# Basics Electrical Formulas

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) / (18Ω)= 0.67 (A)

### Electric Power

P = U I(4a)P = R I^{2}(4b)P = U^{2}/ R(4c)whereP = 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*.- energy to heat water

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)whereμ = efficiencyP_{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_{3-phase}= (U I PF 1.732) / 1,000 (7)whereP_{3-phase}= electrical power 3-phase motor (kW)PF = power factor electrical motor

#### Electrical Motor - Amps

I_{3-phase}= (746 P_{hp}) / (1.732 VμPF) (8)whereI_{3-phase}= electrical current 3-phase motor (amps)PF =power factor electrical motor

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