# Online Energy cost Calculator

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# Energy cost calculator

Electrical energy cost calculator and how to calculate.

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Electrical energy cost calculator and how to calculate.

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- Home
- Amps to Volts
- Amps to Watts
- Amps to KVA
- Amps to Kilowatts
- Joules to Volts
- Joules to Watts
- Volts to Amps
- Volts to Joules
- Volts to Watts
- Voltage Divider
- Watts to Amp
- Watts to Joules
- Watts to Volts
- Amps to VA
- Electricity Bill
- Energy Consumption
- Energy Cost
- Electron-Volts to Volts
- kVA to Amps
- Kilowatts to Amps
- Kilowatts to kilowatt-hours
- Kilowatt-hours to kilowatts
- KVA to kW
- KVA to VA
- KVA to watts
- KW to kVA
- kW to VA
- Kilowatts to Volts
- KWh to Watts
- mAh to Wh
- Ohm's Law
- Power Calculator
- Power Factor
- VA to Amps
- VA to kVA
- VA to kW
- VA to Watts
- Volts to Electron-Volts
- Volts to kilowatts
- Watts to kVA
- Watts to kWh
- Watts to VA
- Watts/Volts/Amps/Ohms
- Wh to mAh

Standard units of measurement are established by the official organization that is tasked with the standardization of international weights and measurements, ensuring that the entire world uses the same standards of weight and measurement. The French organization is called Bureau International des Poids et Mesures, or BIPM, translated to English as International Bureau of Weights and Measures. The definitions on this page are adopted from the official definitions which can be found in BIPM's International System of Units, or SI. References and links are included for each defined term which refer to information provided by BIPM.

A "volt" is a unit of electric potential, also known as electromotive force, and represents "the potential difference between two points of a conducting wire carrying a constant current of 1 ampere, when the power dissipated between these points is equal to 1 watt."[1] Stated another way, a potential of one volt appears across a resistance of one ohm when a current of one ampere flows through that resistance. Volts can be expressed in to SI base units thusly: 1 V = 1 kg times m2 times s-3 times A-1 (kilogram meter squared per second cubed per ampere),

"Voltage" (V) is the potential for energy to move and is analogous to water pressure. The characteristics of voltage are like that of water flowing through pipes. This is known as the "water-flow analogy", which is sometimes used to explain electric circuits by comparing them with a closed system of water-filled pipes, or "water circuit", that is pressurized by a pump. Refer to the image below to visualize how voltage and electric current works.

resistance and is analogous to the water pipe size. Current is proportional to the diameter of the pipe or the amount of water flowing at that pressure.

Voltage is an expression of the available energy per unit charge which drives the electric current around a closed circuit in a direct current (DC) electrical circuit. Increasing the resistance, comparable to decreasing the pipe size in the water circuit, will proportionately decrease the current, or water flow in the water circuit, which is driven through the circuit by the voltage, which is comparable to the hydraulic pressure in a water circuit.

The relationship between voltage and current is defined (in ohmic devices like resistors) by Ohm's Law. Ohm's Law is analogous to the Hagen–Poiseuille equation, as both are linear models relating flux and potential in their respective systems. Electric current (I) is a rate of flow and is measured in amps (A). Ohms (R) is a measure of resistance and is comparable to the water pipe size.

An "amp", short for ampere, is a unit of electrical current which SI defines in terms of other base units by measuring the electromagnetic force between electrical conductors carrying electric current. The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed one metre apart in vacuum, would produce between these conductors a force equal to 2×10−7 newtons per metre of length.

"Amperage" is the strength of a current of electricity expressed in amperes.

"Ohm" is a unit of an electric circuit that is defined as the electrical resistance between two points of a conductor when a constant potential difference of one volt, applied to these points, produces in the conductor a current of one ampere, the conductor not being the seat of any electromotive force.^{[3] }An ohm is expressed as...

A "watt" is a measure of power. One watt (W) is the rate at which work is done when one ampere (A) of current flows through an electrical potential difference of one volt (V). A watt can be expressed as...

It is important to know the terms and formulas on this page because they are helpful in calculating the amount of power and the size of a solar power system, whether it is an off-grid system or one that is grid-connected.

There is also a formula for power. In this formula, *P* is power, measured in watts, *I* is the current, measured in amperes, and *V* is the potential difference (or voltage drop) across the component, measured in volts.A lot of times this is also displayed as W = V * A or watts equals volts multiplied by amps.

