Inactive Tutor answered • 02/20/23
We are given the expression for the capacitance C of an LCR circuit with inductance L, capacitance C, and resistance R:
C = - (1/((D^2) * L))
To show that C has a period T = 2pi sqrt(LC), we can start by using the definition of the period for an oscillating system:
T = 2pi / omega
where omega is the angular frequency. For an LCR circuit, the angular frequency is given by:
omega = sqrt(1/LC - (R/(2L))^2)
Substituting the expression for C in terms of L and D, we get:
omega = sqrt(-1/((D^2)LC) - (R/(2L))^2)
We can simplify the first term using the identity L = 1/C(D^2), which gives:
omega = sqrt(-1/LC - (R/(2L))^2)
Now, to show that the period T = 2pi sqrt(LC), we can square both sides of the equation for omega and rearrange:
omega^2 = 1/LC - (R/(2L))^2
omega^2 LC = 1 - (R/(2L))^2 LC
T^2 / (4 pi^2) LC = 1 - (R/(2L))^2 LC
Solving for T, we get:
T = 2 pi sqrt(LC) / sqrt(1 - (R/(2L))^2 LC)
Assuming the resistance R is small compared to the inductance L and capacitance C, we can neglect the second term under the square root. This gives:
T = 2 pi sqrt(LC)
which is the desired result.
The resonant frequency of the LCR circuit is the frequency at which the circuit exhibits maximum impedance, and can be found by setting the impedance to its maximum value and solving for the frequency. At resonance, the impedance is given by:
Z = sqrt((R^2) + (omegaL - 1/(omegaC))^2)
To find the resonant frequency, we take the derivative of the impedance with respect to the angular frequency and set it equal to zero:
dZ/domega = (L - (1/C) / omega^2) (omegaL - 1/(omegaC)) / sqrt((R^2) + (omegaL - 1/(omegaC))^2) = 0
This gives two solutions:
omega = 1 / sqrt(LC) (series resonance) omega = sqrt(1/LC - (R/L)^2) (parallel resonance)
The resonant frequency is therefore either 1 / sqrt(LC) for a series circuit, or the value of omega that satisfies the second equation for a parallel circuit.
