Understanding Reactive Power in AC Circuits: Impact of Capacitors on Total Q"

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In a series AC circuit with a given impedance, both real and reactive power are present. Adding a capacitor in parallel supplies reactive power (vars), which does not change the total real power (P). The total reactive power (Q) is effectively reduced by the capacitor, leading to an improved power factor. This indicates that the capacitor acts as a static reactive power generator, enhancing circuit efficiency. Understanding power factor correction is crucial, as a poor power factor can lead to increased real power losses due to higher current demands.
noumed
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Assume a single-phase AC voltage applied to a series circuit with a certain impedance. Because of the real and imaginary part of the impedance, we get both real and reactive power. Now, if we were to connect a capacitor in parallel with the circuit, and if this capacitor supplies a certain amount of vars, what happens to the total reactive power of the circuit? My intuition tells me that:

P(total) is unchanged because the capacitor is purely reactive.
Q(total) = Q(source_before) = Q(source) + Q(cap)

So basically by adding that capacitor, you're reducing the the reactive power supplied by the voltage source, and thus increasing the power factor. Am I right?
 
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Yes, indeed capacitor is a static reactive power generator.:smile:


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Thanks! =]
 
You need to look up 'Power Factor Correction".

Poor power factor will increase real power loss because a higher current will flow in the wiring and the generator has to be capable of supplying the higher current.
 
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