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2007-04-04 15:19:28 · 2 answers · asked by dolce00 1 in Science & Mathematics Engineering

2 answers

it is the average flow of current in a wire in an AC circuit.

It is made of two component reactive current and effective current.

Apparent current = sqrt(reactive current^2 + effective current^2)

2007-04-04 16:35:44 · answer #1 · answered by thatoneguy 4 · 0 0

'Apparent Current' is the one in charge at Ohm....

Sorry I couldn't resist.

While DC power is straight forward in Voltage, Current and Power, AC plays tricks due to it's cyclic nature.

A Capacitor blocks electron flow but will pass the affect of voltage by building up a charge on one side of the dielectric and a depletion on the other. Relative polarity changes when the current flow changes. (DC is totally blocked.)

An Inductor, coil gladly passes DC but the current flow will create a magnetic field. The same happens with AC but when the cycle reverses direction, the field collapses, aiding the current flow.

These two phenomenon cause the voltage or current to; lead or lag. ELI the ICE man.
Inductor; 'L' the voltage 'E' leads current 'I'.
Capacitor, 'C' the current 'I' leads voltage 'E'
Phase relationship is deduced through Root Mean Square, 'RMS', which I believe is a Trig function.

For better explanation go to:

http://www.allaboutcircuits.com/vol_2/chpt_11/2.html
We know that reactive loads such as inductors and capacitors dissipate zero power, yet the fact that they drop voltage and draw current gives the deceptive impression that they actually do dissipate power. This “phantom power” is called reactive power, and it is measured in a unit called Volt-Amps-Reactive (VAR), rather than watts.

http://www.atis.org/tg2k/_apparent_power.html
Apparent power: In alternating-current power transmission and distribution, the product of the rms voltage and amperage. Note 1: When the applied voltage and the current are in phase with one another, the apparent power is equal to the effective power, i.e., the real power delivered to or consumed by the load. If the current lags or leads the applied voltage, the apparent power is greater than the effective power. Note 2: Only effective power, i.e., the real power delivered to or consumed by the load, is expressed in watts. Apparent power is properly expressed only in volt-amperes, never watts.

2007-04-04 23:47:49 · answer #2 · answered by Caretaker 7 · 0 0

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