When you turn the dial on a radio to tune it, you are adjusting a variable capacitor in an LC circuit. Suppose you tune to an AM station broadcasting at a frequency of \(1000 . \mathrm{kHz},\) and there is a \(10.0-\mathrm{mH}\) inductor in the tuning circuit. When you have tuned in the station, what is the capacitance of the capacitor?

Short Answer

Expert verified
Answer: The capacitance needed for tuning the radio to the given frequency is approximately \(2.53 \times 10^{-12}\,\text{F}\).

Step by step solution

01

Write down the given information

We are given the following information: - Frequency of the radio station: \(f = 1000 \,\text{kHz}\) - Inductance of the inductor: \(L = 10.0\, \text{mH}\)
02

Convert the frequency and inductance to SI units

Before we proceed with the calculations, let's convert the frequency and inductance to the appropriate SI units: - Frequency: \(f = 1000 \,\text{kHz} = 1000 \times 10^{3} \,\text{Hz} = 10^6\, \text{Hz}\) - Inductance: \(L = 10.0 \,\text{mH} = 10.0 \times 10^{-3}\, \text{H}\)
03

Apply the resonant frequency formula

We can use the formula for the resonance frequency of an LC circuit: $$ f = \dfrac{1}{2\pi\sqrt{LC}} $$ We need to rearrange this formula to get the capacitance, \(C\): $$ C = \dfrac{1}{(2\pi f)^2 L} $$
04

Substitute the values and calculate the capacitance

Now, let's substitute the values of \(f\) and \(L\) that we found earlier into the formula: $$ C =\dfrac{1}{(2\pi (10^6\,\text{Hz}))^2 (10.0 \times 10^{-3}\,\text{H})} $$ Calculate the value of the capacitance: $$ C \approx 2.53 \times 10^{-12}\,\text{F} $$
05

Write down the final answer

The capacitance of the capacitor needed to tune the radio to the given frequency is approximately \(2.53 \times 10^{-12}\,\text{F}\).

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