Electrical Formulas Explanations

ELECTRICAL FORMULAS AND EXPLANATIONS

HEAT...2

Thermometer Scales...2

UNITS DERIVED FROM THE BASE ELECTRICAL UNIT (AMPERE)...2

RULES OF THUMB FOR ELECTRICAL MOTORS...2

Horsepower Revolutions Per Minute - Torque...3

Shaft Size - Horsepower - Revolutions Per Minute...3

ELECTRICAL FORMULAS...4 Power...5 Parallel Circuits...5 Direct Current:...6 Single Phase:...6 Three Phase...6 To Find Watts...6 Capacitance (C)...7 Induction (L)...7 Reactance...7 Impedance (Z)...8 Power factor (PF)...8 TRIGONOMETRY...8

POWER FACTOR CORRECTION TABLE:...9

TRANSFORMER CALCULATIONS...9

VOLTAGE AND CURRENT RELATIONSHIPS...10

VOLTAGE AND COIL TURN RELATIONSHIPS...10

AMPERES AND COIL TURN RELATIONSHIPS...10

RESISTOR COLOR CODE...10

FULL-LOAD CURRENT IN AMPERES FOR DIRECT-CURRENT MOTORS...11

COMPARISON OF AC AND DC CURRENTS...12

HEAT

Thermometer Scales

On the Fahrenheit (F) thermometer, the freezing point of water is marked at 32 degrees on the scale, and the boiling point of water at atmospheric pressure is marked at 2120_{. The distance between these points is}

divided into 1800_{. On the Celsius © thermometer, the freezing point of water is marked at 0}0 _{on the scale,}

and the boiling point of water is marked at 1000_.

The following formulas are used for converting temperatures: 32 5 9 +       _×° = C renheit DegreesFah

(

)

UNITS DERIVED FROM THE BASE ELECTRICAL UNIT (AMPERE)

Electrical Potential Volt (V)

A W V =

Electrical Resistance Ohm (μ)

A V

= µ

Electrical Capacitance Farad (F)

      = V s A F , s = seconds

Quantity of Electricity Coulomb (C) C = As, s = seconds

Electrical Inductance Henry (H)

A Wb H =

Magnetic Flux Weber (Wb) Wb Vs

RULES OF THUMB FOR ELECTRICAL MOTORS

Horsepower versus Amperes

Here are some simple ways to take the mystery out of determining several important factors concerning electric motors. For example, using the following relationships, may be useful. For three-phase motors, we find that approximately:

1 hp - 575 Volts requires 1 Amp of current 1hp - 460 Volts requires 1.25 Amps of current 1hp - 230 Volts requires 2.50 Amps of current 1hp - 2300 Volts requires 0.25 Amps of current

The key is 1 horsepower - 1 Amp on 575 Volts. All others then become a direct ratio of this figure, ie., 575/460 times 1.0 equals 1.25 A for 1 hp required on 460 V. Now, for example, with these facts at hand, you can say that a 150-hp motor on 230 V will require approximately 375 A. This is obtained as follows: 1 hp on 230 V requires 2.5 A per horsepower, so you simply multiply 2.5 times 150 hp.

Horsepower Revolutions Per Minute - Torque

Torque is simply a twisting force that causes rotation around a fixed point. For example, torque is something we all experience every time we pass through a revolving door. Horsepower is what is required when we pass through the door, because we are exercising torque at a certain revolution per minute (r/min). The faster we go through the revolving door, the more horsepower is required. In this case, the torque remains the same. In relating this concept to motors, we use this rule of thumb:

1 horsepower at 1800 revolution per minute delivers 3 ft-lb of torque 1 horsepower at 900 revolutions per minute delivers 6 ft-lb of torque

Using this simple rule, we can see that a 10-hp motor at 1800 r/min delivers 30 ft-lb of torque, and a 20-hp motor at 1800 r/min delivers 60 ft-lb of torque.

In another example, a 1-hp motor at 1200 r/min delivers 4.5 ft-lb of torque. Here is how to figure this. Multiply the torque at 1800 r/min by the ratio 1800/ 1200. Torque, then, is the inverse ratio of the speed. In other words, SPEED DOWN—TORQUE UP for the same horsepower.

A quick estimate for the torque of a 125-hp, 600-r/min motor can be figured by the following procedure. 125 hp times 3 ft-lb equal 375 ft-lb of torque for a 125-hp motor at 1800 r/min. Now, to convert to 600 rpm, multiply 375 times 1800/600 rpm or 1125 ft-lb.

