*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?*  

*Correct Answer is  A: 7.12 MHz*  

### AI Response: (may contain errors)  

To find the resonant frequency of an RLC circuit, we use the formula for the resonant frequency of an LC circuit, which is given by:

$$
f_0 = \frac{1}{2\pi\sqrt{LC}}
$$

where:
- $f_0$ is the resonant frequency in hertz (Hz),
- $L$ is the inductance in henries (H),
- $C$ is the capacitance in farads (F).

Given:
- $L = 50$ microhenries = $50 \times 10^{-6}$ H,
- $C = 10$ picofarads = $10 \times 10^{-12}$ F.

Substitute these values into the formula:

$$
f_0 = \frac{1}{2\pi\sqrt{(50 \times 10^{-6})(10 \times 10^{-12})}}
$$

Calculate the expression inside the square root:

$$
LC = (50 \times 10^{-6}) \times (10 \times 10^{-12}) = 500 \times 10^{-18} = 5 \times 10^{-16}
$$

Now, calculate the square root:

$$
\sqrt{5 \times 10^{-16}} = \sqrt{5} \times 10^{-8}
$$

Then, calculate the resonant frequency:

$$
f_0 = \frac{1}{2\pi \times \sqrt{5} \times 10^{-8}}
$$

Approximating $\sqrt{5} \approx 2.236$ and $\pi \approx 3.1416$, we get:

$$
f_0 \approx \frac{1}{2 \times 3.1416 \times 2.236 \times 10^{-8}}
$$

$$
f_0 \approx \frac{1}{14.049 \times 10^{-8}}
$$

$$
f_0 \approx 7.12 \times 10^6 \text{ Hz} = 7.12 \text{ MHz}
$$

### Explanation of Choices:
- **A: 7.12 MHz** - This is the correct answer, as calculated above.
- **B: 23.5 kHz** - This is incorrect because it is much lower than the calculated resonant frequency.
- **C: 7.12 kHz** - This is incorrect because it is a factor of 1000 too low compared to the correct frequency.
- **D: 23.5 MHz** - This is incorrect because it is higher than the calculated resonant frequency.

The correct answer is **A: 7.12 MHz** because it matches the calculated resonant frequency for the given values of $L$ and $C$.