In physics, power (symbol: P) is the rate at which work is performed or energy is transmitted, or the amount of energy required or expended for a given unit of time. As a rate of change of work done or the energy of a subsystem, power is:

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

where P is power, W is work and t is time.

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 Δt 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 (usually measured in seconds).


The units of power are units of energy divided by time. The SI unit of power is the watt (W), which is equal to one joule per second. Non-SI units of power include ergs per second (erg/s), horsepower (hp), metric horsepower (Pferdestärke (PS) or cheval vapeur (CV)), and foot-pounds per minute. One unit of horsepower is equivalent to 33,000 foot-pounds per minute, or the power required to lift 550 pounds one foot in one second, and is equivalent to about 746 watts. Other units include dBm, a logarithmic measure with 1 milliwatt as reference; (food) calories per hour (often referred to as kilocalories per hour); Btu per hour (Btu/h); and tons of refrigeration (12,000 Btu/h).

Mechanical powerEdit

In mechanics, the work done on an object is related to the forces acting on it by

W = F \cdot \Delta s \,


F is force
Δs is the displacement 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—the power put out by an engine is equal to the force it exerts times its velocity.

In rotational systems, power is related to the torque (τ) and angular velocity (ω):

P(t) = \mathbf{\tau}(t) \cdot \mathbf{\omega}(t) \,

The average power is therefore

P_\mathrm{avg}=\frac{1}{\Delta t}\int\mathbf{\tau} \cdot \mathbf{\omega}\;\mathrm{d}t \,

Electrical powerEdit

Main article: Electric power

Instantaneous electrical power Edit

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

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


P(t) is the instantaneous power, measured in watts (joules per second)
V(t) is the potential difference (or voltage drop) across the component, measured in volts
I(t) is the current flowing through it, measured in amperes

If the component is a resistor, then:

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


R = V/I \,

is the resistance, measured in ohms.

If the component is reactive (e.g. a capacitor or an inductor), 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.

Average electrical power for sinusoidal voltages Edit

The average power consumed by a sinusoidally-driven linear two-terminal electrical device is a function of the root mean square (rms) values of the voltage across the terminals and the current passing through the device, and of the phase angle between the voltage and current sinusoids. That is,

P = I \cdot V \cdot \cos \varphi \,


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
φ is the phase angle 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 apparent power which is expressed in volt-amperes (VA) and does not include the cos φ term due to the current and voltage being out of phase. For simple domestic appliances or a purely resistive network, the cos φ term (called the power factor) 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.

Average electrical power for AC Edit

P = {1 \over T} \int_{0}^{T} i(t) \cdot 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.

Peak power and duty cycle Edit


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.

Power in opticsEdit

Main article: Optical power

In optics, or radiometry, the term power sometimes refers to radiant flux, the average rate of energy transport by electromagnetic radiation, measured in watts. The term "power" is also, however, used to express the ability of a lens or other optical device to focus light. It is measured in dioptres (inverse metres), and equals the inverse of the focal length of the optical device.

See alsoEdit


External links Edit


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