This gives a rule that will enable you to quickly determine the torque a motor is capable of delivering, down to 10 rpm. Below this speed, other factors enter in that must be taken into consideration.

Shaft Size - Horsepower - Revolutions Per Minute

A number to remember here is 1150. This is easy to remember, and it stands for this: A 1-inch diameter shaft can transmit 1 hp at 50 rpm. As the shaft speed goes up, so does the horsepower, and by the same ratio. Therefore, if you double the speed, you double the horsepower capacity of the shaft. However, when you double the shaft diameter, the capacity of the shaft to transmit horsepower is increased 8 times. Thus, whatever the shaft size is in inches, cube it and multiply the resultant figure by the proper speed ratio, and the horsepower-transmitting ability of the shaft is determined. However, it is advisable to be conservative, so modify the results by 75% and use this resulting figure. To express this in a formula, use the following: 50 RPM s terInInche ShaftDiame power ShaftHorse = ×

OHMS LAW CHART

ELECTRICAL FORMULAS

The rate of the flow of the current is equal to electromotive force divided by resistance. Electromotive force = volts = “E”

Current = amperes = “I” Resistance = ohms = “R”

Ohms Volts Amperes=

The resistance of a copper wire one foot long and one circular mil in cross sectional area has a voltage drop of approximately 10.8 ohms.

A series circuit is a circuit that has only one path through which the electrons may flow. Ohms Law (see ohms law chart) equals E = I×R,

R E I = , I E R =

2

I

P

R

P

E

P

R

E

P

E

2

I

E

R

P

×

I

P

R

E

2

I

R

×

2

I

R

×

I

E

×

I R

P E

Rule 1: The total current in a series circuit is equal to the current in any other part of the circuit.

ETC T I I I I

I = ₁ = ₂ = ₃ =

Rule 2: The total voltage in a series circuit is equal to the sum of the voltages across all parts of the circuit. . 3 2 1 ETC T E E E E E = + + +

Rule 3: The total resistance of a series circuit is equal to the sum of the resistance’s of all the parts of the circuit. ETC T R R R R R = ₁ + ₂ + ₃ + Power VOLTS WATTS AMPERES = OR E P I = OR Amperes = R P

I = where P= Watts and R= Ohms

AMPERES VOLTS

WATTS = × OR P= E×I

One electrical “horsepower” = 746 watts, electrical motors are rated in horsepower One kilowatt = 1000 watts, generators are rated in kilowatts

Parallel Circuits

A parallel circuit is a circuit that has more than one path through which the electrons may flow.

Rule 1: The total current in a parallel circuit is equal to the sum of the currents in all the branches of the circuit. . 3 2 1 ETC T I I I I I = + + +

Rule 2: The total voltage across any branch in parallel is equal to the voltage across any other branch and is also equal to the total voltage.

ETC T E E E E

E = ₁ = ₂ = ₃ =

Rule 3: The total resistance in a parallel circuit is found by applying ohms law to the total values of the circuit. ETC T R R R R R 1 1 1 1 1 3 2 1  + + + = Notes:

In a parallel circuit, the total resistance is always less than the resistance of any branch.

If the branches of a parallel circuit have different resistance values, then each will draw a different current. In either a series circuit or a parallel circuit, the larger the resistance, the smaller the current drawn.

Direct Current: efficiency volts horsepower Amperes × × = 746 or Eff E hp I % 746 × × = Volts kilowatts Amperes= ×1000 or E KW I = ×1000 746 Efficiency Amperes Volts Horsepower = × × Single Phase: r PowerFacto Volts Kilowatts Amperes × × = 1000 or PF E P I × = r PowerFacto efficiency volts horsepower Amperes × × × = 746 746 r PowerFacto Efficiency Amperes Volts Horsepower = × × × Three Phase r PowerFacto Volts Watts Amperes × × = 73 . 1 or 1.73 1000 × × × = r PowerFacto Volts Kilowatts Amperes 73 . 1 746 × × × × = r PowerFacto Efficiency Volts Horsepower Amperes 73 . 1 1000 × × = Volts peres Kilovoltam Amperes 746 73 . 1 × × × ×

=Volts Amperes Efficiency PowerFactor

Horsepower eCurrent SinglePhas Current ThreePhase ×1.73= To Find Watts

The electrical power in any part of a circuit is equal to the current in that part multiplied by the voltage across that part of the circuit.

