Capacitance and Inductance.

[airkapp]

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

 

[Davorak]

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. $2.12*10^{-4}\ F$

 

[thepatientmental]

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!