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

4 0

2 years ago

How do you expand polynomials?

d1i1m1o1n [39]

Answer:

Answer is on the pic

Step-by-step explanation:

I hope it's helpful!

8 0

2 years ago