Lets re-order this formula for an example:

- W = V * A
- V = W / A
- A = W / V

This example will show why a higher DC voltage is best in large solar systems.

Lets say you have 1000 watts of loads to run. This is equal to:

- 83.3 amps at 12 volts
- 41.6 amps at 24 volts
- 20.8 amps at 48 volts
- 8.3 amps at 120 volts
- 4.1 amps at 240 volts

Knowing how much current is flowing to your load is very important in selecting the correct wire. We take the distance into consideration to calculate the voltage loss. Ideally we don't want to exceed a 3% voltage loss. The other half of this calculation is the current. You need a larger wire to move more current. If you have a choice the higher voltage is best.

These formulas are also useful in calculating AC (alternating current) wattage to determine the size of an inverter, which converts the DC electricity from a solar array to AC that can then be used to power lights and appliances in homes and businesses. Appliances include a face plate which contains all of its electrical data. Lets suppose you have a microwave oven. The manufacturer will list an amp requirement on the electrical data of the face plate, which is usually attached to the back of the oven. Let's say that the rating on the face plate is 8.3 amps. To calculate the watts, multiply 8.3 amps by the home's voltage of 120 volts. This equals 996 watts.

Now, lets calculate how much power the microwave will use in one day. If you use the microwave for 2 hours a day, then multiply the hours per day by the watts to get watt-hours per day. So, you have 996 watts multiplied by 2 hours, which equals 1992 watt-hours per day.

When sizing a solar power system, this formula is necessary in determining the total power you use per day.

**Watts = Amps x Volts**

**Volt = Watts / Amps**

**Amps = Watts / Volts**

A volt-ampere (VA) is the voltage times the current feeding an electrical load. A kilovolt-ampere (kVA) is 1000 volt-amperes. Electrical power is measured in watts (W): The voltage times the current measured each instant.

The apparent power S in kilovolt-amps is equal to current I in amps, times the voltage V in volts, divided by 1000:

*S*_{(kVA)} = * I*_{(A)}* *×* V*_{(V)}* */ 1000

The apparent power S in kilovolt-amps is equal to phase current I
in amps, times the line to line RMS voltage V_{L-L} in volts, divided by 1000:

*S*_{(kVA)} = *√*3
× * I*_{(A)}* *×* V*_{L-L(V)}* */ 1000

The apparent power S in kilovolt-amps is equal to phase current I
in amps, times the line to neutral RMS voltage V_{L-N} in volts, divided by 1000:

*S*_{(kVA)} = 3 × * I*_{(A)}* *×* V*_{L-N(V)}* */ 1000

The power *P* in kilowatts (kW) is equal to the current *I* in amps (A), times the voltage *V* in volts (V) divided by 1000:

*P*_{(kW)} = *I*_{(A)}* *×* V*_{(V)} / 1000

The power *P* in kilowatts (kW) is equal to the
power factor *PF* times the phase current *I* in amps (A), times the RMS voltage *V* in volts (V) divided by 1000:

*P*_{(kW)} = * PF *×* I*_{(A)}* *×* V*_{(V)} / 1000

The power *P* in kilowatts (kW) is equal to square root of 3 times the
power factor *PF* times the phase current *I* in amps (A), times the line to line RMS voltage *V*_{L-L} in volts (V) divided by 1000:

*P*_{(kW)} = *√*3
× * PF *×* I*_{(A)}* *×* V*_{L-L (V)} / 1000

The power *P* in kilowatts (kW) is equal to 3 times the
power factor *PF* times the phase current *I* in amps (A), times the line to neutral RMS voltage *V*_{L-N} in volts (V) divided by 1000:

*P*_{(kW)} = 3 × * PF *×* I*_{(A)}* *×* V*_{L-N (V)} / 1000

The voltage V in volts (V) is equal to the power P in watts (W), divided by the current I in amps (A):

*V*_{(V)} = *P*_{(W)}* */* I*_{(A)}

The voltage *V* in volts (V) is equal to the current* I* in amps (A), times the resistance *R* in ohms (Ω):

*V*_{(V)} = * I*_{(A)} × *R*_{(Ω)}

The energy E in joules (J) is equal to the power P in watts (W), times the time period t in seconds (s):

*E*_{(J)} = *P*_{(W)}* *×* t*_{(s)}

A voltage divider is a simple circuit which turns a large voltage into a smaller one. Using just two series resistors and an input voltage, we can create an output voltage that is a fraction of the input. Voltage dividers are one of the most fundamental circuits in electronics.

