Electrical Formulas
Commonly used electrical formulas like Ohms Law and more.
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
Electrical Potential  Ohm's Law
Ohm's law can be expressed as:
U = R I (1a)
U = P / I (1b)
U = (P R)^{1/2} (1c)
Electric Current  Ohm's Law
I = U / R (2a)
I = P / U (2b)
I = (P / R)^{1/2} (2c)
Electric Resistance  Ohm's Law
R = U / I (3a)
R = U^{2}/ P (3b)
R = P / I^{2} (3c)
Example  Ohm's law
A 12 volt battery supplies power to a resistance of 18 ohms.
I = (12 V) / (18 Ω)
= 0.67 (A)
Electric Power
P = U I (4a)
P = R I^{2} (4b)
P = U^{2}/ R (4c)
where
P = power (watts, W, J/s)
U = voltage (volts, V)
I = current (amperes, A)
R = resistance (ohms, Ω)
Electric Energy
Electric energy is power multiplied with time:
W = P t (5)
where
W = energy (Ws, J)
t = time (s)
Alternative  power can be expressed
P = W / t (5b)
Power is consumption of energy by consumption of time.
Example  Energy lost in a Resistor
A 12 V battery is connected in series with a resistance of 50 ohm. The power consumed in the resistor can be calculated as
P = (12 V)^{2} / (50 ohm)
= 2.9 W
The energy dissipated in 60 seconds can be calculated
W = (2.9 W) (60 s)
= 174 Ws, J
= 0.174 kWs
= 4.8 10^{5} kWh
Example  Electric Stove
An electric stove consumes 5 MJ of energy from a 230 V power supply when turned on in 60 minutes.
The power rating  energy per unit time  of the stove can be calculated as
P = (5 MJ) (10^{6} J/MJ) / ((60 min) (60 s/min))
= 1389 W
= 1.39 kW
The current can be calculated
I = (1389 W) / (230 V)
= 6 ampere
Electrical Motors
Electrical Motor Efficiency
μ = 746 P_{hp} / P_{input_w} (6)
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) (6b)
Electrical Motor  Power
P_{3phase} = (U I PF 1.732) / 1,000 (7)
where
P_{3phase} = electrical power 3phase motor (kW)
Electrical Motor  Amps
I_{3phase} = (746 P_{hp}) / (1.732 VμPF) (8)
where
I_{3phase} = electrical current 3phase motor (amps)
Related Topics

Electrical
Electrical units, amps and electrical wiring, wire gauge and AWG, electrical formulas and motors.
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