The capacitive reactance is reduced by a factor of 2.
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?
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