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 Î©)

= 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, Î©)

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) (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

Î¼ = 746 Php / Pinput_w (6)

where

Î¼ = efficiency

Php = output horsepower (hp)

Pinput_w = input electrical power (watts)

or alternatively

Î¼ = 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)

PF = power factor electrical motor

Electrical Motor - Amps

I3-phase = (746 Php) / (1.732 V Î¼ PF) (8)

where

I3-phase = electrical current 3-phase motor (amps)

PF = power factor electrical motor

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 Î©)

= 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, Î©)

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) (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

Î¼ = 746 Php / Pinput_w (6)

where

Î¼ = efficiency

Php = output horsepower (hp)

Pinput_w = input electrical power (watts)

or alternatively

Î¼ = 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)

PF = power factor electrical motor

Electrical Motor - Amps

I3-phase = (746 Php) / (1.732 V Î¼ PF) (8)

where

I3-phase = electrical current 3-phase motor (amps)

PF = power factor electrical motor

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