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Answer :
Final answer:
The capacitance required to tune into the FM station broadcasting at 99.7 MHz is approximately 1.59 pF. The range of capacitances needed to cover the entire FM band from 88.0 MHz to 108 MHz is approximately 2.03 pF to 1.66 pF.
Explanation:
To find the capacitance required to tune into the FM station broadcasting at 99.7 MHz, we can use the resonance formula:
f = 1 / (2π√(LC))
Given:
- f = 99.7 MHz = 99.7 × 106 Hz
- R = 10.00 Ω
- L = 2.00 pH = 2.00 × 10-12 H
Substituting the given values into the formula:
99.7 × 106 = 1 / (2π√(C × 2.00 × 10-12))
Simplifying the equation:
2π × 99.7 × 106 × √(C × 2.00 × 10-12) = 1
√(C × 2.00 × 10-12) = 1 / (2π × 99.7 × 106)
Squaring both sides:
C × 2.00 × 10-12 = (1 / (2π × 99.7 × 106))2
Simplifying further:
C = (1 / (2π × 99.7 × 106))2 / (2.00 × 10-12)
Calculating the value of C:
C ≈ 1.59 pF
To find the range of capacitances needed to cover the entire FM band from 88.0 MHz to 108 MHz, we can calculate the resonance frequencies for the minimum and maximum frequencies:
For the minimum frequency (88.0 MHz):
f = 88.0 × 106 Hz
Using the resonance formula:
88.0 × 106 = 1 / (2π√(C × 2.00 × 10-12))
Solving for C:
C ≈ 2.03 pF
For the maximum frequency (108 MHz):
f = 108 × 106 Hz
Using the resonance formula:
108 × 106 = 1 / (2π√(C × 2.00 × 10-12))
Solving for C:
C ≈ 1.66 pF
Therefore, the range of capacitances needed to cover the entire FM band is approximately 2.03 pF to 1.66 pF.
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