## AC Power Calculator

[wp-powercal-ac]

The formula for AC power is:

**P = Irms * Vrms**

Where:

**P** is the power in watts**Irms** is the root-mean-square current in amperes**Vrms** is the root-mean-square voltage in volts

The root-mean-square **(RMS)** value of a quantity is the square root of the mean of the squares of its values. In other words, it is a way of measuring the average value of a quantity that varies over time.

The RMS current and voltage are utilized in the AC power estimation since they address the power or energy that is being utilized in the circuit. The current and voltage might differ over the long haul, yet the RMS values are consistent.

For instance, assuming the RMS current is 2 amperes and the RMS voltage is 200 volts, then, at that point, the power is 400 watts.

The AC power formula can be utilized to ascertain the power present in any AC circuit, no matter what the sort of burden on it. For instance, it very well may be utilized to work out the power in a resistor or the power the resistor disperses during heat misfortune, a capacitor, or an inductor.

The AC power formula can likewise be utilized to compute the power factor of a circuit. The power factor is a proportion of how productive the circuit is in utilizing the power that is being provided to it. A power factor of 1 ampere that the circuit is utilizing the power efficiently. A power factor of under 1 ampere that the circuit isn’t utilizing the power really.

The power factor of a circuit can be determined utilizing the accompanying formula:

**pf = cos(phi)**

Where:

**pf **is the power factor

phi is the phase angle between the current and the voltage

The phase angle is the angle between the current waveform and the voltage waveform. A phase angle of 0 degrees implies that the current and the voltage are in phase. A phase angle of 90 degrees implies that the current and the voltage are 90 degrees out of phase.

The power factor can be improved by involving capacitors or inductors in the circuit. Capacitors and inductors can be utilized to move the current waveform so it is more in phase with the voltage waveform. This will further develop the power factor and make the circuit more productive.

## DC Power Calculator

[wp-powercal-dc]

**P = V * I**

Where:

**P is the power in wattsV is the voltage in voltsI is the current in amperes**DC power is the rate at which electrical energy is converted to another form of energy, such as heat, light, or mechanical work.

The DC power formula can be used to calculate the power in any DC circuit, regardless of the type of load. For example, it can be used to calculate the power in a resistor, a capacitor, or an inductor.

For example, if the voltage is 12 volts and the current is 2 amperes, then the power is 24 watts.

The DC power formula can also be used to calculate the amount of energy that is used in a circuit. The amount of energy is calculated by multiplying the power by the time.

For example, if the power is 24 watts and the circuit is on for 1 hour, then the amount of energy used is 24 watt-hours.

The DC power formula is a simple and useful tool for calculating the power in any DC circuit.

## Energy and Power calculator

[wp-powercal-energy]

formulas for calculating energy and power in circuits with voltage, current, resistance, and time:

**Energy:****Electrical energy:***E*=*V*⋅*I*⋅*t***Heat energy:***E*=*I*2*R*⋅*t*

**Power:****Electrical power:***P*=*V*⋅*I***Heat power:***P*=*I*2*R*

Where:

*E*is the energy in joules*P*is the power in watts*V*is the voltage in volts*I*is the current in amperes*R*is the resistance in ohms*t*is the time in seconds

Let’s look at an example. Suppose we have a light bulb with a resistance of 200 ohms and a voltage of 120 volts. We want to calculate the amount of energy that the light bulb will use in 1 hour.

First, we need to find the current through the light bulb. We can do this using Ohm’s law:

**I = V / R**

**I = 120 V / 200 Ω**

**I = 0.6 A**

we know the current, we can calculate the energy used by the light bulb:

**E = I^2 R t
**

**E = (0.6 A)^2 (200 Ω) (1 h)
**

**E = 720 J**

Therefore, the light bulb will use 720 joules of energy in 1 hour.

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