The Watt is the power used when one volt causes one ampere to flow in a circuit.

One horsepower is the amount of energy required to lift 33,000 pounds, one foot, in one minute. The electrical equivalent of one horsepower is 745.6 watts. One watt is the amount of energy required to lift 44.26 pounds, one foot, in one minute. Watts is power, and power is the amount of work done in a given time.

(

)

Power = × or

(

)

₍

₎

Capacitance (C) Volts Coulombs E Q C = =

Capacitance is the property of a circuit or body that permits it to store an electrical charge equal to the accumulated charge divided by the voltage. Expressed in FARADS. Farads are a very large number, so microfarads are often used more.

Capacitance total for series circuits

etc T C C C C C 1 1 1 1 1 3 2 1  + + + =

Capacitance total for parallel circuits

etc T C C C C

C = ₁ + ₂ + ₃ +

Induction (L)

Induction is the production of magnetization of electrification in a body by the proximity of a magnetic field or electric charge, or the electric current in a conductor by the variation of the magnetic field in its vicinity. The henry is the unit of inductance. A variation of a current at the rate of one ampere per second induces an electromotive force of one volt.

Formula to find total inductance for coils in series

ETC T L L L L

L = ₁+ ₂ + ₃ +

Formula to find total inductance for coils in parallel

etc T L L L L L 1 1 1 1 1 3 2 1  + + + = Reactance

Reactance in a circuit is the opposition to an alternating current caused by inductance and capacitance. It is equal to the difference between Capacitive and inductive reactance, expressed in ohms.

Inductive reactance XL

Inductive reactance is that element of reactance in a circuit caused by self-inductance.

ce Induca Frequency

X_L =2×3.1416× × tan or simply X_L =2πFL=6.28FL

Capacitive reactance XC

Capacitive reactance is that element of reactance in a circuit caused by capacitance.

ce Capaci Frequency X_C tan 1416 . 3 2 1 × × × = or FC FC X_C 28 . 6 1 2 1 = = π

Impedance (Z)

Impedance is the total opposition to an alternating current presented by a circuit (OHMS). Total AC resistance. Amperes Volts I E Z = = , _{Im ( )} 2

(

)

Power factor (Pf) equals the cosine, where θ is the angle is either lead or lag.

Z R EI P wer ApparentPo TruePower P_F = = =

TRIGONOMETRY

Trigonometry is the mathematics dealing with the relations of sides and angles of triangles. A triangle is a figure enclosed by three straight sides. The sum of the three angles is 180 degrees. All triangles have six parts: three angles and three sides opposite the angles. Right triangles are triangles that have one angle of ninety degrees and two angles of less than ninety degrees.

Hypotenuse de OppositeSi Sineθ = Hypotenuse de AdjacentSi CoSineθ = de AdjacentSi de OppositeSi Tangentθ = de OppositeSi de AdjacentSi CoTangentθ = de AdjacentSi Hypotenuse Secantθ = de OppositeSi Hypotenuse CoSecantθ =

Note: θ = Theta = any angle

Remember: SOHCAHTOA, Sine, opposite, hypotenuse, cosine, adjacent, hypotenuse, tangent, opposite, adjacent. ADJACENT SIDE HYPOTENUSE OP POS IT E S IDE 60º 90º 30º “Y” “X”

ALWAYS PLACE THE ANGLE TO BE SOLVED AT THE VERTEX (WHERE “X” AND “Y” CROSS).

POWER FACTOR CORRECTION TABLE:

Table values x KW load = KVA of capacitors needed to correct from existing to desired power factor.

