =

= 100 HEKTARES =

10.7643 11.96033 119.6033 0.247110 2.471098

SQ. FT. SQ. YDS. SQ. YDS. ACRES ACRES

0.38611

SQ. M I L E

THE UNIT I S THE "STERE" = 61.025.38659 CU.

1 DECISTERE 1 STERE 1 DEKASTERE

= = =

10 CENTISTERES = 10 DECISTERES = = 10 STERES

INS.:

3.531662 CU. FT. 1.307986 CU. YDS. 13.07986 CU. YDS.

IN. INS. FT. YDS

METRIC SYSTEM

CUBIC MEASURE: THE UNIT I S THE "METER" = 3 9 . 3 7 INS.: 1

CU. CENTIMETER=

1 CU. DECIMETER = 1

= =

CU. METER

1000 CU. MILLIMETERS = 1 0 0 0 CU. CENTIMETERS = 1000 CU. DECIMETERS = 1 STERE =

0.06125 61.1250 35.3156 1.30797

= =

1 GRAM

CU. CENTIMETER (WATER) 1000 CU CENTIMETERS (WATER) = 1 LITER 1 CU. METER ( 1 0 0 0 LITERS) 1

CU. CU. CU. CU.

IN. INS. FT. YOS.

1 KILOGRAM 1 METRIC TON

=

MEASURES OF WEIGHT: THE UNIT r S THE GRAM = 0.035274 OUNCES: 1 1 1 1 1

MILLIGRAM CENTIGRAM DECIGRAM GRAM DEKAGRAM

= = =

1 5 7

I

1

MYRIAGRA QUINTAL METRIC TON

=

= = =

1 GRAM 1 DRAM

=

1 METRIC TON

= =

10 10 10 10 10 10 10 10 10

= = = =

MILLIGRAMS CENTIGRAMS DECIGRAMS GRAMS DEKAGRAMS HECTOGRAM! KILOGRAMS MYRIAGRAMS QUINTAL

=

=

=

=

0.015432 0.15432 1.5432 15.4323 5.6438

GRAINS GRAINS GRAINS GRAINS DRAMS

22.046223 POUND 1.986412 CWT. 2.204.622 POUNDS

0.56438 DRAMS 1.77186 GRAMS 27.3438 GRAINS 2.204.6223 POUNDS

MEASURE OF CAPACITY: THE UNIT I S THE "LITER" = 1.0567 LIOUID OUARTS: 1 CENTILITER

1 DECILITER 1 LITER 1 DEKALITER 1 HECTOLITER 1 KILOLITER

= = = =

= =

10 10 10 10 10 10

= 8

=

MILLILITERS CENTILITERS DECILITERS LITERS DEKALITERS HECTOLITERS MLIES

= =

= = =

0.338 3.38 33.8 0.284 2.84 264.2

9X 5

8

=

FLUID OUNCES FLUID OUNCES FLUID OUNCES BUSHEL BUSHELS GALLONS K1WKTERS

*

I

CONVERSION TABLES

ATMOSPHERES

g T J

=

=

X X

33.9 29.92 14.7

X

F T . OF WATER I N S . OF MERCURY L B S . PER S Q . I N .

252 777.5 0.0003927 1054 0.0002928

X X

C A L O R I E S (GRAM) F T . LBS. HORSEPOWER-HOURS JOULES KILOWATT-HOURS

l 2 96 0 02356 0 05686

X X X

F T . L B S . PER S E C . HORSEPOWER WATTS

X X

X

BTU (PER M I N . )

=

CALORIES

=

0.003968

X

BTU

DYNE

=

GRAMS

X

CM/SEC/SEC

=

9.48 X lo-" 1.0 7 . 3 7 x loe8

X X

BTU DYNE C E N T I M E T E R S FT. LBS.

0.02950 0.8826 0.4335

X X

0.07717 0.001818 0.001356

X

X

E T U PER M I N . HORSEPOWER KILOWATTS

F E E T OF WATER

FOOT POUNDS PER SECOND

=

=

,

X

X

X

ATMOSPHERES I N S . OF MERCURY F T . L B S . PER SQ.

FOOT CANDLE

=

10.765

X

LUX

HORSEPOWER

=

42.44 33,000 550

X X

B T U . PER M I N . F T . L B S . PER M I N . F T . L B S . PER S E C .

2547 1.98 X lo6 2.68 X lo6

X

.HCIRSFPCIWFR , - ..- - ..HOURS

=

-

. ....

