Difficulty: Medium
Correct Answer: C_app = C / (1 - ω^2 L C)
Explanation:
Introduction / Context:
Real capacitors are not ideal. Their leads and electrode geometry introduce a small series inductance L. At high frequencies, this parasitic inductance alters the net reactance, and it is convenient to describe the combined series L–C as an equivalent “apparent capacitance” C_app that reproduces the actual reactance at the operating frequency ω. This concept is vital in RF design and high-speed electronics.
Given Data / Assumptions:
Concept / Approach:
The series reactance is X_series = ωL − 1/(ωC). For an equivalent single capacitor, X_eq = −1/(ω C_app). Equate X_eq to X_series to solve for C_app. This yields a frequency-dependent capacitance that diverges at series resonance (ω^2 L C = 1), beyond which the branch appears inductive (no meaningful positive C_app).
Step-by-Step Solution:
Verification / Alternative check:
Check limits: as ω → 0, ω^2 L C → 0 ⇒ C_app → C (correct). As ω approaches 1/√(LC), denominator → 0 and |C_app| → ∞ (series resonance), matching the expected behavior of a minimum impedance point.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:
C_app = C / (1 - ω^2 L C)
Discussion & Comments