To Find  Single Phase  Two Phase Four Wire  Three Phase  Direct Current 
Amps when “HP” is known  HP x 746
V x %EFF x PF 
HP x 746
V x %EFF x PF x 2 
HP x 746
V x %EFF x PF x 1.73 
HP x 746
V x %EFF 
Amps when “KW”is known  KW x 1000
V x PF 
KW x 1000
V x PF x 2 
KW x 1000
V x PF x 1.73 
KW x 1000
V 
Amps when “kVA”is known  kVA x 1000
V 
kVA x 1000
V x 2 
kVA x 1000
V x 1.73 
– 
Kilowatts  V x A x PF
1000 
V x A x PF x 2
1000 
V x A x PF x 1.73
1000 
V x A
1000 
KilovoltAmps “kVA”  V x A
1000 
V x A x 2
1000 
V x A x 1.73
1000 
– 
HP  V x A x%EFFx PF
746 
VxA x %EFF x PF x 2
746 
VxAx%EFFxPFx1.73
746 
V x A X %EFF
746 
V=VOLTS A=AMPS EFF=EFFICIENCY PF=POWER FACTOR
 volts x amps = watts
Same as P = VI … Power (watts) = V (volts) x I (amps)
Example: 4500 watt element, 240 volt water heater circuit, how many amps?
Amps = 4500 watt ÷ 240 volt = 18.75 amps
 1000 watts = 1 kilowatt Kw
.002931 Kw needed to raise 1 pound of water 1°F
 Resistance Ohms = volts² ÷ watts
Example: test 4500 watt element 240 volt water heater
Correct ohm reading = 240² ÷ 4500 = 12.8 ohms
 Resistance Ohms = volts ÷ amps
E = IR
volts = amps x resistance (ohms)
1 horsepower = 745.6998 watts
Average efficiency of Motor Power Factor
When an actual efficiency and power factor of the motor to be controlled are not known, then the following approximation
Type  Power Factor 
DC motor, 35HP and less  80% to 85% 
DC motor, above 35HP  85% to 90% 
Synchronous Motor at 100% Power factor  92% to 95% 
Three Phase Induction Motors, 25HP and less  70% 
Three Phase Induction Motors, above 25HP  80% 
USEFUL ELECTRICAL FORMULAS
kVA/kW AMPERAGE CHART 80% POWER FACTOR
kVA  kW  208V  220V  240V  380V  400V  440V  450V  480V  600V  2400V  3300V  4160V 
6.3  5  17.5  16.5  15.2  9.6  9.1  8.3  8.1  7.6  6.1  –  –  – 
9.4  7.5  26.1  24.7  22.6  14.3  13.6  12.3  12  11.3  9.1  –  –  – 
12.5  10  34.7  33  30.1  19.2  18.2  16.6  16.2  15.1  12  –  –  – 
18.7  15  52  49.5  45  28.8  27.3  24.9  24.4  22.5  18  –  –  – 
25  20  69.5  66  60.2  38.4  36.4  33.2  32.4  30.1  24  6  4.4  3.5 
31.3  25  87  82.5  75.5  48  45.5  41.5  40.5  37.8  30  7.5  5.5  4.4 
37.5  30  104  99  90.3  57.6  54.6  49.8  48.7  45.2  36  9.1  6.6  5.2 
50  40  139  132  120  77  73  66.5  65  60  48  12.1  8.8  7 
62.5  50  173  165  152  96  91  83  81  76  61  15.1  10.9  8.7 
75  60  208  198  181  115  109  99.6  97.5  91  72  18.1  13.1  10.5 
93.8  75  261  247  226  143  136  123  120  113  90  22.6  16.4  13 
100  80  278  264  240  154  146  133  130  120  96  24.1  17.6  13.9 
125  100  347  330  301  192  182  166  162  150  120  30  21.8  17.5 
156  125  433  413  375  240  228  208  204  188  150  38  27.3  22 
187  150  520  495  450  288  273  249  244  225  180  45  33  26 
219  175  608  577  527  335  318  289  283  264  211  53  38  31 
250  200  694  660  601  384  364  332  324  301  241  60  44  35 
312  250  866  825  751  480  455  415  405  376  300  75  55  43 
375  300  1040  990  903  576  546  498  487  451  361  90  66  52 
438  350  1220  1155  1053  672  637  581  568  527  422  105  77  61 
500  400  1390  1320  1203  770  730  665  650  602  481  120  88  69 
625  500  1735  1650  1504  960  910  830  810  752  602  150  109  87 
750  600  2080  1980  1803  1150  1090  996  975  902  721  180  131  104 
875  700  2430  2310  2104  1344  1274  1162  1136  1052  842  210  153  121 
1000  800  2780  2640  2405  1540  1460  1330  1300  1203  962  241  176  139 
1125  900  3120  2970  2709  1730  1640  1495  1460  1354  1082  271  197  156 
1250  1000  3470  3300  3009  1920  1820  1660  1620  1504  1202  301  218  174 
1563  1250  4350  4130  3765  2400  2280  2080  2040  1885  1503  376  273  218 
1875  1500  5205  4950  4520  2880  2730  2490  2440  2260  1805  452  327  261 
2188  1750  –  –  5280  3350  3180  2890  2830  2640  2106  528  380  304 
2500  2000  –  –  6020  3840  3640  3320  3240  3015  2405  602  436  348 
2812  2250  –  –  6780  4320  4095  3735  3645  3400  2710  678  491  392 
3125  2500  –  –  7520  4800  4560  4160  4080  3765  3005  752  546  435 
3750  3000  –  –  9040  5760  5460  4980  4880  4525  3610  904  654  522 
4375  3500  –  –  10550  6700  6360  5780  5660  5285  4220  1055  760  610 
