Question:

Capacitive reactance of a capacitor in an AC circuit is 6 k\(\Omega\). If the same capacitor is connected to an AC source of double the frequency, find the new capacitive reactance.

Show Hint

Capacitive reactance varies inversely with frequency: \(X_C \propto 1/f\). Doubling frequency halves reactance.
Updated On: Jul 18, 2026
  • 6 k\(\Omega\)
  • 3 k\(\Omega\)
  • 1.5 k\(\Omega\)
  • 8.5 k\(\Omega\)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Step 1: Recall formula for capacitive reactance.
\[ X_C = \frac{1}{2 \pi f C} \]

Step 2: Identify given values.
Initial reactance: \(X_{C1} = 6 \, \text{k}\Omega\), frequency \(f_1\). New frequency \(f_2 = 2 f_1\).

Step 3: Determine new reactance.
\[ X_{C2} = \frac{1}{2 \pi f_2 C} = \frac{1}{2 \pi (2 f_1) C} = \frac{X_{C1}}{2} \]

Step 4: Substitute values.
\[ X_{C2} = \frac{6}{2} = 3 \, \text{k}\Omega \]

Step 5: Verify reasoning.
Doubling frequency halves the capacitive reactance; formula consistent.

Step 6: Final conclusion.
Hence, the capacitive reactance at double frequency is:
\[ \boxed{3 \, \text{k}\Omega} \]
Was this answer helpful?
0
0

Top AP EAPCET Physics Questions

View More Questions