Max and min frequencies of a capacitor

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SUMMARY

The discussion focuses on calculating the maximum and minimum frequencies of a variable capacitor ranging from 10 to 365 picofarads (pF) in an LC circuit designed for radio tuning. The formula used is ω = 1/√(LC), which establishes the relationship between frequency and capacitance. To adjust the frequency range from 0.54 MHz to 1.60 MHz, the addition of a parallel capacitor is necessary, leading to the equation C_eq = C_old + C_new. The challenge lies in determining the equivalent capacitance (C_eq) and the new capacitor value (C_new) while keeping the inductance (L) constant.

PREREQUISITES
  • Understanding of LC circuit theory
  • Familiarity with capacitor values in picofarads (pF)
  • Knowledge of frequency calculations in megahertz (MHz)
  • Proficiency in solving equations with multiple variables
NEXT STEPS
  • Calculate the inductance (L) using the known frequencies and the existing capacitor value (C_old).
  • Research methods for calculating equivalent capacitance in parallel capacitor configurations.
  • Explore the impact of varying capacitance on frequency tuning in LC circuits.
  • Learn about the practical applications of variable capacitors in radio frequency (RF) circuits.
USEFUL FOR

Electronics engineers, radio frequency (RF) technicians, and students studying circuit design will benefit from this discussion, particularly those interested in tuning circuits and frequency modulation.

laminatedevildoll
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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...
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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