Impedance offset from resonance: A 10 Ω resistor, a 90 mH inductor, and a 0.015 µF capacitor are in series. What is the impedance magnitude at a frequency that is 1,200 Hz below the resonant frequency?

Introduction / Context:This question tests your ability to compute the resonant frequency and then evaluate the series RLC impedance at a specified frequency offset. Understanding how impedance varies around resonance is essential in tuning and selectivity applications.

Given Data / Assumptions:

• R = 10 Ω; L = 90 mH = 0.09 H; C = 0.015 µF = 15 × 10^-9 F.
• Frequency f = f_r − 1200 Hz.
• Ideal components; series connection.

Concept / Approach:First compute the resonant frequency f_r = 1 / (2π√(LC)). Then evaluate the net reactance X = X_L − X_C at f = f_r − 1200 Hz and compute |Z| = √(R^2 + X^2).

Step-by-Step Solution:

Verification / Alternative check:Because the frequency is below resonance, X_C > X_L and the magnitude should be large and capacitive, consistent with the computed value (~1.6 kΩ).

Why Other Options Are Wrong:

• 161 Ω: An order-of-magnitude too small.
• 3,387 Ω or 1,771 Ω: These correspond to individual reactances, not the combined magnitude.

Common Pitfalls:

• Mistaking the offset direction (above vs below resonance) and swapping X_L, X_C magnitudes.

Final Answer:1,616 Ω