5000  4000  –  –  12040  7680  7280  6640  6480  6035  4810  1204  872  695 
BASIC ELECTRICAL ENGINEERING FORMULAS
BASIC ELECTRICAL CIRCUIT FORMULAS
E=Irms^{2}R×tENERGY (dissipated on R or stored in L, C)
E=Li^{2}/2
E=Cv^{2}/2
Notes:
R electrical resistance in ohms, L inductance in henrys, C capacitance in farads, f – frequency in hertz, t time in seconds, π≈3.14159;
ω=2πf – angular frequency; j – imaginary unit ( j^{2}=1 )
The most common used electrical formulas – Ohms Law and combinations
Common electrical units used in formulas and equations are:
 Volt – unit of electrical potential or motive force – potential is required to send one ampere of current through one ohm of resistance
 Ohm – unit of resistance – one ohm is the resistance offered to the passage of one ampere when impelled by one volt
 Ampere – units of current – one ampere is the current which one volt can send through a resistance of one ohm
 Watt – unit of electrical energy or power – one watt is the product of one ampere and one volt – one ampere of current flowing under the force of one volt gives one watt of energy
 Volt Ampere – product of volts and amperes as shown by a voltmeter and ammeter – in direct current systems the volt ampere is the same as watts or the energy delivered – in alternating current systems – the volts and amperes may or may not be 100% synchronous – when synchronous the volt amperes equals the watts on a wattmeter – when not synchronous volt amperes exceed watts – reactive power
 Kilovolt Ampere – one kilovolt ampere – KVA – is equal to 1000 volt amperes
 Power Factor – ratio of watts to volt amperes
Electric Power Formulas
P = V I (1a)
P = R I^{2} (1b)
P = V^{2}/ R (1c)
where
P = power (watts, W)
V = voltage (volts, V)
I = current (amperes, A)
R = resistance (ohms, Ω)
Electric Current Formulas
I = V / R (2a)
I = P / V (2b)
I = (P / R)^{1/2} (2c)
Electric Resistance Formulas
R = V / I (3a)
R = V^{2}/ P (3b)
R = P / I^{2} (3c)
Electrical Potential Formulas – Ohms Law
Ohms law can be expressed as:
V = R I (4a)
V = P / I (4b)
V = (P R)^{1/2} (4c)
Example – Ohm’s law
A 12 volt battery supplies power to a resistance of 18 ohms.
I = (12 V) / (18 Ω)
= 0.67 (A)
Electrical Motor Formulas
Electrical Motor Efficiency
μ = 746 P_{hp} / P_{input_w} (5)
where
μ = efficiency
P_{hp} = output horsepower (hp)
P_{input_w} = input electrical power (watts)
or alternatively
μ = 746 P_{hp} / (1.732 V I PF) (5b)
Electrical Motor – Power
P_{3phase} = (V I PF 1.732) / 1,000 (6)
where
P_{3phase} = electrical power 3phase motor (kW)
PF = power factor electrical motor
Electrical Motor – Amps
I_{3phase} = (746 P_{hp}) / (1.732 V μ PF) (7)
where
I_{3phase} = electrical current 3phase motor (amps)
PF = power factor electrical motor
ThreePhase Power Equations
Electrical 3phase equations
Most AC power today is produced and distributed as threephase power where three sinusoidal voltages are generated out of phase with each other. With singlephase AC power there is only one single sinusoidal voltage.
Real Power
W_{applied} = 3^{1/2} U I cos Φ
= 3^{1/2} U I PF (1)
where
W_{applied} = real power (W, watts)
U = voltage (V, volts)
I = current (A, amps)
PF = cos Φ = power factor (0.7 – 0.95)
For pure resistive load: PF = cos Φ = 1
 resistive loads converts current into other forms of energy, such as heat
 inductive loads use magnetic fields like motors, solenoids, and relays
Example – Pure Resistive Load
For pure resistive load and power factor = 1 the real power in a 415 voltage 20 amps circuit can be calculated as
W_{applied} = 3^{1/2} 415 (V) 20 (A) 1
= 14400 W
= 14.4 kW
Total Power
W = 3^{1/2} U I (2)
Brake Horsepower
W_{BHP} = 3^{1/2} U I PF μ / 746 (3)
where
W_{BHP} = brake horse power (hp)
μ = device efficiency
Converting Ampere between Single Phase and 3 Phase
Converting amperage between single phase (120, 240 and 480 Voltage) and three phase (240 and 480 Voltage)
With singlephase AC power there is only one single sinusoidal voltage. Most AC power today is produced and distributed as threephase power where three sinusoidal voltages are generated out of phase with each other.
Sometimes it is necessary to turn between power (VA), voltage (V) and amperage (A). The diagram and table below can be used to convert amperage between single phase and three phase equipment and vice versa.
