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FOWLER CHAPTER 1 LECTURE 2 BASIC CONCEPTS. SCIENTIFIC NOTATION (powers of ten) SEE TABLE 2-1, P33. EXPONENT. ANY NUMBER CAN BE EXPRESSED AS BASE. ( CAN BE POSITIVE OR NEGATIVE) + OR -. OTHER BASES AND EXPONENTS. ANY NUMBER RAISED TO THE ZEROTH POWER IS EQUAL TO ZERO.
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SCIENTIFIC NOTATION (powers of ten) SEE TABLE 2-1, P33 EXPONENT ANY NUMBER CAN BE EXPRESSED AS BASE (CAN BE POSITIVE OR NEGATIVE) + OR - OTHER BASES AND EXPONENTS
ANY NUMBER RAISED TO THE ZEROTH POWER IS EQUAL TO ZERO EXAMPLES: 10 WRITE 1000 IN SCIENTIFIC NOTATION 1000 WRITE AS 1 FOLLOWED BY 3 ZERO’S ANY NUMBER >0 CAN BE EXPRESSED THIS WAY. EXAMPLE: 1,237 CAN BE EXPRESSED AS 1.237X10³ COUNT THE NUMBER OF DECIMAL PLACES TO THE RIGHT OF THE FIRST DIGIT THEN PLACE THE DECIMAL PT. AFTER THE FIRST DIGIT. 1.237 1.237 THE NUMBER OF DECIMAL PLACES TO THE RIGHT OF THE FIRST DIGIT IS THE EXPONENT EXPRESSED AS THE POWER OF TEN, IN THIS CASE IT IS 3 1,237 =1.237X10³
FOR NEGATIVE EXPONENTS(ANY NUMBER <0) FOLLOW THE PROCEDURE ON THE PRIOR SLIDE, BUT IN THE REVERSE DIRECTION. EXAMPLES: 10־¹ =0.1 10־² =0.01 10־³ =0.001 HERE WE COUNT DECEMICAL PLACES TO THE LEFT INSTEAD OF THE RIGHT. EXAMPLE: 10־³ =0.001 0.001= 10־³ COUNT 3 PLACE TO THE LEFT. FROM OUR LAST EXAMPLE LET’S WRITE 0.001237 IN S. N. COUNT FROM THE FIRST DIGIT TO THE LEFT. .001237 -3 IS THE EXPONENT FOR THIS POWER OF TEN. SO .001237= 1.237X10־³ 1,237 CAN ASLO BE EXPRESSED AS 12.37X10²=1,237 OR 123.7X10¹=1,237 DEPENDING ON WERE THE DECIMAL PT. IS PLACED, BOTH RESULTS GIVE THE NUMBER. .001237 CAN ASLO BE EXPRESSED AS 1.237X10־³= .001237 OR 0.1237X10־²= .001237 DEPENDING ON WHERE THE DECIMAL PT. IS PLACED, BOTH RESULTS GIVE THE NUMBER.
ENGINEERING NOTATION IN THIS SYSTEM POWERS OF TEN ARE ALWAYS MULTIPIES OF 3 …ETC. OR EXAMPLE: EXPRESS 27000 IN S.N. AND E.N. S.N. E.N.
