Power
Common symbols	P
SI unit	watt (W)
In SI base units	kg⋅m2⋅s−3
Derivations from other quantities	P = E/t P = F·v P = V·I P = τ·ω
Dimension	${\displaystyle {\mathsf {L))^{2}{\mathsf {M)){\mathsf {T))^{-3))$

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In physics, power is the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity.^[1][2][3] Power is a scalar quantity.

Specifying power in particular systems may require attention to other quantities; for example, the power involved in moving a ground vehicle is the product of the aerodynamic drag plus traction force on the wheels, and the velocity of the vehicle. The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft. Likewise, the power dissipated in an electrical element of a circuit is the product of the current flowing through the element and of the voltage across the element.^[4][5]

Definition

Power is the rate with respect to time at which work is done; it is the time derivative of work:

$${\displaystyle P={\frac {dW}{dt)),}$$

We will now show that the mechanical power generated by a force F on a body moving at the velocity v can be expressed as the product:

$${\displaystyle P={\frac {dW}{dt))=\mathbf {F} \cdot \mathbf {v} }$$

If a constant force F is applied throughout a distance x, the work done is defined as ${\displaystyle W=\mathbf {F} \cdot \mathbf {x} }$. In this case, power can be written as:

$${\displaystyle P={\frac {dW}{dt))={\frac {d}{dt))\left(\mathbf {F} \cdot \mathbf {x} \right)=\mathbf {F} \cdot {\frac {d\mathbf {x} }{dt))=\mathbf {F} \cdot \mathbf {v} .}$$

If instead the force is variable over a three-dimensional curve C, then the work is expressed in terms of the line integral:

$${\displaystyle W=\int _{C}\mathbf {F} \cdot d\mathbf {r} =\int _{\Delta t}\mathbf {F} \cdot {\frac {d\mathbf {r} }{dt))\ dt=\int _{\Delta t}\mathbf {F} \cdot \mathbf {v} \,dt.}$$

From the fundamental theorem of calculus, we know that

$${\displaystyle P={\frac {dW}{dt))={\frac {d}{dt))\int _{\Delta t}\mathbf {F} \cdot \mathbf {v} \,dt=\mathbf {F} \cdot \mathbf {v} .}$$

Units

The dimension of power is energy divided by time. In the International System of Units (SI), the unit of power is the watt (W), which is equal to one joule per second. Other common and traditional measures are horsepower (hp), comparing to the power of a horse; one mechanical horsepower equals about 745.7 watts. Other units of power include ergs per second (erg/s), foot-pounds per minute, dBm, a logarithmic measure relative to a reference of 1 milliwatt, calories per hour, BTU per hour (BTU/h), and tons of refrigeration.

Average power and instantaneous power

As a simple example, burning one kilogram of coal releases more energy than detonating a kilogram of TNT,^[6] but because the TNT reaction releases energy more quickly, it delivers 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

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

Instantaneous power is the limiting value of the average power as the time interval Δt approaches zero.

$${\displaystyle P=\lim _{\Delta t\to 0}P_{\mathrm {avg} }=\lim _{\Delta t\to 0}{\frac {\Delta W}{\Delta t))={\frac {dW}{dt)).}$$

When power P is constant, the amount of work performed in time period t can be calculated as

$${\displaystyle W=Pt.}$$

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:

$${\displaystyle W_{C}=\int _{C}\mathbf {F} \cdot \mathbf {v} \,dt=\int _{C}\mathbf {F} \cdot d\mathbf {x} ,}$$

If the force F is derivable from a potential (conservative), then applying the gradient theorem (and remembering that force is the negative of the gradient of the potential energy) yields:

$${\displaystyle W_{C}=U(A)-U(B),}$$

The power at any point along the curve C is the time derivative:

$${\displaystyle P(t)={\frac {dW}{dt))=\mathbf {F} \cdot \mathbf {v} =-{\frac {dU}{dt)).}$$

In one dimension, this can be simplified to:

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

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

$${\displaystyle P(t)={\boldsymbol {\tau ))\cdot {\boldsymbol {\omega )),}$$

${\displaystyle \cdot }$

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

$${\displaystyle P(t)=pQ,}$$

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

$${\displaystyle P=F_{\text{B))v_{\text{B))=F_{\text{A))v_{\text{A)),}$$

$${\displaystyle \mathrm {MA} ={\frac {F_{\text{B))}{F_{\text{A))))={\frac {v_{\text{A))}{v_{\text{B)))).}$$

The 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

$${\displaystyle P=T_{\text{A))\omega _{\text{A))=T_{\text{B))\omega _{\text{B)),}$$

$${\displaystyle \mathrm {MA} ={\frac {T_{\text{B))}{T_{\text{A))))={\frac {\omega _{\text{A))}{\omega _{\text{B)))).}$$

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.

Electrical power

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

$${\displaystyle P(t)=I(t)\cdot V(t),}$$

• ${\displaystyle P(t)}$ is the instantaneous power, measured in watts (joules per second),
• ${\displaystyle V(t)}$ is the potential difference (or voltage drop) across the component, measured in volts, and
• ${\displaystyle I(t)}$ is the current through it, measured in amperes.

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

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

$${\displaystyle R={\frac {V}{I))}$$

Peak power and duty cycle

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

$${\displaystyle P_{0}=\max[p(t)].}$$

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

$${\displaystyle \varepsilon _{\mathrm {pulse} }=\int _{0}^{T}p(t)\,dt}$$

$${\displaystyle P_{\mathrm {avg} }={\frac {1}{T))\int _{0}^{T}p(t)\,dt={\frac {\varepsilon _{\mathrm {pulse} )){T)).}$$

One may define the pulse length ${\displaystyle \tau }$ such that ${\displaystyle P_{0}\tau =\varepsilon _{\mathrm {pulse} ))$ so that the ratios

$${\displaystyle {\frac {P_{\mathrm {avg} )){P_{0))}={\frac {\tau }{T))}$$

Radiant power

Power is related to intensity at a radius ${\displaystyle r}$; the power emitted by a source can be written as:^{[citation needed]}

$${\displaystyle P(r)=I(4\pi r^{2}).}$$