Power (physics)

For other types of power, see Power (disambiguation).

In physics, power is the rate at which energy is transferred, used, or transformed. For example, the rate at which a light bulb transforms electrical energy into heat and light is measured in watts—the more wattage, the more power, or equivalently the more electrical energy is used per unit time.

Energy transfer can be used to do work, so power is also the rate at which this work is performed. The output power of an electric motor is the product of the torque the motor generates and the angular velocity of its output shaft. The power expended to move a vehicle is the product of the traction force of the wheels and the velocity of the vehicle.

The integral of power over time defines the work done. Because this integral depends on the trajectory of the point of application of the force and torque, this calculation of work is said to be "path dependent."

Ansel Adams photograph of Electrical Wires of the Boulder Dam Power Units, 1941-1942.

Units

The dimension of power is energy divided by time. The SI unit of power is the watt (W), which is equal to one joule per second. Other 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 horsepower is equivalent to 33,000 foot-pounds per minute, or the power required to lift 550 pounds by one foot in one second, and is equivalent to about 746 watts. Other units include dBm, a relative 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).

Average power

As a simple example, burning a kilogram of coal releases much more energy than does detonating a kilogram of TNT, but because the TNT reaction releases energy much more quickly, it delivers far more power than the coal. If ΔW is the amount of work performed during a period of time of duration Δt, the average power P_avg over that period is given by the formula

$P_\mathrm{avg} = \frac{\Delta W}{\Delta t}\,.$

It is the average amount of work done or energy converted per unit of time. The average power is often simply called "power" when the context makes it clear.

The instantaneous power is then the limiting value of the average power as the time interval Δt approaches zero.

$P = \lim _{\Delta t\rightarrow 0} P_\mathrm{avg} = \lim _{\Delta t\rightarrow 0} \frac{\Delta W}{\Delta t} = \frac{dW}{dt}\,.$

In the case of constant power P, the amount of work performed during a period of duration T is given by:

$W = PT\,.$

In the context of energy conversion it is more customary to use the symbol E rather than W.

Mechanical power

Power in mechanical systems is the combination of forces and movement. In particular, power is the product of a force on an object and the object's velocity, or the product of a torque on a shaft and the shaft's angular velocity.

Mechanical power is also described as the time derivative of work. In mechanics, the work done by a force F on an object that travels along a curve C is given by the line integral:

$W_C = \int_{C} \bold{F} \cdot \mathrm{d}\bold{x} = \int_{C}\bold{F}\cdot \bold{v}dt,$

where x defines the path C and v is the velocity along this path. The time derivative of the equation for work yields the instantaneous power,

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

In rotational systems, power is the product of the torque τ and angular velocity ω,

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

where ω measured in radians per second.

In fluid power systems such as hydraulic actuators, power is given by

$P(t) = pQ, \!$

where p is pressure in pascals, or N/m² and Q is volumetric flow rate in m³/s in SI units.

Mechanical advantage

If a mechanical system has no losses then the input power must equal the output power. This provides a simple formula for the mechanical advantage of the system.

Let the input power to a device be a force F_A acting on a point that moves with velocity v_A and the output power be a force F_B acts on a point that moves with velocity v_B. If there are no losses in the system, then

$P = F_A v_A = F_B v_B, \!$

and the mechanical advantage of the system is given by

$MA = \frac{F_B}{F_A} = \frac{v_A}{v_B}.$

A similar relationship is obtained for rotating systems, where T_A and ω_A are the torque and angular velocity of the input and T_B and ω_B are the torque and angular velocity of the output. If there are no losses in the system, then

$P = T_A \omega_A = T_B \omega_B, \!$

which yields the mechanical advantage

$MA = \frac{T_B}{T_A} = \frac{\omega_A}{\omega_B}.$

These relations are important because they define the maximum performance of a device in terms of velocity ratios determined by its physical dimensions. See for example gear ratios.

Power in optics

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.

Electrical power

Main article: 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 (joules per second)

V(t) is the potential difference (or voltage drop) across the component, measured in volts

I(t) is the current through it, measured in amperes

If the component is a resistor with time-invariant voltage to current ratio, then:

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

where

$R = V/I \,$

is the resistance, measured in ohms.

Peak power and duty cycle

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 (see text for definitions).

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.

See also

• Simple machines
• Mechanical advantage
• Motive power
• Orders of magnitude (power)
• Pulsed power
• Intensity — in the radiative sense, power per area
• Power gain — for linear, two-port networks.