The two resistor voltage divider is one of the most common and useful circuits used by engineers. The primary purpose of this circuit is to scale down the input voltage to a lower value based on the ratio of the two resistors. This calculator helps determine the output voltage of the divider circuit given the input (or source) voltage and the resistor values. Take note that the output voltage in actual circuits might be different, since resistor tolerance and load resistance (where the output voltage is connected) become factors.

- V
_{S}is the source voltage, measured in volts (V), - R
_{1}is the resistance of the 1st resistor, measured in Ohms (Ω). - R
_{2}is the resistance of the 2nd resistor, measured in Ohms (Ω). - V
_{out}is the output voltage, measured in volts (V),

Before we can convert volt-amps to amps we need to understand a little bit more about what these measurements are. A *volt-ampere* (VA) is a measure of the apparent electrical power while an *amp* (A) is a measure of electrical current.

Thus, to convert between them we need to use the formula for power.

Power = Voltage × Current

Using the power formula as the starting point, by reversing the formula you can convert power in volt-amps to current in amps:

Current_{(A)} = Power_{(VA)} ÷ Voltage_{(V)}

So, using the formula above, converting volt-amps to amps is as easy as dividing the volt-amps by the voltage.

The volt-amps to amps conversion is a little bit different for a three-phase electrical circuit. To calculate, use the modified three-phase formula.

Current_{(A)} = Power_{(VA)} ÷ (√3 × Voltage_{(V)})

For a three-phase electrical circuit, the current in amps is equal to the power in volt-amps divided by the square root of three times the voltage.

Energy is present in many forms all around us such as electrical, kinetic, thermal, potential, and various other forms. There is also energy in the process of transfer from one body to another i.e. heat and work. Once energy is transferred it’s always designated according to its nature. Therefore, Thermal energy may result from the transfer of heat and work done may take the shape of mechanical energy.

Energy is the fundamental need of our everyday life. So much so, that the quality of life and even its sustenance, is dependent on the availability of energy. Hence, it is imperative for us to have a conceptual understanding of the various sources of energy, the conversion of energy from one form to another and the implications of these conversions.

You must have heard that heard that Energy conversion from one form to another is a well-known phenomenon. The Law of conservation of energy even tells us that the only thing that takes place with energy is the transformation from one form to another. This means that we can convert electrical energy into heat energy and light energy, solar energy can be converted to chemical energy, potential energy can be converted into kinetic energy, Gravitational potential energy can be converted into kinetic energy etc.

Energy Conversion is defined as the process where there is a change in energy from one form to another such as the conversion of nuclear energy into heat energy, the conversion of light energy into heat, thermal energy into work etc.

There are various types of energy all around us and these energy sources can be converted from one form to another as explained below:

- Light energy can be converted to heat energy.
- Electrical energy can be converted to mechanical energy, light energy, heat energy, etc.
- Chemical energy can be converted to electrical energy.
- Thermal energy can be converted to heat energy.
- Mechanical energy can be converted to electrical energy, potential energy, etc.
- Nuclear energy can be converted to light energy and heat energy.
- The Solar energy can be converted to heat energy, chemical energy, and electrical energy.
- The Gravitational potential energy can be converted to kinetic energy.

Some examples of energy conversion are:

- A vehicle moving is an example of chemical energy being converted into kinetic energy.
- Electricity being produced with water is an example of potential energy being converted into kinetic energy.
- A ball from a height towards the ground is an example of potential energy being converted into kinetic energy.
- When you boil water using an electric kettle, it’s an example of electric energy being converted into heat energy.

These were some of the modern examples of energy conversion. Now, let’s take a look at how the energy conversion technology has evolved over the years and how the early humans used energy conversion by harnessing the natural forms of energy.

**The Use of Fire:**The first evidence of energy conversion from the early ages is when the ancient humans discovered the use of fire. They burned dried plant, wood, and animal waste and used the resultant energy for cooking and heating. This was followed by the use of mechanical energy that was used to harness the energy of flowing water and wind. However, this generation of mechanical energy with the invention of simple devices came much later around 2000 years after humans discovered the use of fire.**Waterwheels:**The earliest evidence of a machine being used to grind grain is that of a waterwheel, which was later also used to drive sawmills and pumps, driving tilt hammers for forging iron, etc. Waterwheels were the primary means of mechanical power production, rivaled only occasionally by windmills. Therefore, many industrial towns, especially in early America, were set up at locations that assured water flow all year.**Windmills:**Apart from waterwheels, windmills were also used as a source of power to replace the animal muscle. Windmills were used in various parts of the world to convert the wind energy into mechanical energy for grinding grain, pumping water, and draining the lowland areas.