X

-

X X

0.7376 0.000278 1.0

x X

X

BTU. FT. LBS JOULES FT:-LBS. WATT-HOURS WATT-SECONDS

KILOWATT-HOURS

=

3415 3.6 x 1013

X

x

B T U PER M I N . ERGS

Lux

=

0.929

X

FT.

CANDLES

IN.

CONVERSION TABLES

BTU PER. M I N . = ERGS PER. SEC. = FT. L B S . PER M I N . = = HORSEPOWER

WATTS WATTS WATTS WATTS

1 GRAM CALORIE

=

0 . 0 0 3 9 6 4 BTU 4 . 1 8 4 JOULES

1 GRAM FORCE

=

9 8 0 . 6 DYNES

1 FOOT POUND

=

1 L B S . FORCE X 1 . 3 5 6 JOULES

1 POUND MASS

=

453.6

1 NEWTON

=

1 KILOGRAM X 1 METER/SEC/SEC 1 0 0 . 0 0 0 DYNES 0 . 2 2 4 LBS. FORCE

1 SLUG

=

3 2 . 2 LBS MASS 1 4 . 6 0 6 KILOGRAMS

1 KILOWATT HOUR

=

3.600.000

1 WATT

=

3 . 4 1 2 BTU/HRS. 0 . 2 3 9 GRAM C A L O R I E / S E C .

1 BTU

0.05692 1 . 0 X 10' 44.26 0.00134

X

X X X

1 FOOT

GRAMS

JOULES

R A I S E S ONE POUND OF WATER 1' F

1 GRAM CALORIE 1 CIRCULAR M I L

R A I S E S ONE GRAM OF WATER =

0.7854

1" C

SQ. M I L

1 SOUARE M I L

=

1.27 CIR. MILS

1 MIL

=

0 . 0 0 1 INS.

T O DETERMINE THE CIRCULAR M I L OF A CONDUCTOR 1. ROUND CONDUCTOR

CM =

2.

cM =

BUS BAR

( DIAMETER I N M I L S )'

WIDTH ( M I L S ) X THICKNESS ( M I L S ) 0.7854

NOTES:

1.

1 MILLIMETER

2.

1 CIR.

3.

1 SQ. M I L L I M E T E R

MILLIMETER

39.37

MILS

=

1550 CIR.

MILS

=

1974 CIR.

MILS

METALS

METAL

SYMB.

ALUMINUM ANTIMONY ARSENIC BERYLLIUM BISMUTH B R A S S (70-30) BRONZE (5% S N ) CADMIUM CALCIUM COBALT COPPER ROLLED TUBING GOLD GRAPHITE INDIUM IRIDIUM

AL SB AS BE B I

IRON

CD CA CO CU

I N IR FE

7.20

AU

7.20 7.70

WROUGHT PB MG MN HG MO NI P PT K SE S I AG

11.40 1.74 7.20 13.65 10.20 8.87 8.90 1.82 21.46 0.860 4.81 2.40 10.50 7.84

STEEL (CARBON)

MELT

c0

660 2.710 6.620 630 5.730 - - - 1.830 1280 9.800 271 8.510 900 8.870 1000 8.650 321 1.550 850 8.900 1495 8.890 8,950 19.30 2.25 7.30 22.40

MALLEABLE

LEAD MAGNESIUM MANGANESE MERCURY MOLYBDENUM MONEL (63-37) NICKEL PHOSPHORUS PLATINUM POTASSIUM SELENIUM SILICON SILVER

SPEC. GRAV.

F"

1220 1167

----

2336 520 1652 1382 610 1562 2723

1083

1981

....

....

1063 3500 156 2450 1200

1945 6332 311 4442 2192

TO

TO

1400 1500

2552 2732

TO

TO

1600 1500

2912 2732

TO

1600 327 651 1245 -38.9 2620 1300 1452 44.1 1773 62.3 220 1420 960 1330

TO

2912 621 1204 2273 -37.7 4748 2372 2646 111.4 3221 144.1 428 2588 1760 2436

TO

TO

1380

2516

1500 1520 1500 2900

2732 2768 2732 5414

E L E C . COND. % COPPER

LBS. CU."

