# Video: Finding the Capacitance in an LC Circuit

An 𝐿𝐶 circuit in an AM tuner (in a car stereo) uses a coil with an inductance of 2.5 mH and a variable capacitor. If the natural frequency of the circuit is to be adjustable over the range 540 to 1600 kHz (the AM broadcast band), what range of capacitance is required?

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### Video Transcript

An 𝐿𝐶 circuit in an AM tuner, in a car stereo, uses a coil with an inductance of 2.5 millihenries and a variable capacitor. If the natural frequency of the circuit is to be adjustable over the range 540 to 1600 kilohertz, the AM broadcast band, what range of capacitance is required?

We can name the inductance in this circuit, 2.5 millihenries, capital 𝐿. We’ll call the lower end of the frequency range, 540 kilohertz, 𝑓 sub one. And the upper end of the range, 1600 kilohertz, we’ll call 𝑓 sub two. We want to solve for the range of capacitance values that will allow this frequency range of broadcast. We’ll call the bottom end of this range 𝑐 sub one and the top end 𝑐 sub two. If we solve for these two capacitance values individually, then we’ll know the range.

We can start off by recalling the relationship between frequency, capacitance, and inductance in an 𝐿𝐶 circuit. In this type of circuit, the frequency of the charge as it moves through the circuit is equal to one over two 𝜋 times the square root of the circuit’s inductance times its capacitance.

In this relationship, we want to solve not for frequency 𝑓 but for the capacitance 𝑐. So we can rearrange it algebraically to do that. We find that the capacitance is equal to one over four 𝜋 squared times the frequency squared multiplied by the inductance 𝐿.

This means that if we want to solve for 𝑐 sub one, the bottom end of our capacitance range, we can do so by plugging in our value for inductance 𝐿 and our 𝑓 sub one value for frequency. When we do that, taking care to put our frequency in in units of hertz and our inductance in in units of henries, we’re ready to calculate 𝑐 one on our calculator. We find that, to two significant figures, 𝑐 sub one is 3.5 times 10 to the negative 11th farads. That’s the lower end of the range of our capacitance values.

Now we move on to solving for 𝑐 sub two, the upper range of the capacitance value. To do that, we can use the same equation as for 𝑐 sub one. But we simply change out the frequency and change out our subscript for 𝑐 sub two. With that change made, we then enter these values on our calculator and find that 𝑐 two is 4.0 times 10 to the negative 12th farads. These values for 𝑐 one and 𝑐 two will enable us to broadcast over the AM frequency band.