**Steam Engine:**With the rapid growth of the industry around the mid-18th century (and somewhat later in various other countries), there was a pressing need for new sources of motive power, particularly those independent of geographic location and weather conditions. To meet the demands of the industry, the steam engine was developed as the first device for converting thermal energy into mechanical energy.**Newcomen Engine:**A few years later, a better and more efficient version of the steam pump which consisted of a cylinder fitted with a piston was developed. Although engines of this kind converted only about one percent of the thermal energy in the steam to mechanical energy, they remained unrivaled for more than 50 years.**Watt’s Engine:**Subsequently in the later years, the Newcomen engine was modified by adding a separate condenser to make it unnecessary to heat and cool the cylinder with each stroke. Since the cylinder and piston remained at steam temperature while the engine was operating, the fuel costs dropped by about 75 percent.

Lastly, let’s understand the law of conservation of energy to further strengthen your fundamentals.

The Conservation of energy is the principle that energy can neither be created nor destroyed. It can only be converted from one form to another or transferred from one object to another. The energy transformation is applicable to all types of energy such as kinetic energy, potential energy, light energy, heat energy, sound energy, nuclear energy, gravitational potential energy etc. These transformations would occur again and again till the time that resource is present there. The friction effect and the air-pressure are the other factors which can affect the transformation to some extent.

Thus, we can safely say that in the whole system, the total energy remains constant but only its transformation takes place.

We hope this page proved useful to you in understanding energy conversion and the various aspects related to it.

Decay by internal conversion (IC) results in the emission of an atomic electron. This electron, called the internal conversion electron, is emitted from an atom after absorbing the excited energy of a nucleus.

The power *P* in kilowatts (kW) is equal to the current *I* in amps (A), times the voltage *V* in volts (V) divided by 1000:

*P*_{(kW)} = *I*_{(A)}* *×* V*_{(V)} / 1000

The power *P* in kilowatts (kW) is equal to the power factor *PF* times the phase current *I* in amps (A), times the RMS voltage *V* in volts (V) divided by 1000:

*P*_{(kW)} = * PF *×* I*_{(A)}* *×* V*_{(V)} / 1000

The power *P* in kilowatts (kW) is equal to square root of 3 times the power factor *PF* times the phase current *I* in amps (A), times the line to line RMS voltage *V*_{L-L} in volts (V) divided by 1000:

*P*_{(kW)} = *√*3
× * PF *×* I*_{(A)}* *×* V*_{L-L (V)} / 1000

The power *P* in kilowatts (kW) is equal to 3 times the
power factor *PF* times the phase current *I* in amps (A), times the line to neutral RMS voltage *V*_{L-N} in volts (V) divided by 1000:

*P*_{(kW)} = 3 × * PF *×* I*_{(A)}* *×* V*_{L-N (V)} / 1000

The apparent power S in kilovolt-amps is equal to current I in amps, times the voltage V in volts, divided by 1000:

*S*_{(kVA)} = * I*_{(A)}* *×* V*_{(V)}* */ 1000

The apparent power S in kilovolt-amps is equal to phase current I
in amps, times the line to line RMS voltage V_{L-L} in volts, divided by 1000:

*S*_{(kVA)} = *√*3
× * I*_{(A)}* *×* V*_{L-L(V)}* */ 1000

The apparent power S in kilovolt-amps is equal to phase current I
in amps, times the line to neutral RMS voltage V_{L-N} in volts, divided by 1000:

*S*_{(kVA)} = 3 × * I*_{(A)}* *×* V*_{L-N(V)}* */ 1000

The apparent power S in volt-amps is equal to current I in amps, times the voltage V in volts:

*S*_{(VA)} = * I*_{(A)}* *×* V*_{(V)}

The apparent power S in kilovolt-amps is equal to square root if 3 current I in amps, times the line to line voltage V_{L-L} in volts:

*S*_{(VA)} = *√*3
× * I*_{(A)}* *×* V*_{L-L(V)} = 3
× * I*_{(A)}* *×* V*_{L-N(V)}

The voltage *V* in volts (V) is equal to the energy E in electron-volts (eV), divided by
the electric charge
*Q* in elementary charge or proton/electron charge (e):

*V*_{(V)} = *E*_{(eV)}* */* Q*_{(e)}

The voltage V in volts (V) is equal to 1.602176565×10^{-19} times the energy E in electron-volts (eV), divided by
the electrical charge Q in coulombs (C):

