# Electrical Formulas

## The most common used electrical formulas - Ohms Law and combinations

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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:

V = R I(1a)

V = P / I(1b)

V = (P R)^{1/2}(1c)

### Electric Current - Ohm's Law

I = V / R(2a)

I = P / V(2b)

I = (P / R)^{1/2}(2c)

### Electric Resistance - Ohm's Law

R = V / I(3a)

R = V^{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 = V I(4a)

P = R I^{2}(4b)

P = V^{2}/ R(4c)

where

P = power (watts, W)

V = voltage (volts, V)

I= current (amperes, A)

R= resistance (ohms, Ω)

### Electric Energy

Electric energy is power multiplied time, or

* W = P t (5)*

*where *

*W = energy (Ws, J)*

*t = time (s)*

#### 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

*E = (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 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_{3-phase}= (V I PF 1.732) / 1,000 (7)

where

P_{3-phase}= electrical power 3-phase motor (kW)

#### Electrical Motor - Amps

I_{3-phase}= (746 P_{hp}) / (1.732 VμPF) (7)

where

I_{3-phase}= electrical current 3-phase motor (amps)

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