Question

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?

The capacitive reactants is doubled.
The capacitive reactants is traduced by a factor of 4.
The capacitive reactants remains constant.
The capacitive reactants is quadrupled.
The capacitive reactants is reduced by a factor of 2.

Answer 1

the capacitive reactance is reduced by a factor 2

Step-by-step explanation:

The capacitive reactance is given by:

`X_C = (1)/(2\pi f C)`

where

f is the frequency of the source

C is the capacitance

In this case, the frequency of the source is doubled:

f' = 2f

while the capacitance does not change. So the new capacitive reactance will be

`X_C' = (1)/(2\pi f' C)=(1)/(2\pi (2f)C)=(1)/(2)X_C`

so the capacitive reactance is reduced by a factor 2.