*V*_{(V)} = 1.602176565×10^{-19}
× *E*_{(eV)}* */* Q*_{(C)}

The power P in watts (W) is equal to the energy E in joules (J), divided by the time period t in seconds (s):

*P*_{(W)} = *E*_{(J)}* */* t*_{(s)}

The voltage V in volts (V) is equal to the energy E in joules (J), divided by the charge Q in coulombs (C):

*V*_{(V)} = *E*_{(J)}* */* Q*_{(C)}

The current I in amps is equal to 1000 times the apparent power S in kilovolt-amps, divided by the voltage V in volts:

*I*_{(A)} = 1000 × *S*_{(kVA)}* */ *V*_{(V)}

The phase current I in amps (with balanced loads) is equal to 1000 times the apparent power S in kilovolt-amps, divided by
the square root of 3 times the line to line RMS voltage V_{L-L} in volts:

*I*_{(A)} = 1000 × *S*_{(kVA)}* */ (*√*3
× *V*_{L-L(V)}* *)

The phase current I in amps (with balanced loads) is equal to 1000 times the apparent power
S in kilovolt-amps, divided by 3 times the line to neutral RMS voltage V_{L-N} in volts:

*I*_{(A)} = 1000 × *S*_{(kVA)}* */ (3 × *V*_{L-N(V)}* *)

The real power P in watts (W) is equal to 1000 times the apparent power S in kilovolt-amps (kVA), times the power factor PF:

*P*_{(W)} = 1000 × *S*_{(kVA)} × *PF*

The real power P in kilowatts (kW) is equal to the apparent power S in kilovolt-amps (kVA), times the power factor PF:

*P*_{(kW)} = * S*_{(kVA)} × *PF*

The apparent power S in volt-amps (VA) is equal to 1000 times the apparent power S in kilovolt-amps (kVA):

*S*_{(VA)} = 1000 × *S*_{(kVA)}

The current *I* in amps (A) is equal to 1000 times the power *P* in
kilowatts (kW), divided by the voltage *V* in volts (V):

* I*_{(A)} = 1000* *×* P*_{(kW)} / * V*_{(V)}

The phase current *I* in amps (A) is equal to 1000 times the power *P* in
kilowatts (kW), divided by the
power factor *PF* times the RMS voltage *V* in volts (V):

* I*_{(A)} = 1000* *×* P*_{(kW)} / (*PF* × * V*_{(V)}* *
)

The phase current *I* in amps (A) is equal to 1000 times the power *P* in
kilowatts (kW), divided by square root of 3 times the
power factor *PF* times the line to line RMS voltage *V*_{L-L} in volts (V):

* I*_{(A)} = 1000* *×* P*_{(kW)} / (*√*3
× * PF* × * V*_{L-L(V)}* *)

The phase current *I* in amps (A) is equal to 1000 times the power *P* in
kilowatts (kW), divided by 3 times the
power factor *PF* times the line to neutral RMS voltage *V*_{L-N} in volts (V):

* I*_{(A)} = 1000* *×* P*_{(kW)} / (3 × * PF* × * V*_{L-N(V)}* *)

The voltage *V* in volts (V) is equal to 1000 times the power
*P* in
kilowatts (kW), divided by
the current *I* in amps (A):

*V*_{(V)} = 1000 × *P*_{(kW)}* */*
I*_{(A)}

The voltage *V* in volts (V) is equal to 1000 times the power
*P* in
kilowatts (kW), divided by the
power factor *PF* times the current I in amps (A):

*V*_{(V)} = 1000 × * P*_{(kW)} / (*PF* × * I*_{(A)}* *
)

The line to line RMS voltage *V*_{L-L} in volts (V) is equal to 1000 times the power
*P* in kilowatts (kW), divided by square root of 3 times the power factor
*PF* times the current *I* in amps (A):

*V*_{L-L(V)} = 1000 × * P*_{(kW)} / (*√*3
× * PF* × * I*_{(A)}* *)

≈ 1000 × * P*_{(kW)} / (1.732
× * PF* × * I*_{(A)}* *)

The energy E in kilowatt-hours (kWh) is equal to the power P in kilowatts (kW), times the time period t in hours (h):

*E*_{(kWh)} = *P*_{(kW)}* *×* t*_{(h)}

The apparent power S in volt-amps (VA) is equal to 1000 times the real power P in kilowatts (kW), divided by the power factor PF:

*S*_{(VA)} = * *1000 × *P*_{(kW)} / *PF*