Circuit Theory/LC Tuned Circuits

Series LC

A circuit of one Capacitor and one inductor connected in series

Circuit Impedance

Z = Z_L + Z_C

$Z = j\omega L + \frac{1}{j\omega C}$

$Z = \frac{1}{j\omega C} (j\omega^2 + 1)$

Z = LC

Natural Response

At equilibrium , the total volatge of the two components are equal to zero

$L\frac{dI}{dt} + IR = 0$

$\frac{dI}{dt} = - I \frac{R}{L}$

$\int \frac{dI}{I} = - \frac{R}{L} \int dt$

$ln I = - \frac{t}{T} + C$

$I = e^(- \frac{t}{T} + C)$

$I = A e^(- \frac{t}{T})$

The Natural Response of the circuit is a Exponential Decrease in time

Resonance Response

Z_L − Z_C = 0 . V_L + V_C = 0

$\omega L = \frac{1}{\omega C}$

$\omega = \sqrt{\frac{1}{LC}}$

V_C = − V_L

In Resonance, Impedance of Inductor and Capacitance is equal and the sum of the Capacitor and Inductor's voltage are equal result in Standing Wave Oscillation . Therefore, Lossless LC series can generate Standing Wave Oscillation

LC in Parallel

A circuit of one Capacitor and one inductor connected in parallel

Circuit Impedance

$\frac{1}{Z} = \frac{1}{Z_L} + \frac{1}{Z_C}$

$Y = \frac{1}{j\omega L} + j \omega C$

$Y = \frac{1}{j\omega L} (j \omega^2 + 1)$