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An RLC circuit is used in a radio to tune into an FM station broadcasting at [tex]f = 99.7 \, \text{MHz}[/tex]. The resistance in the circuit is [tex]R = 10.0 \, \Omega[/tex], and the inductance is [tex]L = 1.20 \, \mu \text{H}[/tex]. What capacitance (in pF) should be used?

Answer :

To find the capacitance needed in an RLC circuit to resonate at a particular frequency, such as an FM radio station's frequency, we need to use the formula for the resonant frequency of the RLC circuit:

[tex]f = \frac{1}{2\pi\sqrt{LC}}[/tex]

Where:

  • [tex]f[/tex] is the resonant frequency (99.7 MHz),
  • [tex]L[/tex] is the inductance (1.20 [tex]\mu[/tex]H),
  • [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 = 1.20 \text{ } \mu\text{H} = 1.20 \times 10^{-6} \text{ H}[/tex]

We need to rearrange the formula to solve for [tex]C[/tex]:

[tex]C = \frac{1}{(2\pi f)^2 L}[/tex]

First, calculate [tex](2\pi f)^2[/tex]:

[tex]2\pi f = 2\pi \times 99.7 \times 10^6 = 6.28318 \times 99.7 \times 10^6 \\
2\pi f \approx 6.28318 \times 99.7 \times 10^6 \approx 626319849.2 \\
(2\pi f)^2 = 626319849.2^2 \approx 3.92 \times 10^{17}[/tex]

Now substitute [tex](2\pi f)^2[/tex] and [tex]L[/tex] into the formula for [tex]C[/tex]:

[tex]C = \frac{1}{3.92 \times 10^{17} \times 1.20 \times 10^{-6}} \approx \frac{1}{4.704 \times 10^{11}} \approx 2.13 \times 10^{-12} \text{ F}\\[/tex]

Converting farads to picofarads (1 F = [tex]10^{12}[/tex] pF):

[tex]C \approx 2.13 \times 10^{-12} \text{ F} = 2.13 \text{ pF}[/tex]

Therefore, the required capacitance is approximately 2.13 picofarads. This is the value needed to tune the radio circuit to the FM station broadcasting at 99.7 MHz.

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