(ENGINEERING NOTATION) POWER OF TEN
ELECTRICAL UNITS AND SYMBOLS TABLE 2-3 P.37 QUANTITY UNIT SYMBOL CURRENT AMPERE(A) I VOLTAGE VOLT(V) V RESISTANCE OHM(Ω) R FREQUENCY HERTZ(Hz) f CAPACITANCE FARAD(F) C INDUCTANCE HENERY(H) L POWER WATTS(W) P
EXAMPLES : USES OF ENGINEERING NOTATION (E.N.) 1,000,000Ω = = 1MΩ 27,340Ω = =27.34X10³Ω =27.43KΩ IN E.N. OR .0274MΩ OR O.OOO274GΩ 0.000546Ω =.546X10־³ Ω = .546mΩ =546uΩ =546000nΩ
QUICK MATH REVIEW 4 OPERATES IN ALL OF MATHEMATICS ADDITION SUBTRACTION MULTIPICATION DIVISION DIVISION AND MULTIPICATION CAN BE DERIVATED FROM ADDITION AND SUBTRACTION. MULTIPICATION IS A SERIES OF REPEATED ADDITIONS EXAMPLE: 2X4=8 OR 2+2+2+2=8 DIVISION IS A SERIES OF REPEATED SUBTRACTIONS EXAMPLE:
SIMPLE ALGEBRA ANY QUANTITY ON BOTH SIDES OF AN EQUATION ARE EQUAL. EXAMPLES V=V EXAMPLES I=I R=R ADDITION V+V=2V 1+1=2 SUBTRACTION V-V=0 1-1=0 2V-V=V
A+B=C ANY QUANTITY CAN BE ADDED OR SUBTRACTED TO BOTH SIDES OF ANY EQUATION. GIVEN A+B=C SOLVE FOR A SINCE -B=- B, ADD THIS TO BOTH SIDES OF THE EQUATION. A+B-B=C-B SINCE B-B=0 A+0=C-B A=C-B SOLVE FOR B A+B=C ADD –A TO BOTH SIDES. A+B-A=C-A B+A-A=C-A B+0=C-A B=C-A
LAWS OF EXPONENTS V CAN ALSO BE EXPRESSED AS V¹, SO V=V¹ ANY QUANTITY DIVIDED BY ITSELF= 1 SINCE 1/V=1/V V=V V(1/V) =V(1/V) V(1/V) =V(1/V) 1=1 1/V CAN BE WRITTEN AS V־¹ 1/V=V־¹ OR
V/V = 1 OR V¹/V¹ =V¹־¹ =Vº =1 EXAMPLE: V²/V =V²/V¹ = V²־¹ =V¹ =V OR VxV/V = VxV/V =V EXAMPLE: EXAMPLE: OR
OR EXAMPLE:
SQUARE ROOTS WE CAN RAISE A BASE NUMBER TO ANY POWER 8² =64 LETS REVERSE THIS PROCESS FIND IS DEFINED AS A RADIAL SIGN ANOTHER WAY OF SHOWING THE SAME THING INDEX INDEX: HOW MANY TIMES WAS THIS NUMBER X MULTILED BY ITSELF TO GET 64
ANOTHER WAY TO EXPRESS THIS EXAMPLE:
OHM’S LAW R IS A PROPORTIONIALITY CONSTANT THE PRODUCT IR CAN BE WRITTEN SEVERAL WAYS
V=IR SOLVE FOR I MULTIPLIE BOTH SIDES BY 1/R (1/R)V=IR(1/R) V/R=IR/R V/R=I R/R V/R=I(1) V/R=I OR I=V/R
HOW CAN WE INCREASE I ONE WAY IS TO INCREASE V OR DECREASE R I ISINVERSELY PROPORTIONAL TO R. AS R ↓, I↑ IF WE WANT TO DECREASE I, ↑R I IS INVERSELY PROPORTIONAL TO 1/R. AS R↑, I↓
POWER,CURRENT, RESISTANCE, VOLTAGE WHEEL ANY VARIABLE ON THE POWER WHEEL CAN BE FOUND USING THE FOLLOWING TWO EQUATIONS. 1. V=IR 2. P=IV EXAMPLE: P=V²/R WHERE DID THIS COME FROM? SOLVE EQ. 1. FOR I V=IR V/R=IR/R I=V/R SUBSITUTE I=V/R INTO EQ. 2 P=IV P=(V/R)V =V²/R P=V²/R
DERIVE P=I²R FROM EQUATIONS 1. AND 2. P=IV SUB. FOR V=IR IN P=IV P=I(IR) P=I²R ONE MORE TO TORTURE YOU!!! SOLVE SUB. FOR I=V/R
APPLE APP ANDRIOD APP
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