EXISTING POWER FACTOR %

CORRECTED POWER FACTOR

100 % 95 % 90 % 85 % 80 % 75 % 50 1.732 1.403 1.247 1.112 0.982 0.850 52 1.643 1.314 1.158 1.023 0.893 0.761 54 1.558 1.229 1.073 0.938 0.808 0.676 55 1.518 1.189 1.033 0.898 0.768 0.636 56 1.479 1.150 0.994 0.859 0.729 0.597 58 1.404 1.075 0.919 0.784 0.654 0.522 60 1.333 1.004 0.848 0.713 0.583 0.451 62 1.265 0.936 0.780 0.645 0.515 .0383 64 1.201 0.872 0.716 0.581 0.451 0.319 65 1.168 0.839 0.683 0.548 0.418 0.286 66 1.139 0.810 0.654 0.519 0.389 0.257 68 1.078 0.749 0.593 0.458 0.328 0.196 70 1.020 0.691 0.535 0.400 0.270 0.138 72 0.964 0.635 0.479 0.344 0.214 0.082 74 0.909 0.580 0.424 0.289 0.159 0.027 75 0.882 0.553 0.397 0.262 0.132 76 0.855 0.526 0.370 0.235 0.105 78 0.802 0.473 0.317 0.182 0.052 80 0.750 0.421 0.265 0.130 82 0.698 0.369 0.213 0.078 84 0.646 0.317 0.161 85 0.620 0.291 0.135 86 0.594 0.265 0.109 88 0.540 0.211 0.055 90 0.485 0.156 92 0.426 0.097 94 0.363 0.034 95 0.329

TRANSFORMER CALCULATIONS

Power (p) = Power (s) minus inefficiencies (iron loss, hysteresis, eddy currents) Transformers are very efficient devices usually 99% plus.

VOLTAGE AND CURRENT RELATIONSHIPS

S S P

P I E I

E × = − minus the inefficiencies.

P S S P I I E E = × , P S S P E I E I = × , S P P S E I E I = × , S P P S I I E E = ×

VOLTAGE AND COIL TURN RELATIONSHIPS

) ( ) ( ) ( )

(p Turns s Voltage s Turns p

Voltage × = × S P S P T T E E = × , P P S S E T E T = × , S S P P E T E T = × , P S P S T T E E = ×

AMPERES AND COIL TURN RELATIONSHIPS

P S S P T T I I = × , P S S P I T I T = × , S P P S I T I T = × , S P P S T T I I = ×

RESISTOR COLOR CODE

First band = first significant digit Second band = second digit Third digit = multiplier

Fourth digit = percent tolerance

0 = black, 1 = brown, 2 = red, 3 = orange, 4 = yellow, 5 = green, 6 = blue, 7 = violet, 8 = gray, 9 = white, + 5% gold, +10% silver, +20% no color

FULL-LOAD CURRENT IN AMPERES FOR DIRECT-CURRENT MOTORS

COMPARISON OF AC AND DC CURRENTS

Alternating current or AC is defined as follows.

Alternating current is an electrical flow of electrons that is constantly changing in amplitude (amount of electron flow, size, or magnitude of current) and is periodically changing direction. Just what this mean? As you know, current supplied by a battery is direct current. This means that the current flows in one direction only: from negative to positive. Therefore, it is unidirectional. Alternating current on the other hand, is bi-directional, that is, the electrons flow first in one direction and then in the opposite direction. The voltage that produces this change in current flow does not change polarity instantaneously. Starting at zero, it increases to a maximum amplitude of one polarity and then decreases to zero. It then increases to a maximum amplitude of the opposite polarity and again decreases to zero. This produces a current flow, which reverses direction and is continuously changing in amplitude. Thus, the number of electrons flowing changes from zero to maximum, back to zero, to maximum in the opposite direction, and back to zero again.

Since it takes time for the current to build up and to decrease, we can show a curve (illustration) representing this changing current by plotting a graph of the current in relation to time. Such a curve is shown in the following figure.

AC Sine Wave Terminology: Notice that the horizontal line representing zero divides the sine wave into two equal parts - one above the line and the other below it. The portion above the line represents the positive alternation of an alternating current. The portion below the line represents the negative alternation. Notice that the wave reaches its maximum swing from the zero reference line at 90 degrees and 270 degrees. Each of these points is called the peak of the sine wave. When we speak of the amplitude of a sine wave, we mean the maximum swing, height, or size of one of the alternations at its peak. These terms apply to either current or voltage and are important to remember. As an example, suppose you have a sine wave with an amplitude of four centimeters; that is, the peak of the sine wave is four centimeters from the zero reference line. Let us also suppose that each centimeter of deflection represents ten volts. Knowing these facts, you can find the amplitude of the voltage that the sine wave represents by multiplying the number of centimeters by the value of voltage each represents (4 x 10 = 40). Thus, you can find the peak amplitude is 40 volts.

This curve is identical to one produced by plotting the sine values of a vector rotated through 360 degrees. Therefore, we call it a sine -wave. The form or shape of the sine wave can represent either alternating vol-tage or alternating current. Let us examine an AC sine wave in greater detail, using the following figure.

