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 = U2/ P (3b)
R = P / I2 (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 I2 (4b)
P = U2/ 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) (106 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 Php / Pinput_w (6)
where
μ = efficiency
Php = output horsepower (hp)
Pinput_w = input electrical power (watts)
or alternatively
μ = 746 Php / (1.732 V I PF) (6b)
Electrical Motor - Power
P3-phase = (U I PF 1.732) / 1,000 (7)
where
P3-phase = electrical power 3-phase motor (kW)
Electrical Motor - Amps
I3-phase = (746 Php) / (1.732 V μ PF) (8)
where
I3-phase = electrical current 3-phase motor (amps)