64.9 4.42 4.9 9.32 1.50 28.0 18.0 22.7 50.1 17.8

.0978 .2390 ,2070 .0660 .3540 .3070 ,3200 ,3120 .0560 ,3210

100.00 100.00 71.2 20.6 32.5

.3210 .3230 .6970 ,0812 ,2640 .a090

17.6

,2600

10

,2600

10

.2780

8.35 38.7 0.9 1.80 36.1 3.0 25.0 lo-'' 17.5 28 14.4 10 106

,4120 .0628 ,2600 .4930 ,3680 ,3200 .3210 ,0657 ,7750 ,0310 .I740 .0866 .3790

10

,2830

2.5 3.5 3.0 13.9

.2860 .2810 ,2790 .5990

STAINLESS

(18-8) (13-CR)

(18-CR) TANTALUM

TA

7.92 7.78 7.73 16.6

METALS

METAL TELLURIUM THORIUM TIN TITANIUM TUNGSTEN URANIUM VANADIUM ZINC ZIRCONIUM

SYMB.

SPEC. GRAV.

TE TH SN T I W U V ZN ZR

6.2 11.70 7.30 4.50 19.30 18.70 5.96 7.14 6.40

-

c0

F"

450 1845 232 1800 3410 1130 1710 419 1700

846 3353 449 3272

'

E L E C . CONO. % COPPER

- - -

2066 3110 786 3092

LBS. CU:

1 0 ~ ~,2240 9.10 .422 15.00 ,264 2.10 .I62 31.50 ,697 2.80 .675 6.63 ,215 29.10 .258 ,231 4.20

SPECIFIC RESISTANCE (K)

THE S P E C I F I C R E S I S T A N C E ( K ) OF A M A T E R I A L I S THE R E S I S T A N C E O F F E R E D B Y A WIRE OF THIS MATERIAL WHICH IS ONE FOOT LONG WITH D I A M E T E R OF ONE M I L .

MATERIAL

-

MELT PO1NT

"K"

MATERIAL

"K"

BRASS

43.0

ALUMINUM

17.0

CONSTANTAN

295

MONEL

253

COPPER

10.8

NICHROME

600

GERMAN S I L V E R 18%

200

NICKEL

947

GOLD

14.7

TANTALUM

93.3

IRON (PURE)

60.0

TIN

69.0

MAGNESIUM

276

TUNGSTEN

34.0

MANGANIN

265

SILVER

9.7

NOTE:

1. T H E R E S I S T A N C E OF A W I R E I S D I R E C T L Y P R O P O R T I O N A L TO T H E S P E C I F I C R E S I S T A N C E OF T H E M A T E R I A L .

2. " K " = S P E C I F I C R E S I S T A N C E

A

L .

CENTIGRADE AND FAHRENHEIT THERMOMETER SCALES

1.

TEMP.

C"

=

5/9 X

2.

TEMP.

F'

=

(9/5

3.

A M B I E N T T E M P E R A l U R E I S THE TEMPERATURE OF THE SURROUNDING C O O L I N G M E D I U M

4.

RATED TEMPERATURE R I S E I S THE P E R M I S S I B L E R I S E I N TEMPERATURE ABOVE A M B I E N T WHEN O P E R A T I N G UNDER L O A D .

(TEMP. X

TEMP.

F'

- 32)

C o ) + 32

USEFUL MATH FORMULAS

OBTUSE T R I A N G L E

RIGHT TRIANGLE

h "C"

"A"

"B"

S O L V E A S TWO R I G H T TRIANGLES

SPHERE AREA = D ' X 3 . 1 4 1 6 VOLUME = D3 x 0 . 5 2 3 6

CYLINDRICAL VOLUME = AREA O F END X H E I G H T

Q -- -

fl CONE

VOLUME = AREA OF END X H E I G H T / 3

--"A"

"W"

ELLIPTICAL SOLVE T H E SAME A S C Y L I N D R I C A L

M

=

A

X

B

X

C

THE CIRCLE

DEFINITION:

A C L O S E D P L A N E CURVE H A V I N G EVERY P O I N T AN EQUAL D I S T A N C E FROM A F I X E D P O I N T W I T H I N THE C U R V E .

CIRCUMFERENCE: : DIAMETER : RADIUS

T H E D I S T A N C E AROUND A C I R C L E . T H E D I S T A N C E ACROSS A C I R C L E THROUGH T H E C E N T E R T H E D I S T A N C E FROM THE CENTER TO THE EDGE OF A CIRCLE. A PART OF THE C I R C U M F E R E N C E . A S T R A I G H T L I N E CONNECTING THE ENDS O F AN A R C . AN AREA BOUNDED BY AN ARC A N 0 A CHORD. A P A R T OF C I R C L E E N C L O S E 0 BY TWO R A D I I AND THE ARC W H I C H THEY CUT O F F .

