What is the reactance of a 25 mf capacitor when the applied frequency is 400 hz?

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

  (1)

Where:

is the Reactance in ohms (Ω)

is the frequency of the alternating current in Hertz ()

is the Capacitance of the Capacitor in Farad ()

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

Note

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

Then:

XC=0.0159 Ω   This is the Reactance of the Capacitor

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?