Question:
In an a.c. circuit, if the frequency is doubled, what happens to the capacitive reactance (\(X_{c}\)) and the current (\(I\))?
In an a.c. circuit, if the frequency is doubled, what happens to the capacitive reactance (\(X_{c}\)) and the current (\(I\))?
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Capacitive reactance is inversely proportional to frequency, and current is inversely proportional to reactance.
Updated On: Nov 12, 2026
- \(X_{c}\) is halved and \(I\) is doubled
- \(X_{c}\) is doubled, \(I\) is halved
- Both \(X_{c}\) and \(I\) are halved
- Both \(X_{c}\) and \(I\) are doubled
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
Capacitive reactance is \(X_{c}=\dfrac{1}{2\pi fC}\), so \(X_{c}\propto\dfrac{1}{f}\).
Step 2: Effect on Reactance
When \(f\) is doubled, \(X_{c}\) becomes half.
Step 3: Effect on Current
In a purely capacitive circuit, \(I=\dfrac{V}{X_{c}}\), so halving \(X_{c}\) doubles \(I\).
Step 4: Compare the options
\(X_{c}\) is halved and \(I\) is doubled, option (A).
Final Answer:
The reactance is halved and the current is doubled. \[ \boxed{\text{(A) }X_{c}\text{ is halved and }I\text{ is doubled}} \]
Capacitive reactance is \(X_{c}=\dfrac{1}{2\pi fC}\), so \(X_{c}\propto\dfrac{1}{f}\).
Step 2: Effect on Reactance
When \(f\) is doubled, \(X_{c}\) becomes half.
Step 3: Effect on Current
In a purely capacitive circuit, \(I=\dfrac{V}{X_{c}}\), so halving \(X_{c}\) doubles \(I\).
Step 4: Compare the options
\(X_{c}\) is halved and \(I\) is doubled, option (A).
Final Answer:
The reactance is halved and the current is doubled. \[ \boxed{\text{(A) }X_{c}\text{ is halved and }I\text{ is doubled}} \]
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