Power (VA) 
Amperage (A)  
Volts Single Phase  Volts 3 Phase Balanced Load 

120  240  480  240  480  
100 150 200 250 300 
0.83 1.25 1.67 2.08 2.50 
0.42 0.63 0.83 1.04 1.25 
0.21 0.31 0.42 0.52 0.63 
0.24 0.36 0.49 0.61 0.73 
0.13 0.18 0.25 0.30 0.37 
350 400 450 500 600 
2.92 3.33 3.75 4.17 5.00 
1.46 1.67 1.88 2.08 2.50 
0.73 0.84 0.93 1.04 1.25 
0.85 0.97 1.10 1.20 1.45 
0.43 0.49 0.55 0.60 0.73 
700 750 800 900 1,000 
5.83 6.25 6.67 7.50 8.33 
2.92 3.13 3.33 3.75 4.17 
1.46 1.56 1.67 1.87 2.10 
1.70 1.81 1.93 2.17 2.41 
0.85 0.91 0.97 1.09 1.21 
1,100 1,200 1,250 1,300 1,400 
9.17 10.0 10.4 10.8 11.7 
4.58 5.00 5.21 5.42 5.83 
2.30 2.51 2.61 2.71 2.91 
2.65 2.90 3.10 3.13 3.38 
1.33 1.45 1.55 1.57 1.69 
1,500 1,600 1,700 1,750 1,800 
12.5 13.3 14.2 14.6 15.0 
6.25 6.67 7.08 7.29 7.50 
3.12 3.34 3.54 3.65 3.75 
3.62 3.86 4.10 4.22 4.34 
1.82 1.93 2.05 2.10 2.17 
1,900 2,000 2,200 2,500 2,750 
15.8 16.7 18.3 20.8 23.0 
7.92 8.33 9.17 10.4 11.5 
3.96 4.17 4.59 5.21 5.73 
4.58 4.82 5.30 6.10 6.63 
2.29 2.41 2.65 3.05 3.32 
3,000 3,500 4,000 4,500 5,000 
25.0 29.2 33.3 37.5 41.7 
12.5 14.6 16.7 18.8 20.8 
6.25 7.30 8.33 9.38 10.42 
7.23 8.45 9.64 10.84 12.1 
3.62 4.23 4.82 5.42 6.1 
6,000 7,000 8,000 9,000 10,000 
50.0 58.3 66.7 75.0 83.3 
25.0 29.2 33.3 37.5 41.7 
12.50 14.59 16.67 18.75 20.85 
14.50 16.9 19.3 21.7 24.1 
7.25 8.5 9.65 10.85 12.1 
Asynchronous Motors – Electrical Data
Electrical motor data – nominal current, fuse, start ampere, contactor and circuit breaker of asynchronous motors
The table below can used to determine electrical data for asynchronous 380 Voltage motors.
380 Voltage 50 Hz motors have been commonly used in Europe. Note that the nominal voltage of existing 220/380 V and 240/415 V systems shall evolve toward the IEC recommended value of 230/400 V.
Rated Power  Nominal current – I_{n} – _{ }(A) 
Directly Fused (A) 
Star – Delta Started (A) 
Star – Delta contactor – I_{n} –_{ }(A) 
Circuit Breaker – I_{n} – (A) 

kW  HP  
0.2  0.3  0.7  2  2  16  
0.33  0.5  1.1  2  2  16  
0.5  0.7  1.4  2  2  16  
0.8  1.1  2.1  4  4  16  
1.1  1.5  2.6  4  4  16  
1.5  2  3.6  6  4  (16)22  16 
2.2  3  5.0  10  6  (16)22  16 
3  4  6.6  16  10  (16)22  16 
4  5.5  8.5  20  16  (16)22  16 
5.5  7.5  11.5  25  20  (16)22  16 
7.5  10  15.5  35  25  (25)22  25 
11  15  22.2  35  35  (40)30  40 
15  20  30  50  35  (40)30  40 
22  30  44  63  50  (63)/60  60 
30  40  57  80  63  (63)/60  60 
45  66  85  125  100  90  100 
55  75  104  160  125  110  100 
75  100  140  200  160  150  200 
90  125  168  225  200  220  200 
110  150  205  300  250  220  200 
132  180  245  400  300  300  400 
160  220  290  430  300  300  400 
200  270  360  500  430  480  400 
240  325  430  630  500  480  480 
Fullvoltage, singlespeed motor starters
Fullvoltage starters (manual and magnetic) apply full voltage directly to motor terminals.
Reducedvoltage, singlespeed motor starters
Some machines or loads may require a gentle start and smooth acceleration up to full speed.
Many starters apply reduced voltage to motor windings, primary resistor, primary reactor, autotransformer and solid state. Part winding and wyedelta starters can also provide reducedvoltage starting, although technically they are not reducedvoltage starters.
Motor Protection
Motors should have protection for themselves, in the branch circuit, and in the feeder line. Protection provided by fuses and circuit breakers guards against fault conditions caused by short circuits or grounds and over currents exceeding lockedrotor values.
Electrical Units
Definition of common electrical units – like Ampere, Volt, Ohm, Siemens
Ampere – A
The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to 2 x 10^{7} Newton per meter of length.
Electric current is the same as electric quantity in movement, or quantity per unit time, expressed like
I = dq / dt
where
I = electric current (ampere, A)
dq = electric quantity (coulomb, C)
dt = time (s)
Coulomb – C
The standard unit of quantity in electrical measurements. It is the quantity of electricity conveyed in one second by the current produced by an electromotive force of one volt acting in a circuit having a resistance of one ohm, or the quantity transferred by one ampere in one second.
Farad – F
The farad is the standard unit of capacitance. Reduced to base SI units one farad is the equivalent of one second to the fourth power ampere squared per kilogram per meter squared (s^{4 }A^{2}/kg m^{2}).
When the voltage across a 1 F capacitor changes at a rate of one volt per second (1 V/s) a current flow of 1 Aresults. A capacitance of 1 F produces 1 V of potential difference for an electric charge of one coulomb (1 C).
In common electrical and electronic circuits units of microfarads μF (1 μF = 10^{6} F) and picofarads pF (1 pF = 10^{12}F) are used.
Ohm – Ω
The derived SI unit of electrical resistance – the resistance between two points on a conductor when a constant potential difference of 1 volt between them produces a current of 1 ampere.
Henry – H
The Henry is the unit of inductance. Reduced to base SI units one henry is the equivalent of one kilogram meter squared per second squared per ampere squared (kg m^{2} s^{2} A^{2}).
Inductance
An inductor is a passive electronic component that stores energy in the form of a magnetic field.
The standard unit of inductance is the henry abbreviated H. This is a large unit and more commonly used units are the microhenry abbreviated μH (1 μH =10^{6}H) and the millihenry abbreviated mH (1 mH =10^{3} H). Occasionally, thenanohenry abbreviated nH (1 nH = 10^{9} H) is used.
