Current in Electric Circuit/L, R, C in Series

Theorem
Consider the electrical circuit $K$ consisting of:
 * a resistance $R$
 * an inductance $L$
 * a capacitance $C$

in series with a source of electromotive force $E$ which is a function of time $t$.


 * [[File:CircuitRLCseries.png]]

The electric current $I$ in $K$ is given by the linear second order ODE:
 * $L \dfrac {\mathrm d^2 I} {\mathrm d t^2} + R \dfrac {\mathrm d I} {\mathrm d t} + \dfrac 1 C I = \dfrac {\mathrm d E} {\mathrm d t}$

Proof
Let:
 * $E_L$ be the drop in electromotive force across $L$
 * $E_R$ be the drop in electromotive force across $R$
 * $E_C$ be the drop in electromotive force across $C$.

From Kirchhoff's Voltage Law:
 * $E - E_L - E_R - E_C = 0$

From Ohm's Law:
 * $E_R = R I$

From Drop in EMF caused by Inductance is proportional to Rate of Change of Current:
 * $E_L = L \dfrac {\mathrm d I} {\mathrm d t}$

From Drop in EMF caused by Capacitance is proportional to Accumulated Charge:
 * $E_C = \dfrac 1 C Q$

where $Q$ is the electric charge $Q$ that has accumulated on $C$.

Thus:
 * $E - L \dfrac {\mathrm d I} {\mathrm d t} - R I - \dfrac 1 C Q = 0$

which can be rewritten:
 * $L \dfrac {\mathrm d I} {\mathrm d t} + R I + \dfrac 1 C Q = E$

Differentiating $t$ gives:
 * $L \dfrac {\mathrm d^2 I} {\mathrm d t^2} + R \dfrac {\mathrm d I} {\mathrm d t} + \dfrac 1 C I = \dfrac {\mathrm d E} {\mathrm d t}$