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reactance dissipate power

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zhi_yi

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the reactance component such as capacitor and inductr can not dissipate power, what is the reason?? is there any equations that show the proof?

thank you
 

General definition POWER = VI, this is also called apparent power when the circuit contains active devices.

Since your circuit contains reactive devices such as inductor and capacitor you power no longer be a real number but rather a combination of TRUE POWER and REACTIVE POWER. From the power triangle, your REactive Power is located on teh y - axis and the True Power is located on the X - axis and the vector sum of these two power is generalized P = VI or Apparent Power

Thats why in dealing with active circuits you get

Power = VI cos (θ). where your theta is the angle difference between your current and voltage. This is the actual Power being consumed in the circuit. It only calculate the power consumed on passive devices such as resistors.
 

ooo. so, in the inductive and capacitive devices, the current and voltage phase shifted 90 degree, for example, in the inductor, the current lagging voltage 90 degree, so the power would be p = v.i.cos90, and cos 90 = 0, so, the reactive devices can not dissipate power, am i right? please correct me, if it is correct, my next question is how does it come? the 90 degree, why not 100 degree, or 180 degree, or anything else, where does it come?

thank you very much :)
 

zhi_yi
,
Consider an ideal capacitor with a sinusoidal voltage input. The current through a capcitor is proportional to the rate of change of voltage, with the constant of proportionality being the Capacitance. Hence;
~ I = C dv/dt.
For a sinusiodal input voltage of 1 volt;
~ let w = radian frequency, t = time
~ I = C d[sin(wt)]/dt
~ I = [Cw]cos(wt)
cos is 90 deg out of phase with the sin.
~
For an inductor, the voltage is proportionl to the rate of change of current, with the constant of proportionality being the Inductance.
~
A similar development yields the same sin-cos relationship.
~
Remember, there is no such thing as an ideal capacitor or inductor, so in reality, real reactive components do dissipate power from there resistive components.
regards,
Kral
 

ooo.. okay, thank you :)

and then, since the reactive component cannot dissipate power, when we connect a resistor in parallel with reactive component for example capacitor, is it the thermal noise would become loss?
 

NO resistance by itself dissipates power.Noise has nothing to do with it .
Physically it can be explained like this..

when some potential is applied , the electrons in random motion in the resistance
begin to get accelerated in a direction opposite to the electric field ..but this does not keep continuing..they experience inelastic collisions with the lattice atoms ...
so they lose energy in this collisions ...this is dissipated as power ( given mathematically as I²R)

this is similar to a ball rolling down slowly in staircase ...to an extent
 

the current of capacitor and inductor don't have component parallel to the voltage
 

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