:

ARC CHORD SEGMENT SECTOR

:

: :

CIRCUMFERENCE OF A C I R C L E = 3.1416 X 2 X RADIUS 3 . 1 4 1 6 X RADIUS X RADIUS AREA OF A C I R C L E = DEGREES I N ARC X R A O I U S X 0 . 0 1 7 4 5 ARC L E N G T H = ONE H A L F L E N G T H OF D I A M E T E R RADIUS LENGTH = SECTOR AREA = ONE H A L F L E N G T H OF ARC X R A D I U S CHORD LENGTH SEGMENT AREA

2 SECTOR AREA M I N U S T R I A N G L E A R E A .

= =

m: 3.1416

X

2 X R =

3 6 0 DEGREES

v w 0.0087266 0.01745

X 2 X

R =

X

R

OR

1 DEGREE

T H I S G I V E S U S THE ARC FORMULA DEGREES X

RADIUS X 0.01745 DEVELOPEO L E N G T H

=

EXAMPLE: FOR A N I N E T Y DEGREE C O N D U I T B E N D , H A V I N G A R A D I U S OF 1 7 . 2 5 " : 9 0 X 17.25" 27"

X

0.01745

= DEVELOPED LENGTH

=

DEVELOPED LENGTH

1

FRACTIONS

. I

DEFINITIONS: A.

A F R A C T I O N I S A Q U A N T I T Y L E S S THAN A U N I T

8.

A NUMERATOR I S THE TERM OF A F R A C T I O N I N D I C A T I N G HOW MANY OF T H E P A R T S OF A U N I T ARE T O B E T A K E N . I N A COMMON F R A C T I O N I T A P P E A R S ABOVE OR TO T H E L E F T OF T H E L I N E .

C.

A O E N O M I N A T O R I S T H E TERM OF A F R A C T I O N I N D I C A T I N G THE NUMBER OF E Q U A L P A R T S I N T O W H I C H THE U N I T I S O I V I D E U . IN A COMMON F R A C T I O N I T A P P E A R S BELOW OR TO T H E R I G H T OF THE LINE.

D.

EXAMPLES:

a

1

1

-

(1.)

2

(2. )

+

= +

NUMERATOR

FRACTION

=

DENOMINATOR

NUMERATOR

+

1/2

OENOMINATOR

+

TO ADD OR SUBTRACT: TOSOLVE

112

-

2/3

-

3/4

+

5/6

+

7/12

=

?

A.

D E T E R M I N E T H E LOWEST COMMON D E N O M I N A T O R T H A T E A C H OF THE D E N O M I N A T O R S 2 . 3 . 4 . 6 . AND 1 2 W I L L D I V I D E I N T O A N EVEN NUMBER OF T I M E S .

B.

WORK ONE F R A C T I O N A T A T I M E U S I N G T H E FORMULA

. I

THE LOWEST COMMON D E N O M I N A T O R I S 1 2

-

T I M E S NUMERATOR OF F R A C T I O N D E N O M I N A T O R OF F R A C T I O N (1.)

12/2

X

1 =

6

X

1

=

(2.)

12/3

X

2

=

4

X

2

(3.)

12/4

X

3

=

3

X

3

(4.)

12/6

X

5

=

2

X

5

(5.)

7/12

REMAINS 7/12

C O N T I N U E D N E X T PAGE

6

112

BECOMES

=

8

2/3

BECOMES

8/12

=

9

3/4

BECOMES

9/12

=

10

5/6

BECOMES 1 0 1 1 2

6/12

FRACTIONS

T O ADD O R S U B T R A C T ( C O N T I N U E D ) : C.

WE CAN NOW CONVERT THE PROBLEM FROM I T S O R I G I N A L FORM TO I T S NEW FORM U S I N G 1 2 AS THE COMMON D E N O M I N A T O R .

112

-

2/3

+

3/4

5/6

-

+

7/12

+

+

=

ORIGINALFORM

=

P R E S E N T FORM

12 +22

-

l8

12 D.

=

=

12

1

REDUCED TO LOWEST FORM

3

TO CONVERT F R A C T I O N S TO D E C I M A L FORM S I M P L Y D I V I D E THE NUMERATOR OF THE F R A C T I O N B Y THE DENOMINATOR OF THE FRACTION. EXAMPLE:

1 D I V I D E D BY 3

=

0.33

=

A.

THE NUMERATOR OF F R A C T I O N # 1 T I M E S THE NUMERATOR OF F R A C T I O N # 2 I S EQUAL TO THE NUMERATOR OF T H E PRODUCT

8.

