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A capacitor is connected across an ac source. Suppose the frequency of the source is doubled. What happens to the capacitive reactance of the inductor?.

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The capacitive reactance is reduced by a factor of 2.

Calculation:

We know the capacitive reactance is given as,

[tex]Xc = \frac{1}{2\pi fC}[/tex]

where,

[tex]Xc[/tex] = capacitive reactance

f = frequency

C = capacitance

It is given that frequency is doubled, i.e.,

f' = 2f

To find,

[tex]Xc'[/tex] =?

[tex]Xc' = \frac{1}{2\pi f'C}[/tex]

      [tex]= \frac{1}{2\pi 2f C}[/tex]

      [tex]= \frac{1}{2} (\frac{1}{2\pi fC} )[/tex]

[tex]Xc'[/tex]  [tex]= \frac{1}{2} Xc[/tex]

Therefore, the capacitive reactance is reduced by a factor of 2.

I understand the question you are looking for is this:

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?

  1. The capacitive reactance is doubled.
  2. The capacitive reactance is traduced by a factor of 4.
  3. The capacitive reactance remains constant.
  4. The capacitive reactance is quadrupled.
  5. The capacitive reactance is reduced by a factor of 2.

Learn more about capacitive reactance here:

https://brainly.com/question/23427243

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