Answer :
To solve the problem of finding the capacitance for an RLC circuit that tunes into a radio station broadcasting at 99.7 MHz, we need to use the formula for the resonant frequency [tex]f[/tex] of an RLC circuit, which is given by:
[tex]f = \frac{1}{2\pi\sqrt{LC}}[/tex]
Where:
- [tex]f[/tex] is the resonant frequency,
- [tex]L[/tex] is the inductance,
- [tex]C[/tex] is the capacitance.
We are given:
- [tex]f = 99.7 \text{ MHz} = 99.7 \times 10^6 \text{ Hz}[/tex]
- [tex]L = 1.58 \text{ } \mu\text{H} = 1.58 \times 10^{-6} \text{ H}[/tex]
We're solving for [tex]C[/tex], the capacitance.
Rearrange the formula to solve for [tex]C[/tex]:
[tex]C = \frac{1}{(2\pi f)^2 L}[/tex]
Now substitute the given values into the formula:
[tex]C = \frac{1}{(2\pi \times 99.7 \times 10^6)^2 \times 1.58 \times 10^{-6}}[/tex]
Calculate the result:
Compute [tex]2\pi \times 99.7 \times 10^6[/tex]:
[tex]2\pi \times 99.7 \times 10^6 \approx 6.267 \times 10^8[/tex]Square the result:
[tex](6.267 \times 10^8)^2 \approx 3.927 \times 10^{17}[/tex]Compute the product [tex]3.927 \times 10^{17} \times 1.58 \times 10^{-6}[/tex]:
[tex]6.203 \times 10^{11}[/tex]Calculate [tex]C[/tex]:
[tex]C \approx \frac{1}{6.203 \times 10^{11}} \approx 1.61 \times 10^{-12} \text{ F}[/tex]
Convert the capacitance to picofarads (pF):
[tex]C \approx 1.61 \text{ pF}[/tex]
So, the capacitance that should be used is approximately [tex]1.61 \text{ pF}[/tex].