Joule – J
The unit of energy work or quantity of heat done when a force of one Newton is applied over a displacement of one meter. One joule is the equivalent of one watt of power radiated or dissipated for one second.
In imperial units the British Thermal Unit (Btu) is used to express energy. One Btu is equivalent to approximately1,055 joules.
Siemens – S
The unit of electrical conductance S = A / V
Watt
The watt is used to specify the rate at which electrical energy is dissipated, or the rate at which electromagnetic energy is radiated, absorbed, or dissipated.
The unit of power W or Joule/second
Weber – Wb
The unit of magnetic flux.
The flux that when linking a circuit of one turn, produces an Electro Motive Force – EMF – of 1 volt as it is reduced to zero at a uniform rate in one second.
 1 Weber is equivalent to 10^{8} Maxwells
Tesla – T
The unit of magnetic flux density the Tesla is equal to 1 Weber per square meter of circuit area.
Volt
The Volt – V – is the Standard International (SI) unit of electric potential or electromotive force. A potential of one volt appears across a resistance of one ohm when a current of one ampere flows through that resistance.
Reduced to SI base units,
1 (V) = 1 (kg m^{2} / s^{3} A)
Electromotive Force – e.m.f
Change in electrical potential between two points
A change in electrical potential between two points in an electric circuit is called a “potential difference”. The “electromotive force (e.m.f)” provided by a source of energy – such as a generator or battery – is measured in volts.
One volt i one joule per coulomb and is defined as the difference in potential between two points in a conductor which, when carrying a current of one ampere, dissipates a power of one watt:
volts = watts / amperes = (joules/sec) / amperes = joules / ampere seconds = joules / coulombs
Resistivity, Conductivity and Temperature Coefficients for some Common Materials
Resistivity, conductivity and temperature coefficients for some common materials as silver, gold, platinum, iron and more – Including a tutorial explanation of resistivity and conductivity
The factor in the resistance which takes into account the nature of the material is the resistivity.
Resistivity is the
 resistance of a unit cube of the material measured between the opposite faces of the cube
Material  Resistivity Coefficient ^{2)
}– ρ – (ohm m) 
Temperature Coefficient ^{2)} (per degree C, 1/^{o}C) 
Conductivity – σ – (1 /Ωm) 
Aluminum  2.65 x 10^{8}  3.8 x 10^{3}  3.77 x 10^{7} 
Animal fat  14 x 10^{2}  
Animal muscle  0.35  
Antimony  41.8 x 10^{8}  
Barium (0^{o}C)  30.2 x 10^{8}  
Beryllium  4.0 x 10^{8}  
Bismuth  115 x 10^{8}  
Brass – 58% Cu  5.9 x 10^{8}  1.5 x 10^{3}  
Brass – 63% Cu  7.1 x 10^{8}  1.5 x 10^{3}  
Cadmium  7.4 x 10^{8}  
Caesium (0^{o}C)  18.8 x 10^{8}  
Calcium (0^{o}C)  3.11 x 10^{8}  
Carbon (graphite)^{1)}  3 – 60 x 10^{5}  4.8 x 10^{4}  
Cast iron  100 x 10^{8}  
Cerium (0^{o}C)  73 x 10^{8}  
Chromel (alloy of chromium and aluminum)  0.58 x 10^{3}  
Chromium  13 x 10^{8}  
Cobalt  9 x 10^{8}  
Constantan  49 x 10^{8}  3 x 10^{5}  0.20 x 10^{7} 
Copper  1.724 x 10^{8}  4.29 x 10^{3}  5.95 x 10^{7} 
Dysprosium (0^{o}C)  89 x 10^{8}  
Erbium (0^{o}C)  81 x 10^{8}  
Eureka  0.1 x 10^{3}  
Europium (0^{o}C)  89 x 10^{8}  
Gadolium  126 x 10^{8}  
Gallium (1.1K)  13.6 x 10^{8}  
Germanium^{1)}  1 – 500 x 10^{3}  50 x 10^{3}  
Glass  1 – 10000 x 10^{9}  10^{12}  
Gold  2.24 x 10^{8}  
Graphite  800 x 10^{8}  2.0 x 10^{4}  
Hafnium (0.35K)  30.4 x 10^{8}  
Holmium (0^{o}C)  90 x 10^{8}  
Indium (3.35K)  8 x 10^{8}  
Iridium  5.3 x 10^{8}  
Iron  9.71 x 10^{8}  6.41 x 10^{3}  1.03 x 10^{7} 
Lanthanum (4.71K)  54 x 10^{8}  
Lead  20.6 x 10^{8}  0.45 x 10^{7}  
Lithium  9.28 x 10^{8}  
Lutetium  54 x 10^{8}  
Magnesium  4.45 x 10^{8}  
Manganese  185 x 10^{8}  1.0 x 10^{5}  
Mercury  98.4 x 10^{8}  8.9 x 10^{3}  0.10 x 10^{7} 
Mica  1 x 10^{13}  
Mild steel  15 x 10^{8}  6.6 x 10^{3}  
Molybdenum  5.2 x 10^{8}  
Neodymium  61 x 10^{8}  
Nichrome (alloy of nickel and chromium)  100 – 150 x 10^{8}  0.40 x 10^{3}  
Nickel  6.85 x 10^{8}  6.41 x 10^{3}  
Nickeline  50 x 10^{8}  2.3 x 10^{4}  
Niobium (Columbium)  13 x 10^{8}  
Osmium  9 x 10^{8}  
Palladium  10.5 x 10^{8}  
Phosphorus  1 x 10^{12}  
Platinum  10.5 x 10^{8}  3.93 x 10^{3}  0.943 x 10^{7} 
Plutonium  141.4 x 10^{8}  
Polonium  40 x 10^{8}  
Potassium  7.01 x 10^{8}  
Praseodymium  65 x 10^{8}  
Promethium  50 x 10^{8}  
Protactinium (1.4K)  17.7 x 10^{8}  
Quartz (fused)  7.5 x 10^{17}  
Rhenium (1.7K)  17.2 x 10^{8}  
Rhodium  4.6 x 10^{8}  
Rubber – hard  1 – 100 x 10^{13}  
Rubidium  11.5 x 10^{8}  
Ruthenium (0.49K)  11.5 x 10^{8}  
Samarium  91.4 x 10^{8}  
Scandium  50.5 x 10^{8}  
Selenium  12.0 x 10^{8}  
Silicon^{1)}  0.160  70 x 10^{3}  
Silver  1.59 x 10^{8}  6.1 x 10^{3}  6.29 x 10^{7} 
