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In `There was an error working with the wiki: Code[2]`

is performed or Energy is transferred. In the SI system of measurement, power is measured in Watts (symbol: W). As a rate of change of work done or the energy of a subsystem, power is:

: P=\frac{\mathrm{}W}{\mathrm{}t}

where

:P is power

:W is Mechanical work

:t is `There was an error working with the wiki: Code[14]`

.

The average power (often simply called "power" when the context makes it clear) is the average amount of work done or energy transferred per unit time. The instantaneous power is then the limiting value of the average power as the time interval &Deltat approaches zero.

: P=\lim_{\Delta t\rightarrow 0} \frac{\Delta W}{\Delta t} = \lim_{\Delta t\rightarrow 0} P_\mathrm{avg}

When the rate of energy transfer or work is constant, all of this can be simplified to

: P=\frac{W}{t} = \frac{E}{t} ,

where

:W and E are, respectively, the work done or energy transferred in time t.

The units of power are units of energy divided by time. The SI unit of power is the watt, which is equal to one joule per second.

Non-SI units of power include `There was an error working with the wiki: Code[3]`

(PS), `There was an error working with the wiki: Code[4]`

(CV) and `There was an error working with the wiki: Code[5]`

per minute. One unit of horsepower is equivalent to 33,000 foot-pounds per minute, or the power required to lift 550 `There was an error working with the wiki: Code[6]`

one foot in one second, and is equivalent to about 746 watts. Other units include `There was an error working with the wiki: Code[15]`

, a logarithmic measure with 1 milliwatt as reference and (food) `There was an error working with the wiki: Code[16]`

s per hour (often referred to as `There was an error working with the wiki: Code[17]`

s per hour).

In `There was an error working with the wiki: Code[7]`

done on an object is related to the forces acting on it by

:W = \mathbf{F} \cdot \mathrm{}\mathbf{s}

where

:F is Force

:s is the `There was an error working with the wiki: Code[8]`

of the object.

This is often summarized by saying that work is equal to the force acting on an object times its displacement (how far the object moves while the force acts on it). Note that only motion that is along the same axis as the force "counts", however motion in the same direction as force gives positive work, and motion in the opposite direction gives negative work, while motion perpendicular to the force yields zero work.

Differentiating by time gives that the instantaneous power is equal to the force times the object's Velocity v(t):

:P(t) = \mathbf{F}(t) \cdot \mathbf{v}(t).

The average power is then

:P_\mathrm{avg} = \frac{1}{\Delta t}\int\mathbf{F} \cdot \mathbf{v}\\mathrm{d}t.

This formula is important in characterizing Engines&mdashthe power put out by an engine is equal to the force it exerts times its velocity.

Main article: PowerPedia:Electric power

The instantaneous electrical power P delivered to a component is given by

: P(t) = I(t) \cdot V(t) \,\!

where

:P(t) is the instantaneous power, measured in Watts (`There was an error working with the wiki: Code[18]`

s per `There was an error working with the wiki: Code[19]`

)

:V(t) is the `There was an error working with the wiki: Code[20]`

(or voltage drop) across the component, measured in Volts

:I(t) is the Current (electricity) flowing through it, measured in Amperes

If the component is a `There was an error working with the wiki: Code[21]`

, then:

: P=I^2 \cdot R = \frac{V^2}{R}

where

:R = V/I is the Electrical resistance, measured in `There was an error working with the wiki: Code[9]`

s.

If the component is reactive (e.g. a Capacitor or an `There was an error working with the wiki: Code[22]`

), then the instantaneous power is negative when the component is giving stored energy back to its environment, i.e., when the current and voltage are of opposite signs.

The average power consumed by a `There was an error working with the wiki: Code[10]`

-driven linear two-terminal electrical device is a function of the `There was an error working with the wiki: Code[11]`

passing through the device, and of the phase angle between the voltage and current sinusoids. That is,

:P=I \cdot V \cdot \cos\phi \,\!

where

:P is the average power, measured in Watts

:I is the root mean square value of the sinusoidal alternating current (AC), measured in Amperes

:V is the root mean square value of the sinusoidal alternating voltage, measured in Volts

:&phi is the `There was an error working with the wiki: Code[23]`

between the voltage and the current sine functions.

The amplitudes of sinusoidal voltages and currents, such as those used almost universally in mains electrical supplies, are normally specified in terms of root mean square values. This makes the above calculation a simple matter of multiplying the two stated numbers together.

This figure can also be called the Effective power, as compared to the larger `There was an error working with the wiki: Code[24]`

which is expressed in `There was an error working with the wiki: Code[25]`

(VAR) and does not include the cos &phi term due to the current and voltage being out of phase. For simple domestic appliances or a purely resistive network, the cos &phi term (called the `There was an error working with the wiki: Code[26]`

) can often be assumed to be unity, and can therefore be omitted from the equation. In this case, the effective and apparent power are assumed to be equal.

:

P = {1 \over T} \int_{0}^{T} i(t) v(t)\, dt

Where v(t) and i(t) are, respectively, the instantaneous voltage and current as functions of time.

For purely resistive devices, the average power is equal to the product of the rms voltage and rms current, even if the waveforms are not sinusoidal. The formula works for any waveform, periodic or otherwise, that has a mean square that is why the rms formulation is so useful.

For devices more complex than a resistor, the average effective power can still be expressed in general as a power factor times the product of rms voltage and rms current, but the power factor is no longer as simple as the cosine of a phase angle if the drive is non-sinusoidal or the device is not linear.

In a train of identical pulses, the instantaneous power is a periodic function of time. The ratio of the pulse duration to the period is equal to the ratio of the average power to the peak power. It is also called the duty cycle.

In the case of a periodic signal s(t) of period T, like a train of identical pulses, the instantaneous power p(t) = |s(t)|^2 is also a periodic function of period T. The peak power is simply defined by:

: P_0 = \max (p(t))

The peak power is not always readily measurable, however, and the measurement of the average power P_\mathrm{avg} is more commonly performed by an instrument. If one defines the energy per pulse as:

: \epsilon_\mathrm{pulse} = \int_{0}^{T}p(t) \mathrm{d}t

then the average power is:

: P_\mathrm{avg} = \frac{1}{T} \int_{0}^{T}p(t) \mathrm{d}t = \frac{\epsilon_\mathrm{pulse}}{T}

One may define the pulse length \tau such that P_0\tau = \epsilon_\mathrm{pulse} so that the ratios

: \frac{P_\mathrm{avg}}{P_0} = \frac{\tau}{T}

are equal. These ratios are called the duty cycle of the pulse train.

`There was an error working with the wiki: Code[1]`

In `There was an error working with the wiki: Code[12]`

or other optical device to `There was an error working with the wiki: Code[13]`

light. It is measured in `There was an error working with the wiki: Code[27]`

s (inverse Metres), and is equal to one over the `There was an error working with the wiki: Code[28]`

of the optical device.

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`There was an error working with the wiki: Code[31]`

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— in the radiative sense, power per area

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