Practical Electronics/Series RLC

A circuit of 3 components connected in series

At Equilibrium, the sum of all voltages equal to zero

${\displaystyle v_{L}+v_{C}+v_{R}=0}$

${\displaystyle L{\frac {di}{dt}}+{\frac {1}{C}}\int idt+iR=0}$

${\displaystyle {\frac {d^{2}i}{dt^{2}}}+{\frac {R}{L}}{\frac {di}{dt}}+{\frac {1}{LC}}i=0}$

The equation above can be written as below

${\displaystyle {\frac {d^{2}i}{dt^{2}}}=-2\alpha {\frac {di}{dt}}-\beta i}$

With

${\displaystyle \alpha ={\frac {R}{2L}}=\beta \gamma }$

${\displaystyle \beta ={\frac {1}{LC}}={\frac {1}{T}}}$

${\displaystyle T=LC}$

${\displaystyle \gamma =RC}$

Roots of 2nd ordered differential equation above

• ${\displaystyle \alpha =\beta }$

${\displaystyle i(t)=Ae^{-\alpha t}}$

• ${\displaystyle \alpha >\beta }$

${\displaystyle i(t)=Ae^{(-\alpha \pm \lambda )t}}$

• ${\displaystyle \alpha <\beta }$

${\displaystyle i(t)=Ae^{(-\alpha \pm j\omega )t}=A(\alpha )\sin \omega t}$

The total impedance of the circuit

${\displaystyle Z=Z_{R}+Z_{L}+Z_{C}=R+0=R}$

${\displaystyle i={\frac {V}{R}}}$

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

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

${\displaystyle \omega _{o}=\pm j{\sqrt {\frac {1}{T}}}}$

${\displaystyle T=LC}$

At ${\displaystyle \omega _{o}=\pm j{\sqrt {\frac {1}{T}}}}$ the total impedance of the circuit is Z = R . Therefore, current is equal to ${\displaystyle i={\frac {V}{R}}}$

At ${\displaystyle \omega =0.Z_{C}=oo}$ , Capacitor opens circuit . Therefore, current is equal to zero

At ${\displaystyle \omega =oo.Z_{L}=oo}$ , Inductor opens circuit . Therefore, current is equal to zero