Circuit Theory/LC Tuned Circuits

A circuit of one Capacitor and one inductor connected in series

${\displaystyle Z=Z_{L}+Z_{C}}$

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

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

${\displaystyle Z=LC}$

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

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

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

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

${\displaystyle lnI=-{\frac {t}{T}}+C}$

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

${\displaystyle I=Ae^{(}-{\frac {t}{T}})}$

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

${\displaystyle Z_{L}-Z_{C}=0}$ . ${\displaystyle V_{L}+V_{C}=0}$

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

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

${\displaystyle V_{C}=-V_{L}}$

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

A circuit of one Capacitor and one inductor connected in parallel

${\displaystyle {\frac {1}{Z}}={\frac {1}{Z_{L}}}+{\frac {1}{Z_{C}}}}$

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

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