Sodium  4.2 x 10^{8}  
Soil, typical ground  10^{2} – 10^{4}  
Solder  15 x 10^{8}  
Stainless steel  10^{6}  
Strontium  12.3 x 10^{8}  
Sulfur  1 x 10^{17}  
Tantalum  12.4 x 10^{8}  
Terbium  113 x 10^{8}  
Thallium (2.37K)  15 x 10^{8}  
Thorium  18 x 10^{8}  
Thulium  67 x 10^{8}  
Tin  11.0 x 10^{8}  4.2 x 10^{3}  
Titanium  43 x 10^{8}  
Tungsten  5.65 x 10^{8}  4.5 x 10^{3}  1.79 x 10^{7} 
Uranium  30 x 10^{8}  
Vanadium  25 x 10^{8}  
Water, distilled  10^{4}  
Water, fresh  10^{2}  
Water, salt  4  
Ytterbium  27.7 x 10^{8}  
Yttrium  55 x 10^{8}  
Zinc  5.92 x 10^{8}  3.7 x 10^{3}  
Zirconium (0.55K)  38.8 x 10^{8} 
^{1)} Note! – the resistivity depends strongly on the presence of impurities in the material.
^{2)} Note! – the resistivity depends strongly on the temperature of the material. The table above is based on 20^{o}C reference.
The electrical resistance of a wire is greater for a longer wire and less for a wire of larger cross sectional area. The resistance depend on the material of which it is made and can be expressed as:
R = ρ L / A (1)
where
R = resistance (ohm, Ω)
ρ = resistivity coefficient (ohm m, Ω m)
L = length of wire (m)
A = cross sectional area of wire (m^{2})
The factor in the resistance which takes into account the nature of the material is the resistivity. Since it is temperature dependent, it can be used to calculate the resistance of a wire of given geometry at different temperatures.
The inverse of resistivity is called conductivity and can be expressed as:
σ = 1 / ρ (2)
where
σ = conductivity (1 / Ω m)
Example – Resistance in an Aluminum Cable
Resistance of an aluminum cable with length 10 m and cross sectional area of 3 mm^{2} can be calculated as
R = (2.65 10^{8} Ω m) (10 m) / ((3 mm^{2}) (10^{6} m^{2}/mm^{2}))
= 0.09 Ω
Resistance
The electrical resistance of a circuit component or device is defined as the ratio of the voltage applied to the electric current which flows through it:
R = V / I (3)
where
R = resistance (ohm)
V = voltage (V)
I = current (A)
Ohm’s Law
If the resistance is constant over a considerable range of voltage, then Ohm’s law,
I = V / R (4)
can be used to predict the behavior of the material.
Temperature Coefficient of Resistance
The electrical resistance increases with temperature. An intuitive approach to temperature dependence leads one to expect a fractional change in resistance which is proportional to the temperature change:
dR / R_{s} = α dT (5)
where
dR = change in resistance (ohm)
R_{s} = standard resistance according reference tables (ohm)
α = temperature coefficient of resistance
dT = change in temperature (K)
(5) can be modified to:
dR = α dT R_{s} (5b)
Example – Resistance of a Carbon resistor when changing Temperature
A carbon resistor with resistance 1 kΩ is heated 100 ^{o}C. With a temperature coefficient 4.8 x 10^{4} (1/^{o}C) the resistance change can be calculated as
dR = (4.8 x 10^{4} 1/^{o}C) (100 ^{o}C) (1 kΩ)
= – 0.048 (kΩ)
The resulting resistance for the resistor
R = (1 kΩ) – (0.048 kΩ)
= 0.952 (kΩ)
= 952 (Ω)
Power Factor ThreePhase Electrical Motors
Power Factor definition for threephase electrical motors
The power factor of an AC electric power system is defined as the ratio of the active (true or real) power to theapparent power
where
 Active (Real or True) Power is measured in watts and is the power drawn by the electrical resistance of a system doing useful work
 Apparent Power is measured in voltamperes (VA) and is the voltage on an AC system multiplied by all the current that flows in it. It is the vector sum of the active and the reactive power
 Reactive Power is measured in voltamperes reactive (VAR). Reactive Power is power stored in and discharged by inductive motors, transformers and solenoids
Reactive power is required for the magnetization of a motor but doesn’t perform any action. The reactive power required by inductive loads increases the amounts of apparent power – measured in kilovolt amps (kVA) – in the distribution system. Increasing of the reactive and apparent power will cause the power factor – PF – to decrease.
Power Factor
It is common to define the Power Factor – PF – as the cosine of the phase angle between voltage and current – or the “cosφ”.
PF = cos φ
where
PF = power factor
φ = phase angle between voltage and current
The power factor defined by IEEE and IEC is the ratio between the applied active (true) power – and theapparent power, and can in general be expressed as:
PF = P / S (1)
where
PF = power factor
P = active (true or real) power (Watts)
S = apparent power (VA, volts amps)
A low power factor is the result of inductive loads such as transformers and electric motors. Unlike resistive loads creating heat by consuming kilowatts, inductive loads require a current flow to create magnetic fields to produce the desired work.
