Question E5A02:
What is the resonant frequency of an RLC circuit if R is 22 ohms, L is 50 microhenries, and C is 40 picofarads?  

Choices:  
A: 44.72 MHz  
B: 22.36 MHz  
C: 3.56 MHz  
D: 1.78 MHz    

Correct Answer is  C: 3.56 MHz  

To find the resonant frequency of a simple RLC circuit, use:
$$f_0 = \frac{1}{2\pi\sqrt{LC}}$$
Note: In an ideal RLC, $R$ does not affect the resonant frequency (it affects bandwidth/Q), so 22 Ω is a distractor.

Compute:
- $L = 50\,\mu\text{H} = 50\times10^{-6}\,\text{H}$
- $C = 40\,\text{pF} = 40\times10^{-12}\,\text{F}$
- $LC = (50\times10^{-6})(40\times10^{-12}) = 2\times10^{-15}$
- $\sqrt{LC} = \sqrt{2\times10^{-15}} \approx 4.472\times10^{-8}$
- $$f_0 = \frac{1}{2\pi(4.472\times10^{-8})} \approx 3.56\times10^6\ \text{Hz} = 3.56\ \text{MHz}$$

Why the correct answer (C: 3.56 MHz) is correct:
- It follows the correct formula including the $2\pi$ factor.

Why the others are incorrect:
- A (44.72 MHz): Roughly $2/\sqrt{LC}$; this incorrectly doubles the result that already omitted $2\pi$.
- B (22.36 MHz): Uses $f = 1/\sqrt{LC}$, forgetting the $2\pi$ factor.
- D (1.78 MHz): Uses $f = 1/(4\pi\sqrt{LC})$, mistakenly using $4\pi$ instead of $2\pi$.

Useful formulas / rules of thumb:
- $$f_0 = \frac{1}{2\pi\sqrt{LC}}$$
- With $L$ in µH and $C$ in pF (for $f$ in MHz):  
  $$f_{\text{MHz}} \approx \frac{159}{\sqrt{L_{\mu H}\,C_{pF}}}$$
  For this problem: $\sqrt{50\times40}=\sqrt{2000}\approx 44.72$, so $f \approx 159/44.72 \approx 3.56$ MHz.