Tuning a parallel resonant circuit: To shift the resonant frequency to a higher value, how should the capacitance be adjusted?

Difficulty: Easy

Correct Answer: decreased

Explanation:


Introduction / Context:
Tuning resonant circuits (LC tanks) is central in communication systems and filters. The resonant frequency depends on L and C, and understanding how each parameter affects frequency enables practical tuning.


Given Data / Assumptions:

  • Parallel resonant LC circuit (losses neglected for the concept).
  • Goal: increase resonant frequency f0.
  • Inductance L held fixed; only C is adjusted.


Concept / Approach:

The resonance formula is f0 = 1 / (2 * π * √(L * C)). For fixed L, f0 varies inversely with √C. Therefore, reducing C increases f0; increasing C lowers f0.


Step-by-Step Solution:

Start: f0 = 1 / (2π√(LC)).To make f0 larger, reduce √(LC). With L fixed, reduce C.Hence, set a smaller capacitance value to raise the tuned frequency.


Verification / Alternative check:

Example: If C is reduced by a factor of 4, √C halves and f0 doubles, illustrating the inverse-square-root relationship.


Why Other Options Are Wrong:

Increasing C decreases f0. Leaving C unchanged does not shift frequency. Replacing C with an inductor changes the topology rather than tuning it upward.


Common Pitfalls:

Confusing series vs. parallel resonance; thinking f0 is inversely proportional to C (it is inversely proportional to √C).


Final Answer:

decreased

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