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 1,000 volt amperes
Power Factor - ratio of watts to volt amperes
Electric Power Formulas
W = E I (1a)
W = R I2 (1b)
W = E2/ R (1c)
where
W = power (Watts)
E = voltage (Volts)
I = current (Amperes)
R = resistance (Ohms)
Electric Current Formulas
I = E / R (2a)
I = W / E (2b)
I = (W / R)1/2 (2c)
Electric Resistance Formulas
R = E / I (3a)
R = E2/ W (3b)
R = W / I2 (3c)
Electrical Potential Formulas - Ohms Law
Ohms law can be expressed as:
E = R I (4a)
E = W / I (4b)
E = (W R)1/2 (4c)
Example - Ohm's law
A 12 volt battery supplies power to a resistance of 18 ohms.
I = (12 Volts) / (18 ohms)
= 0.67 Ampere
Electrical Motor Formulas
Electrical Motor Efficiency
μ = 746 Php / Winput (5)
where
μ = efficiency
Php = output horsepower (hp)
Winput = input electrical power (Watts)
or alternatively
μ = 746 Php / (1.732 E I PF) (5b)
Electrical Motor - Power
W3-phase = (E I PF 1.732) / 1,000 (6)
where
W3-phase = electrical power 3-phase motor (kW)
PF = power factor electrical motor
Electrical Motor - Amps
I3-phase = (746 Php) / (1.732 E μ PF) (7)
where
I3-phase = electrical current 3-phase motor (Amps)
PF = power factor electrical motor
taken from=http://www.engineeringtoolbox.com/electrical-formulas-d_455.html
B
E = Voltage / I = Amps /W = Watts
PF = Power Factor / Eff = Efficiency / HP = Horsepower
| AC Efficiency and Power Factor Formulas |
| To Find |
Single Phase |
Three Phase |
| Efficiency |
746 x HP
E x I x PF |
746 x HP
E x I x PF x 1.732 |
| Power Factor |
Input Watts
V x A |
Input Watts
E x I x 1.732 |
| Power - DC Circuits |
| Watts = E xI |
| Amps = W / E |
| Ohm's Law / Power Formulas |
 |
P = watts
I = amps
R = ohms
E = Volts
|
| Voltage Drop Formulas |
Single Phase
(2 or 3 wire) |
VD = |
2 x K x I x L
CM |
K = ohms per mil foot
(Copper = 12.9 at 75°)
(Alum = 21.2 at 75°)
| Note: |
K value changes with temperature.
See Code chapter 9, Table 8 |
L = Length of conductor in feet
I = Current in conductor (amperes)
CM = Circular mil area of conductor
|
| CM= |
2K x L x I
VD |
| Three Phase |
VD= |
1.73 x K x I x L
CM |
| CM= |
1.73 x K x L x I
VD |
C
| |
|
| BASIC ELECTRICAL CIRCUIT FORMULAS |
CIRCUIT
ELEMENT |
IMPEDANCE |
VOLT-AMP EQUATIONS |
ENERGY
(dissipated on R or stored in L, C) |
absolute
value |
complex
form |
instantaneous
values |
RMS values for sinusoidal signals |
| RESISTANCE |
R |
R |
v=iR |
Vrms=IrmsR |
E=Irms2R×t |
| INDUCTANCE |
2πfL
|
jωL |
v=L×di/dt |
Vrms=Irms×2πfL |
E=Li2/2 |
| CAPACITANCE |
1/(2πfC) |
1/jωC |
i=C×dv/dt |
Vrms=Irms/(2πfC) |
E=Cv2/2 |
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 ( j2=-1 )
|
| EQUATIONS FOR SERIES AND PARALLEL CONNECTIONS |
CIRCUIT
ELEMENT |
SERIES
CONNECTION |
PARALLEL
CONNECTION |
| RESISTANCE |
 |
Rseries=
R1+R2+... |
 |
Rparallel=
1/
(1/R1+1/R2+...) |
| INDUCTANCE |
 |
Lseries=
L1+L2+... |
 |
Lparallel=
1/(1/L1+1/L2+...) |
| CAPACITANCE |
 |
Cseries=
1/
(1/C1+1/C2+...) |
 |
Cparallel=
C1+C2+... |
| RLC IMPEDANCE FORMULAS |
| CIRCUIT CONNECTION |
COMPLEX FORM |
ABSOLUTE VALUE |
Series  |
Z=R+jωL+1/jωC |
|
Parallel  |
Z=
1/(1/R+1/jωL+jωC) |
 |
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Ohms Law Calculator
Enter any two known values and press "Calculate" to solve for the others. For example, a 100 watt light bulb operating on 120 volts AC will have 144 ohms of resistance and will draw 0.833 Amps. Enter 100 in the Watts field and 120 in the Voltage field and press Calculate to find the resistance and current. Fields should be reset to 0 before each new calculation.
Voltage (E) = Current (I) * Resistance (R)
Power (watts) = Current Squared (I^2) * Resistance (R)
Power = I*E = E^2 / R
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