What is the reactance of a 25 mF capacitor when the applied frequency is 400 Hz?

1 answer:

4 0

The Reactance of a Capacitor is t<u>he resistance it has when alternating current (A/C) with a specific frequency (f) passes through it. </u>This is measured in ohms (Ω) and its formula is:

$X_{C}=\frac{1}{2\pi fC}$ (1)

Where:

$X_{C}$ is the Reactance in ohms (Ω)

$f$ is the frequency of the alternating current in Hertz ( $Hz$ )

$C$ is the Capacitance of the Capacitor in Farad ( $F$ )

In this case we have a 25 mF capacitor with an applied frequency of 400 Hz.

Note $1mF=1*10^{-3}F=0.001F$

Well, with the given data we have to solve (1):

$X_{C}=\frac{1}{2\pi(400Hz)(25*10^{-3}F)}$

Then:

XC=0.0159 Ω This is the Reactance of the Capacitor

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1) $65\textdegree$

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