Question:medium

In an L-C-R circuit, which one of the following statements is correct?

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$XL$ increases with frequency, $XC$ decreases with frequency.
Updated On: May 24, 2026
  • L and R oppose each other
  • R value increase with frequency
  • The inductive reactance increases with frequency
  • The capacitive reactance increases with frequency
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The Correct Option is C

Solution and Explanation

The question involves understanding the behavior of elements in an L-C-R circuit, particularly how reactances change with frequency. Let's break down each statement:

  1. Statement: L and R oppose each other
    • This statement is incorrect. In an L-C-R circuit, L (inductance) and C (capacitance) are reactive components, while R (resistance) is a resistive component. Resistance does not "oppose" inductance; instead, resistive and reactive components contribute to the total impedance of the circuit through different mechanisms.
  2. Statement: R value increases with frequency
    • This statement is incorrect. The resistance (R) of a material is typically independent of frequency in normal operation. It's the reactances (inductive and capacitive) that vary with frequency.
  3. Statement: The inductive reactance increases with frequency
    • This statement is correct. The inductive reactance \(X_L\) is given by \(X_L = 2\pi f L\), where \(f\) is the frequency and \(L\) is the inductance. As frequency \(f\) increases, \(X_L\) increases proportionally. This is why inductors oppose changes in current more strongly at higher frequencies.
  4. Statement: The capacitive reactance increases with frequency
    • This statement is incorrect. Capacitive reactance \(X_C\) is given by \(X_C = \frac{1}{2\pi f C}\), where \(f\) is the frequency and \(C\) is the capacitance. As frequency \(f\) increases, \(X_C\) actually decreases. Therefore, capacitors are more effective at higher frequencies.

Based on this analysis, the correct statement is: The inductive reactance increases with frequency.

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