Max and min frequencies of a capacitor

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A variable capacitor ranging from 10 to 365 pF is used in an LC circuit to tune radio frequencies. The ratio of maximum to minimum frequencies can be calculated using the formula ω = 1/√(LC). To achieve a frequency range of 0.54 MHz to 1.60 MHz, an additional capacitor in parallel is necessary to adjust the capacitance. The equivalent capacitance (C_eq) must be calculated using the known frequencies and the existing inductance (L), which remains unchanged. This results in three equations with three unknowns: L, C_eq, and C_new, which need to be solved to find the required capacitance.
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A variable capacitor with a range from 10 to 365 p is used with a coil to form a variable frequency LC circuit t0 tune the input to a radio.

a. What ratio of max and min frequencies may be obtained with such capcitor.

I used that \omega=\frac 1{\sqrt(LC)} to get the ratio of max to min frequencies.

b. If this circuit is to obtain frequencies from 0.54 MHz to 1.60 mHz, the rato computed is too large. By adding a capacitor in parallel, this range may be adjused. What should the capacitance of this added be?

I said that C_{eq} = C_{old}+C_{new}

I have to compute what C_{new}. From part 1, I know my C_{old}. But, how do I find C_{eq} first? I am thinking that I have to compute that using the frequencies, but how do I get L?
 
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You're not doing anything to the coil, are you? Why would L change from what it was before adding the second capacitor?
 
I thought that the L canceled out when I was doing the ratio before. So, would I use L from part A? Do I solve for L?
 
Use the earlier inductance and the frequencies the circuit is supposed to be able to deal with. Solve for C_eq and solve for C_new.
 
You are given \omega_1, ~\omega _2. You know C_{old}. The unknowns are L,~C_{eq},~C_{new}.

You have three equations in 3 unknowns...
 
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