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The volt-ampere (SI symbol: VA, [1] sometimes V⋅A or V A) is the unit of measurement for apparent power in an electrical circuit. It is the product of the root mean square voltage (in volts) and the root mean square current (in amperes). [2] Volt-amperes are usually used for analyzing alternating current (AC) circuits.
The watt-second is the energy equivalent to the power of one watt sustained for one second. While the watt-second is equivalent to the joule in both units and meaning, there are some contexts in which the term "watt-second" is used instead of "joule", such as in the rating of photographic electronic flash units. [36]
Electric power is the rate of transfer of electrical energy within a circuit.Its SI unit is the watt, the general unit of power, defined as one joule per second.Standard prefixes apply to watts as with other SI units: thousands, millions and billions of watts are called kilowatts, megawatts and gigawatts respectively.
In physics and chemistry, it is common to measure energy on the atomic scale in the non-SI, but convenient, units electronvolts (eV). 1 eV is equivalent to the kinetic energy acquired by an electron in passing through a potential difference of 1 volt in a vacuum. It is common to use the SI magnitude prefixes (e.g. milli-, mega- etc) with ...
Symbol [1] Name of quantity Unit name Symbol Base units E energy: joule: J = C⋅V = W⋅s kg⋅m 2 ⋅s −2: Q electric charge: coulomb: C A⋅s I electric current: ampere
The watt-second is a unit of energy, equal to the joule. One kilowatt hour is 3,600,000 watt seconds. One kilowatt hour is 3,600,000 watt seconds. While a watt per hour is a unit of rate of change of power with time, [ iii ] it is not correct to refer to a watt (or watt-hour) as a watt per hour.
An electronvolt is the amount of energy gained or lost by a single electron when it moves through an electric potential difference of one volt. Hence, it has a value of one volt, which is 1 J/C, multiplied by the elementary charge e = 1.602 176 634 × 10 −19 C. [2] Therefore, one electronvolt is equal to 1.602 176 634 × 10 −19 J. [1]
The useful power of this linear motor is =, being the power, the useful voltage (applied voltage minus back-EMF voltage), and the current. But, since power is also equal to force multiplied by speed, the force F {\displaystyle F} of the linear motor is F = P / ( V K v ) {\displaystyle F=P/(VK_{\text{v}})} or F = I / K v {\displaystyle F=I/K ...