How Does Changing Frequency Affect Capacitance in an AM Radio Tuner?

AI Thread Summary
The variable capacitor in an AM radio tuner adjusts capacitance based on frequency, with an initial capacitance of 1800 pF at 550 kHz. To find the capacitance at 1600 kHz, the relationship f = 1/(2π√LC) is used, allowing for rearrangement to solve for capacitance. By calculating the inductance first, the new capacitance can be determined using the formula C = 1/(4π²f²L). After performing the calculations, the new capacitance for the 1600 kHz station is approximately 1.8 pF. This demonstrates how frequency changes directly affect capacitance in radio tuners.
airkapp
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The variable capacitor in the tuner of an AM radio has capacitance of 1800pF when the radio is tuned to a station at 550 kHz. a) What must be the capacitance for a station at 1600 kHz?

I found the frequency first and then am I supposed to setup the 1600 equal to the frequency to find the capitance.

f = f = w / 2π = 1 / (2 π √ LC )

550000 Hz = 1 / (2 π √L * 1.8E-9 Hz)

= H

Do I just use algebra to isolate for L and then use my new inductance to my find my new C.

Am I going to need to use my period? None of these methods seem to be working..
T = 1/f

thanks
air
 
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airkapp said:
Do I just use algebra to isolate for L and then use my new inductance to my find my new C.
This is what I would do given the information.

What are you getting? What is the right answer I am getting. 2.12*10^{-4}\ F
 


Thank you for your response. The frequency and capacitance are related by the equation f = 1/(2π√LC). So if we know the capacitance and frequency for one station, we can use this equation to find the capacitance for a different station. In this case, we can rearrange the equation to solve for C and then plug in the new frequency to find the new capacitance:

C = 1/(4π²f²L)

Plugging in the values, we get:

C = 1/(4π²(1600000 Hz)²L)

Now we need to find the inductance for this new frequency. We can use the same equation but rearrange it to solve for L:

L = 1/(4π²f²C)

Plugging in the values, we get:

L = 1/(4π²(1600000 Hz)²(1800 pF))

Simplifying, we get:

L = 1.24E-11 H

So the new capacitance for a station at 1600 kHz would be approximately 1.8 pF. Hope this helps!
 
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