Question E5A10:
What is the resonant frequency of an RLC circuit if R is 33 ohms, L is 50 microhenries, and C is 10 picofarads?  

Choices:  
A: 7.12 MHz  
B: 23.5 kHz  
C: 7.12 kHz  
D: 23.5 MHz    

Correct Answer is  A: 7.12 MHz  

To find the resonant frequency (ignoring the small effect of R for frequency), use:
$$f_0 = \frac{1}{2\pi\sqrt{LC}}$$
Given $L = 50\,\mu\text{H} = 50\times 10^{-6}\ \text{H}$ and $C = 10\,\text{pF} = 10\times 10^{-12}\ \text{F}$:
- $LC = 50\times 10^{-6} \cdot 10\times 10^{-12} = 5\times 10^{-16}$
- $\sqrt{LC} \approx \sqrt{5}\times 10^{-8} \approx 2.236\times 10^{-8}$
- $f_0 \approx \dfrac{1}{2\pi \cdot 2.236\times 10^{-8}} \approx 7.12\times 10^{6}\ \text{Hz} = 7.12\ \text{MHz}$

Why 7.12 MHz is correct:
- The calculation above yields 7.12 MHz (Choice A).
- The series resistance $R = 33\ \Omega$ mainly affects the Q, not $f_0$. Here, $X_L = 2\pi f_0 L \approx 2235\ \Omega$, so $Q \approx X_L/R \approx 2235/33 \approx 68$; with such a high Q, the shift in resonant frequency due to R is negligible.

Why the other choices are incorrect:
- 23.5 MHz (D): Too high by about a factor of 3. This is what you’d get if you mistakenly used $C \approx 1\ \text{pF}$ instead of 10 pF (a unit/prefix error).
- 7.12 kHz (C): Off by a factor of 1000. This comes from forgetting the micro prefix on L (treating 50 μH as 50 H).
- 23.5 kHz (B): Orders of magnitude too low; with such small L and C, resonance must be in the RF MHz range. This would require a much larger $LC$ product than given.

Useful formulas and rules of thumb:
- Resonant frequency:
  $$f_0 = \frac{1}{2\pi\sqrt{LC}}$$
- Handy quick form (with L in μH and C in pF):
  $$f_{\text{MHz}} \approx \frac{159}{\sqrt{L_{\mu H}\, C_{pF}}}$$
  For $L=50\ \mu\text{H}$ and $C=10\ \text{pF}$: $f \approx 159/\sqrt{500} \approx 7.1\ \text{MHz}$.