Power factor is an important measurement in electrical AC systems because
 an overall power factor less than 1 indicates that the electricity supplier need to provide more generating capacity than actually required
 the current waveform distortion that contributes to reduced power factor is caused by voltage waveform distortion and overheating in the neutral cables of threephase systems
International standards such as IEC 6100032 have been established to control current waveform distortion by introducing limits for the amplitude of current harmonics.
Example – Power Factor
A industrial plant draws 200 A at 400 V and the supply transformer and backup UPS is rated 200 A × 400 V = 80 kVA.
If the power factor – PF – of the loads is only 0.7 – only
80 kVA × 0.7
= 56 kW
of real power is consumed by the system. If the power factor is close to 1 (purely resistive circuit) the supply system with transformers, cables, switchgear and UPS could be made considerably smaller.
Any power factor less than 1 means that the circuit’s wiring has to carry more current than what would be necessary with zero reactance in the circuit to deliver the same amount of (true) power to the resistive load.
A low power factor is expensive and inefficient and some utility companies may charge additional fees when the power factor is less than 0.95. A low power factor will reduce the electrical system’s distribution capacity by increasing the current flow and causing voltage drops.
“Leading” or “Lagging” Power Factors
Power factors are usually stated as “leading” or “lagging” to show the sign of the phase angle.
 With a purely resistive load current and voltage changes polarity in step and the power factor will be 1. Electrical energy flows in a single direction across the network in each cycle.
 Inductive loads – transformers, motors and wound coils – consumes reactive power with current waveform lagging the voltage.
 Capacitive loads – capacitor banks or buried cables – generates reactive power with current phase leading the voltage.
Inductive and capacitive loads stores energy in magnetic or electric fields in the devices during parts of the AC cycles. The energy is returned back to the power source during the rest of the cycles.
Power Factor for a ThreePhase Motor
The total power required by an inductive device as a motor or similar consists of
 Active (true or real) power (measured in kilowatts, kW)
 Reactive power – the nonworking power caused by the magnetizing current, required to operate the device (measured in kilovars, kVAR)
The power factor for a threephase electric motor can be expressed as:
PF = P / [(3)^{1/2} U I] (2)
where
PF = power factor
P = power applied (W, watts)
U = voltage (V)
I = current (A, amps)
Typical Motor Power Factors
Power (hp)  Speed (rpm)  Power Factor  
1/2 load  3/4 load  full load  
0 – 5  1800  0.72  0.82  0.84 
5 – 20  1800  0.74  0.84  0.86 
20 – 100  1800  0.79  0.86  0.89 
100 – 300  1800  0.81  0.88  0.91 
 1 hp = 745.7 W
Active, Apparent and Reactive Power
Apparent, True Applied and Reactive Power – kVA
Total electrical power consumption depends on real power – electrical energy consumption, and reactive power – imaginary power consumption, and can be expressed (power triangle or Pythagorean relationship).
S = (Q^{2} + P^{2})^{1/2} (1)
where
S = apparent power (kilovolt amps, kVA)
Q = reactive power (kilovolt amps reactive, kVAR)
P = active power (kilowatts, kW)
Apparent Power
Apparent Power is measured in voltamperes (VA) and is the voltage on an AC system multiplied by all the current that flows in it. It is the vector sum of the active and the reactive power.
Single Phase Current
S = U I (2a)
where
U = electric potential (V)
I = current (A)
Three Phase Current
S = 1.732 U I (2b)
Active Power
Active (Real or True) Power is measured in watts and is the power drawn by the electrical resistance of a system doing useful work.
Single Phase Current
P = U I cos φ (3a)
where
φ = phase angle
Three Phase Current
P = 1.732 U I cos φ (3b)
Direct Current
P = U I (3c)
Reactive Power
Reactive inductive Power – Q – is measured in voltamperes reactive (VAR) and is the power stored in and discharged by the inductive motors, transformers or solenoids.
Reactive power required by inductive loads increases the amount of apparent power – measured in kilovolt amps (kVA) – in the distribution system. Increasing the reactive and apparent power causes the power factor – PF – to decrease.
Single Phase Current
Q = U I sin φ (4a)
where
φ = phase angle
Three Phase Current
Q = 1.732 U I sin φ (4b)
Synchronous Speed of Electrical Motors
The speed at which an induction motor will operate depends on the input power frequency and the number of electrical magnetic poles in the motor
The synchronous speed for an electric induction motor is determined by the power supply frequency and the number of poles in the motor winding and can be expressed as:
ω = 2 · 60 · f / n (1)
where
ω = pump shaft rotational speed (rev/min, rpm)
f = frequency (Hz, cycles/sec)
n = number of poles
Synchronous rotation speed at different frequencies and number of poles
Shaft rotation speed – ω – (rev/min, rpm)  
Frequency – f – (Hz) 
Number of poles – n  
2  4  6  8  10  12  
10  600  300  200  150  120  100 
20  1200  600  400  300  240  200 
30  1800  900  600  450  360  300 
40  2400  1200  800  600  480  400 
50^{1)}  3000  1500  1000  750  600  500 
60^{2)}  3600  1800  1200  900  720  600 
70  4200  2100  1400  1050  840  700 
80  4800  2400  1600  1200  960  800 
90  5400  2700  1800  1350  1080  900 
100  6000  3000  2000  1500  1200  1000 
 Motors designed for 50 Hz are common outside U.S
 Motors designed for 60 Hz are common in U.S
Variable Frequency Drive
A variable frequency drive modulates the speed of an electrical motor by changing the frequency of the power supplied.
