Question

In an a.c. circuit, if the frequency is doubled, what happens to the capacitive reactance (\(X_{c}\)) and the current (\(I\))?

• A. \(X_{c}\) is halved and \(I\) is doubled
• B. \(X_{c}\) is doubled, \(I\) is halved
• C. Both \(X_{c}\) and \(I\) are halved
• D. Both \(X_{c}\) and \(I\) are doubled

Hint:

Capacitive reactance is inversely proportional to frequency, and current is inversely proportional to reactance.

Answer 1

The Correct Option is A

Step 1: Substitute the New Frequency
$X_{c}'=\dfrac{1}{2\pi(2f)C}=\dfrac{X_{c}}{2}$

Step 2: Current
$I'=\dfrac{V}{X_{c}'}=\dfrac{2V}{X_{c}}=2I$

Step 3: Result
$X_{c}$ is halved and $I$ is doubled, option (A).

Final Answer:
Doubling the frequency halves $X_{c}$ and doubles the current. \[ \boxed{\text{(A) }X_{c}\text{ is halved and }I\text{ is doubled}} \]