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crimeas [40]
3 years ago
9

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

Mathematics
1 answer:
LuckyWell [14K]3 years ago
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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EEEE p() = 35/59 * 34/58 * 33/57 * 32/56 =  

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EESE p() = 35/59 * 34/58 * 24/57 * 33/56 =

ESEE p() = 35/59 * 24/58 * 34/57 * 33/56 =

SEEE p() = 24/59 * 35/58 * 34/57 * 33/56 =

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Step-by-step explanation:

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</span>



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