Induction Motors – Synchronous and Full Load Speed
Synchronous and full load speed of amplitude current (AC) induction motors
Synchronous and approximate full load speed of AC electrical motors:
Speed (rpm)  
Number of Poles  Frequency (Hz, cycles/sec)  
60  50  
Synchronous  Full Load  Synchronous  Full Load  
2  3600  3500  3000  2900 
4  1800  1770  1500  1450 
6  1200  1170  1000  960 
8  900  870  750  720 
10  720  690  600  575 
12  600  575  500  480 
14  515  490  429  410 
16  450  430  375  360 
18  400  380  333  319 
20  360  340  300  285 
22  327  310  273  260 
24  300  285  240  230 
26  277  265  231  222 
28  257  245  214  205 
30  240  230  200  192 
Frequency and Speed of Electrical Motors
The velocity of 2, 4, 6 and 8 poles electrical motors at 50 Hz and 60 Hz
The velocity at which an induction motor will operate depends on the input power frequency and the number of electrical magnetic poles in the motor.
Speed (RPM)  
Poles  Frequency 50 Hz  Frequency 60 Hz  
Synchronous  Full Load  Synchronous  Full Load  
2  3000  2850  3600  3450 
4  1500  1425  1800  1725 
6  1000  950  1200  1150 
8  750  700  900  850 
 Slip – in synchronous electrical induction motors
Electrical Motors – Horsepower and Amps
Horsepower rating of electrical motors compared to their ampere rating
Horsepower rating compared to electric motor ampere rating – 115 VAC and 230 VAC – are indicated below:
Ampere Rating (amps)  
Power Rating (hp) 
115 VAC  230 VAC  
Efficiency (%)  
50  70  50  70  
1/4  3.2  2.3  1.6  1.2 
1/3  4.3  3.1  2.2  1.5 
1/2  6.5  4.6  3.2  2.3 
3/4  9.7  7.0  4.9  3.5 
1  13.0  9.3  6.5  4.6 
1 1/2  19.5  13.9  9.7  7.0 
2  25.9  18.5  13.0  9.3 
5  64.9  46.3  32.4  23.2 
 1 hp = 745.7 W
Torques in Electrical Induction Motors
Torques used to describe and classify electrical motors
Torque is the turning force through a radius with the units – Nm – in the SIsystem and – lb ft – in the imperial system.
The torque developed by an asynchronous induction motor varies when the motor accelerates from full stop (or zero speed) to maximum operating speed.
Locked Rotor or Starting Torque
The Locked Rotor Torque or Starting Torque is the torque the electrical motor develop when its starts at rest or zero speed.
A high Starting Torque is more important for application or machines hard to start – as positive displacement pumps, cranes etc. A lower Starting Torque can be accepted for centrifugal fans or pumps where the start load is low or close to zero.
Pullup Torque
The Pullup Torque is the minimum torque developed by the electrical motor when it runs from zero to fullload speed (before it reaches the breakdown torque point)
When the motor starts and begins to accelerate the torque in general decrease until it reach a low point at a certain speed – the pullup torque – before the torque increases until it reach the highest torque at a higher speed – the breakdown torque – point.
The pullup torque may be critical for applications that needs power to go through some temporary barriers achieving the working conditions.
Breakdown Torque
The Breakdown Torque is the highest torque available before the torque decreases when the machine continues to accelerate to the working conditions.
Fullload (Rated) Torque or Braking Torque
The Fullload Torque is the torque required to produce the rated power of the electrical motor at fullload speed.
In imperial units the Fullload Torque can be expressed as
T = 5252 P_{hp} / n_{r} (1)
where
T = fullload torque (lb ft)
P_{hp} = rated horsepower
n_{r} = rated rotational speed (rev/min, rpm)
In metric units the rated torque can be expressed as
T = 9550 P_{kW} / n_{r} (2)
where
T = rated torque (Nm)
P_{kW} = rated power (kW)
n_{r} = rated rotational speed (rpm)
Example – Electrical Motor and Braking Torque
The torque of a 60 hp motor rotating at 1725 rpm can be expressed as:
T_{fl} = (60 hp) 5,252 / (1725 rpm)
= 182.7 lb ft
NEMA Design
NEMA (National Electrical Manufacturers Association) have classified electrical motors in four different designs where torques and startingload inertia are important criteria.
IEC/NEMA Standard Torques (percent of full load torque)  
Power (hp)  2 Pole  4 Pole  
Locked Rotor Torque  Pull Up Torque  Break Down Torque  Locked Rotor Torque  Pull Up Torque  Break Down Torque  
3  170/160  110/110  200/230  180/215  120/150  200/250 
5  160/150  110/105  200/215  170/185  120/130  200/225 
7.5  150/140  100/100  200/200  160/175  110/120  200/215 
10  150/135  100/100  200/200  160/165  110/115  200/200 
15 – 20  140/130  100/100  200/200  150/150  110/105  200/200 
Accelerating Torque
Accelerating Torque = Available Motor Torque – Load Torque
Reduced Voltage Soft Starters
Reduced Voltage Soft Starters are used to limit the start current reducing the Locked Rotor Torque or Starting Torque and are common in applications which is hard to start or must be handled with care – like positive displacement pumps, cranes, elevators and simila
Electrical Motors – Starting Devices
Directonline starters, stardelta starters, frequency drives and soft starters
Commonly used starting methods for squirrel cage motors are
 directonline starters
 stardelta starters
 frequency drives
 soft starters
DirectonLine Starters
The simplest and most common starting device is the directonline starter where the equipment consists of a main contactor and a thermal or electronic overload relay.
The disadvantage of the directonline method is very high starting current (6 to 10 times the rated motor currents) and high starting torque, causing
 slipping belts, heavy wear on bearings and gear boxes
 damaged products in the process
 water hammers in piping systems
StarDelta Starters
A stardelta starting device consists normally of three contactors, an overload relay and a timer for setting the time in the starposition (starting position).
The starting current is about 30 % of the directonline starting device. The starting torque is about 25 % of the directonline starting torque.
The stress on an application is reduced compared to the directonline starting method.
