Answer :
An RLC circuit is used in a radio to tune into an FM station broadcasting at f = 99.7 MHz, the capacitance that should be used in the RLC circuit to tune into the FM station is approximately 1.026 picofarads (pF).
The resonance condition for an RLC circuit may be used to estimate the capacitance (C) required in the RLC circuit to tune into an FM station.
An RLC circuit's resonance frequency (fr) is provided by:
fr = 1 / (2π√(LC))
Here,
f = 99.7 MHz = 99.7 × [tex]10^6[/tex] Hz
f = fr = 1 / (2π√(LC))
Now,
C = 1 / ([tex]4\pi^2f^2L[/tex])
C = 1 / ([tex]4\pi^2 * (99.7 * 10^6 Hz)^2 * 1.62 * 10^{(-6)} H[/tex])
Calculating the result:
C ≈ 1.026 × [tex]10^{(-12)[/tex] F
Thus, the capacitance that should be used in the RLC circuit to tune into the FM station is 1.026 picofarads.
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The capacitance required for the RLC circuit to tune into the FM station is 100 pF.
An RLC circuit is used in a radio to tune into an FM station broadcasting at f = 99.7 MHz. The resistance in the circuit is R = 13.0 Ω, and the inductance is L = 1.62 µH.
The reactance X of the circuit can be calculated as; X = XL - XC
Where XL is the inductive reactance and XC is the capacitive reactance; X = ωL - 1 / ωC
Where ω is the angular frequency. Since f = 99.7 MHz, ω can be calculated as; ω = 2πf= 2π × 99.7 × 10^6 rad/sX = ωL - 1 / ωCFor a resonant circuit, XL = XC. Therefore, ωL = 1 / ωCω^2 LC = 1C = 1 / ω^2 LC
The capacitance C can be obtained by rearranging the above equation as;C = 1 / (ω^2 L) = 1 / [ (2π × 99.7 × 10^6 rad/s)^2 × 1.62 × 10^-6 H] = 99.4 × 10^-12 F ≈ 100 pF.
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