What Determines the Resonant Frequency in a Series RLC Circuit?

[neiks997]

Homework Statement

I will give a circuit of 5 passive components and an AC voltage source producing a sinusoidal voltage at a fixed frequency of omega / (2*pi) Hz.
i will post a picture asking what is being looked for along with the circuit.

Homework Equations

z = R + jwL + 1 / jwc, jw0L + 1 / jw0c = 0, w0 = 1/ sqrt(LC), B = w2 - w1 = R/L = w0/Q, Q = w0L / R = 1/w0RC, w1 = w0 - B/2, w2 = w0 + B/2

The Attempt at a Solution

clueless on how to make an equivalent circuit.

 

[berkeman]

Homework Statement

I will give a circuit of 5 passive components and an AC voltage source producing a sinusoidal voltage at a fixed frequency of omega / (2*pi) Hz.
i will post a picture asking what is being looked for along with the circuit.

Homework Equations

z = R + jwL + 1 / jwc, jw0L + 1 / jw0c = 0, w0 = 1/ sqrt(LC), B = w2 - w1 = R/L = w0/Q, Q = w0L / R = 1/w0RC, w1 = w0 - B/2, w2 = w0 + B/2

The Attempt at a Solution

clueless on how to make an equivalent circuit.

Welcome to the PF.

It says it wants you to find the equivalent circuit impedance Z that is formed by those components in that configuration.

There are no obvious simplifications of the circuit, so you need to write the KCL equations for the circuit, to find the Z = Vin/Iin impedance. Use the complex impedances for the L and C components...

 

[neiks997]

none of the capacitors or inductors are in parallel or series with each other?

 

[berkeman]

none of the capacitors or inductors are in parallel or series with each other?

Nope. That's why you need to write the KCL equations to solve this.

 

[CWatters]

I'd be tempted to do parts 2 and 3 first. What happens to L's and C's at DC and very high frequencies?

It's been many years since I did this but I half remember a trick that relies on the impedance of the source being low to simplify the circuit?

 

[neiks997]

I'd be tempted to do parts 2 and 3 first. What happens to L's and C's at DC and very high frequencies?

It's been many years since I did this but I half remember a trick that relies on the impedance of the source being low to simplify the circuit?

brilliant