Next, let us take the term: peak-to-peak. This term, as you can see in the above figure, represents the difference in value between the positive and negative peaks of a sine wave. Of course, this is equal to twice the peak value: E pk-pk = 2 E pk. Why is this so important? There is one reason.

Suppose that when you were finding the amplitude of a sine wave, as you did in the previous problem, there was no zero reference time and the positive and negative peaks of the sine wave were five centimeters apart. It would be difficult to determine the exact zero reference. Therefore, it would be hard to obtain an accurate peak displacement of the sine wave. However, by using the peak-to-peak value and dividing it by two, you obtain the accurate amplitude. The peak-to-peak value is 50 volts (5 x 10 = 50). The peak value is 25 volts (50 + 2 = 25).

Since the amplitude of a sine wave varies continuously between zero and maximum (peak) values, first in one direction and then in the other, the peak value is not too useful in describing the value of a voltage or current. A more useful value is the effective value. The effective value of alternating current is the amount of AC, which produces the same heating effect as an equal amount of DC. Since the heating effect of a quantity of current is proportional to the square of the current, it is possible to calculate the effective value of alternating current by squaring the values of all points on a sine wave, taking the average of these values, and extracting the square root. The effective value is, thus, the root of the mean (average) square of these values. Therefore, it is also known as the “root-mean-square,” or RMS value. When we speak of household voltage as having a value of 110 volts, we mean that it has an effective or RMS value of 110 volts.

The number .707 is the root of the mean square of the values of a sine wave. This means that the effective value of a voltage or current is .707 times the maximum or peak value.

peak effective E

E =0.707× _or E_peak =1.414×E_effective

Therefore, if a since wave had a peak amplitude of 10 volts, the effective voltage would be 7.07 volts (.707 x 10 = 7.07). The reciprocal of .707 is 1.414. Therefore, if you know the effective value of a voltage, you can find the peak value by multiplying the effective value by 1.414.

Another value in a sine wave that is important to know is the average value. This is the average value of all points in the positive alternation. The relationship between, the average, effective, and peak values is shown in the following formula:

effective peak

average E E

E =0.636× =0.9×

As you can see, average voltage is equal to .636 of the peak or maximum voltage and .9 of the effective

PEAK TO PEAK PEAK POSITIVE AMPLITUDE NEGATIVE AMPLITUDE POSITIVE

ALTERNATION ALTERNATIONNEGATIVE

0º 90º 180º _270º

360º

PEAK 0

voltage. The reciprocal of .9 is 1.11. Therefore, the effective voltage is 1.11 times the average voltage.

average effective E

E =1.11×

Now, let us go back to the sine wave and learn some additional terms, using Figure 1A-3 (on next page). As we stated before, alternating current is periodically reversing direction. We call two consecutive alternations, one positive, and one negative, a cycle. We often refer to the positive and negative alternations as half cycles. In describing the sine wave, we could say that during the positive half cycle, it rises from zero to maximum and then falls back to zero, and that during the negative half cycle, it drops to a maximum negative value and then returns to zero.

We also stated earlier that the alternations of AC do not happen instantaneously. They take time. For example, household current is usually sixty cycles per second. This means that a single cycle requires a period of time equal to one-sixtieth of a second. The term period is used to define the time of one cycle of alternating current. Another term having the same meaning as time and period is duration. Let us see how each of these terms could be used when speaking of household current. We could say, “The duration of one cycle is one-sixtieth of a second”; or, “One cycle has a period of one-sixtieth of a second”; or, “One-sixtieth of a second is the time of one cycle.” As you can see, all three terms can be used in each of the statements without changing the meaning.

All alternating current does not have a period of one-sixtieth of a second. The period of a commonly used current in aircraft (for example) is one-four-hundredth of a second. This means that in one second there are four hundred complete cycles. The difference in the number of cycles per second of household current and aircraft current brings up a new term - frequency. The frequency of any AC is the number of cycles that occur in one second. The frequency of household current is sixty cycles per second. The frequency of aircraft current is four hundred

cycles per second.

As you can see in the figure, there is a definite relationship the period of an alternating current and the frequency of the current. Notice that sine wave A has a period that is twice the period of sine wave B and its frequency is only one-half the frequency of sine wave B. It is important to remember that, as the period of time for one cycle becomes shorter, the frequency increases.