THE OENOMINATOR OF F R A C T I O N # 1 T I M E S THE D E N O M I N A T O R OF F R A C T I O N # 2 I S E Q U A L TO THE OENOMINATOR OF THE PRODUCT.

C.

EXAMPLE: FRACTION #2

PRODUCT

1 NUMERATORS

-2 = -

1-1' -

1

2-3

DENOMINATORS

TO CHANGE 1 / 3 TO D E C I M A L FORM.

I

-

ANS.

T O MULTIPLY:

FRACTION # 1

-

DIVIDE 1 BY 3 =

0.33

-

FRACTIONS

1

-

TO DIVIDE:

-

A.

THE NUMERATOR OF F R A C T I O N # 1 T I M E S T H E O E N O M I N A T O R OF F R A C T I O N # 2 I S EQUAL TO THE NUMERATOR OF THE Q U O T I E N T .

8.

THE D E N O M I N A T O R OF F R A C T I O N # I T I M E S THE NUMERATOR OF F R A C T I O N # 2 I S EQUAL TO THE O E N O M I N A T O R OF THE Q U O T I E N T

C.

EXAMPLE: F R A C T I O N #1

FRACTION #2

L-

QUOTIENT

3

DENOMINATORS

TO CHANGE 3 / 4

TO D E C I M A L FORM. D I V I D E 3 B Y 4 =

0.75

EQUATIONS

T H E W O R D E Q U A T I O N M E A N S E Q U A L O R T H E S A M E AS. EXAMPLE:

2

10 = 20 =

X

5

4 X 20

w: A.

THE SAME NUMBER MAY B E ADOEO TO B O T H S I D E S OF A N E Q U A T I O N WITHOUT CHANGING I T S VALUES. EXAMPLE:

B.

(2

X

10) +

3 = 23 =

(2

X

10) -

3

17

= =

(4 X 17

5)

-

-

3

2 x 1 0 4 x 5 = 20

20

B O T H S I D E S OF AN E Q U A T I O N MAY B E M U L T I P L I E D BY T H E SAME NUMBER W I T H O U T C H A N G I N G I T S V A L U E S . EXAMPLE:

E.

+ 3

BOTH S I D E S OF AN E Q U A T I O N MAY BE D I V I D E D B Y T H E SAME NUMBER WITHOUT CHANGING I r s V A L U E S . EXAMPLE:

0.

5)

THE SAME NUMBER MAY B E S U B T R A C T E D FROM B O T H S I D E S OF AN EQUATION WITHOUT CHANGING I T S VALUES. EXAMPLE:

C.

(4 X 23

3

X

(2

X 10) 60

= =

3 X 60

(4

X 5)

TRANSPOSITION: THE PROCESS OF M O V I N G A Q U A N T I T Y FROM ONE S I D E OF A N E Q U A T I O N TO THE OTHER S I D E OF AN E Q U A T I O N BY C H A N G I N G I T S S I G N OF O P E R A T I O N I S T R A N S P O S I N G . 1.

A TERM MAY B E TRANSPOSED I F I T S S I G N I S C;iANGED FROM P L U S ( + ) TO M I N U S ( - ) . OR FROM M I N U S ( - ) TO P L U S ( + ) . EXAMPLES

-

EQUATIONS

w: 1

E.

TRANSPOSITION: 2.

A M U L T I P L I E R MAY B E REMOVED FROM ONE S I D E OF AN E Q U A T I O N B Y M A K I N G I T A D I V I S O R I N THE O T H E R . OR A D I V I S O R MAY B E REMOVED FROM ONE S I D E OF AN E Q U A T I O N BY M A K I N G I T A M U L T I P L I E R I N THE O T H E R . EXAMPLE:

EXAMPLE:

M U L T I P L I E R FROM ONE S I D E OF E Q U A T I O N BECOMES D I V I S O R I N OTHER S I D E OF THE E Q U A T I O N . 4X = 4 0

X

BECOMES

=

40

4

D I V I S O R FROM ONE S I D E OF E Q U A T I O N BECOMES M U L T I P L I E R I N OTHER S I D E OF THE E Q U A T I O N . EXAMPLE:

1

10

=

BECOMES

X =

4 X

10

4

w: A.

ADDITION:

1

1.

RULE:

U S E THE S I G N OF THE LARGER AND ADD

EXAMPLES:

+ 3

- 2 -

-- 2

8.

3

+

3

9

2

SUBTRACTION: RULE:

CHANGE THE S I G N OF THE SUBTRAHEND AND PROCEED AS I N A D D I T I O N :

EXAMPLES:

+ 3

-

2

- 2

+

3

3

+

+

2

- 3