Frequency Drives
With a variable frequency drive the electrical frequency to the motor is modulated between typical 0250 Hz.
 the rated motor torque is available at lower speed
 the starting current is low – ranging 0.5 – 1 times the rated motor current
The frequency drive can also be used for soft stops.
Soft Starters
With soft starters thyristors are used to reduce starting voltage. With lower motor voltage
 the starting current and starting moment can be very low compared to other methods
Stress on the application can be close to minimum compared to other methods.
Electrical Motor Efficiency
Calculating electric motor efficiency
Electrical motor efficiency is the ratio between shaft output power – and electrical input power.
Electrical Motor Efficiency when Shaft Output is measured in Watt
If power output is measured in Watt (W), efficiency can be expressed as:
η_{m} = P_{out} / P_{in} (1)
where
η_{m} = motor efficiency
P_{out} = shaft power out (Watt, W)
P_{in} = electric power in to the motor (Watt, W)
Electrical Motor Efficiency when Shaft Output is measured in Horsepower
If power output is measured in horsepower (hp), efficiency can be expressed as:
η_{m} = P_{out} 746 / P_{in} (2)
where
P_{out} = shaft power out (horsepower, hp)
P_{in} = electric power in to the motor (Watt, W)
Primary and Secondary Resistance Losses
The electrical power lost in the primary rotor and secondary stator winding resistance are also called copper losses. The copper loss varies with the load in proportion to the current squared – and can be expressed as
P_{cl} = R I^{2} (3)
where
P_{cl} = stator winding – copper loss (W, watts)
R = resistance (Ω)
I = current (A, amps)
Iron Losses
These losses are the result of magnetic energy dissipated when when the motors magnetic field is applied to the stator core.
Stray Losses
Stray losses are the losses that remains after primary copper and secondary losses, iron losses and mechanical losses. The largest contribution to the stray losses is harmonic energies generated when the motor operates under load. These energies are dissipated as currents in the copper winding, harmonic flux components in the iron parts, leakage in the laminate core.
Mechanical Losses
Mechanical losses includes friction in the motor bearings and the fan for air cooling.
NEMA Design B Electrical Motors
Electrical motors constructed according NEMA Design B must meet the efficiencies below:
Power (hp) 
Minimum Nominal Efficiency^{1)} 
1 – 4  78.8 
5 – 9  84.0 
10 – 19  85.5 
20 – 49  88.5 
50 – 99  90.2 
100 – 124  91.7 
> 125  92.4 
^{1)} NEMA Design B, Single Speed 1200, 1800, 3600 RPM. Open Drip Proof (ODP) or Totally Enclosed Fan Cooled (TEFC) motors 1 hp and larger that operate more than 500 hours per year.
Potential Divider
Output voltage from a potential divider
The output voltage from a potential divider can be calculated as
V_{out} = V_{in }R_{2 }/ (R_{1} + R_{2}) (1)
where
V_{out} = output voltage (V)
R = resistance (Ohms, Ω)
V_{in} = input voltage (V)
Example – Potential Divider
The output voltage from a potential divider with two resistors R_{1} = 10 Ohms and R_{2} = 20 Ohms, and input voltage12 V can be calculated as
V_{out} = (12 V) (20 Ω) / ((10 Ω) + (20 Ω))
= 8 (V)
Capacitors
Capacitors and capacitance – charge and unit of charge
A capacitor is a device capable to store electrical energy.
The plates of a capacitor is charged and there is an electric field between them. The capacitor will be discharged if the plates are connected together through a resistor.
Charge of a Capacitor
The charge of a capacitor can be expressed as
Q = I t (1)
where
Q = charge (coulomb, C)
I = current (amp, A)
t = time (s)
The quantity of charge (number of electrons) is measured in the unit Coulomb – C – where
1 coulomb = 6.24 10^{18} electrons
The smallest charge that exists is the charge carried by an electron, equal to 1.602 10^{19} coulomb.
Example – Quantity of Electricity Transferred
If a current of 5 amp flows for 2 minutes, the quantity of electricity – coulombs – can be calculated as
Q = (5 A) (2 min) (60 s/min)
= 600 C
Electric Field Strength
The charged plates are separated with a dielectric – an insulating medium. The electric field strength – the ratio between the potential difference or voltage and the thickness of the dielectric can be expressed as
E = V / d (2)
where
E = electric field strength (volt/m)
V = potential difference (volt)
d = thickness of dielectric (m)
Electric Flux Density
Electric flux density is the ratio between the charge of the capacitor and the surface area of the capacitor plates and can be expressed as
D = Q / A (3)
where
D = electric flux density (coulomb/m^{2})
A = surface area of the capacitor (m^{2})
Charge and Applied Voltage
Charge on a capacitor is proportional to the applied voltage and can be expressed as
Q = C V (4)
where
C = constant of proportionality or capacitance (farad, F)
Capacitance
From (4) the capacitance can be expressed as
C = Q / V (5)
One farad is defined as the capacitance of a capacitor when there is a potential difference across the plates of one volt when holding a charge of one coulomb.
It is common to use µF (10^{6}^{ }F).
Absolute Permittivity
The ratio of electric flux density – D – to the electric field – E – is called absolute permittivity – ε – of a dielectric and can be expressed as
ε = D / E (6)
where
ε = absolute permittivity (F/m)
The absolute permittivity of free space or vacuum – also called the electric constant – ε_{0} = 8.85 10^{12} F/m.
Relative Permittivity
Relative permittivity – also called dielectric constant – is the ratio between the flux density of the field in an actual dielectric and the flux density of the field in absolute vacuum.
The actual permittivity can be calculated by multiplying the relative permittivity by ε_{0}.
ε = ε_{r} ε_{0 }_{ } (7)
Energy Stored in Capacitors
The energy stored in a capacitor can be expressed as
W = 1/2 C V^{2} (8)
W = energy stored – or work done in establishing the electric field (joules, J)