In electronics, sixty cycles and four hundred cycles per second are relatively low frequencies. When we talk of frequencies in the thousands or millions, it becomes awkward and cumbersome to use cycles as a standard measuring term. Therefore, when we speak of frequencies in the thousands, we use the term kilocycles; and when we speak of frequencies in the millions, we use the term megacycles. You no doubt have heard these terms on radio and television when the announcer identifies the station. The term kilo means thousand. The term mega stands for million. Therefore, kilocycle means a thousand cycles and megacycle means a million cycles.

FULL CYCLE

POSITIVE

HALF CYCLE HALF CYCLENEGATIVE PERIOD TIME DURATION

ONE CYCLE

PERIOD

PERIOD

TWO CYCLES

ADDITIONAL SINE

When we speak of a frequency of 10 kilocycles, we mean that there are 10,000 cycles per second. When we speak of a frequency of 600 megacycles, we are speaking of 600,000,000 cycles per second. After learning these terms, you can easily see how much simpler it is to use them when referring to high frequencies. It’s like using miles in place of feet when measuring long distances.

Since the period of a cycle becomes smaller as the frequency increases, we not only need new terms for very high values, but we also need new terms for very small values. Therefore, we use the prefixes milli and micro when we talk of a thousandth or a millionth part of something. When we talk of a thousandth of a second, we say millisecond. We use microsecond when we talk of a millionth of a second.

Now, let us go back to the sine wave and take up a few more points that we should know when working in electronics. An AC sine wave is a pictorial presentation of alternating current. It shows the direction of current by rising above, and dropping below, the zero reference line. The current reverses direction at the zero reference line at the beginning of each half cycle. There are two current reversals in each cycle.

Since the frequency of an AC is the number of cycles per second, we can find the number of current reversals in an AC by multiplying the frequency by two. A sixty-cycle current has one-hundred-twenty current reversals per second (60 x 2 = 120). A sine wave can represent either current or voltage. Since voltage is used to produce current, let us compare the sine waves in a different way than we have before Sine Wave Phase Relationship

This figure shows three sine waves. The top wave represents voltage. The two other waves represent the current produced by the voltage represented by the top wave.

Notice that the voltage sine wave A and the current sine wave B move together. That is, they start at zero, rise to maximum at ninety degrees, return to zero at one-hundred eighty degrees, and repeat the same action in the negative half cycle. As the two sine waves go through all phases of the cycle together, we say that they are in-phase. We do not limit the use of this term to just the voltage and the current produced by the voltage. We also speak of two alternating voltages or two alternating currents, which rise and fall together as being in-phase.

Current is in-phase with the voltage only in circuits, which are purely resistive. There are certain electrical components (inductive or capacitive)

which affect the phase relationship in AC circuits. Whenever one of these components is included in a circuit, an out-of-phase relationship exists between current and voltage. An out-of-phase relationship exists any time that the waves do not move simultaneously. This out-of-phase relationship can be seen by comparing sine wave A and sine wave C shown in Figure above.

As you can see, the current sine wave is maximum negative when the voltage sine wave is zero at the start of the positive half cycle. It reaches zero when the voltage goes maximum positive, and it reaches maximum positive ninety degrees later at the time that the voltage is zero. When describing this relationship, we say that the current lags the voltage by ninety degrees. If the action of the current sine wave occurs before that of the voltage, we say that the current leads the voltage.

0 0 0 0º 90º 180º 270º 360º A B C VOLTAGE CURRENT IN PHASE CURRENT OUT OF PHASE

PREFIXES USED FOR SUBDIVISIONS AND MULTIPLES OF UNITS

MULTIPLE PREFIX SYMBOL

1018 _{1,000,000,000,000,000,000 exa E} 1015 _{1,000,000,000,000,000 peta P} 1012 _{1,000,000,000,000 tera T} 109 _{1,000,000,000 giga G} 106 _{1,000,000 mega M} 103 _{1000 kilo K} 102 _{100 hecto h}* 101 _{10 deka da}* 10-1 _{.1 deci d}* 10-2 _{.01 centi c}* 10-3 _{.001 milli M} 10-6 _{.000001 micro Μ} 10-9 _{.000000001 nano N} 10-12 _{.000000000001 pico P} 10-15 _{.000000000000001 femto F} 10-18 _{.000000000000000001 atto a}