An RLC circuit (pictured above) is governed by two equations: \begin{align} -I_1 R &= -L \frac{dI_2}{dt} = \frac{q}{C}+V(t) \\ \frac{dq}{dt} &= I_1+I_2 \, . \end{align}

$q$ satisfies the equation $$\frac{d^2q}{dt^2}+\frac{1}{RC}\frac{dq}{dt}+\frac{1}{LC}q=-\frac{1}{R}\frac{dV}{dt}-\frac{1}{L}V \, .$$ The system is held in a steady state (i.e. $dq/dt=0$ and $V(t) = Q/C$) for negative time. At $t=0$ the voltage is switched off and $V(t) = 0$ for $t \geq 0$.

How does one derive the initial conditions for the system, i.e. $q(0)=Q$ and $\dot{q}(0)=Q/RC$?

My attempt: to calculate the charge in the steady state (just before $t=0$), I can set all derivatives with respect to time to 0. Then I get $V=-C/q$ and I can define $Q=-C/V$. I don't know how to handle the discontinuity at $t=0$ to obtain $\dot{q}(0)$ though.

  • $\begingroup$ Yep, the last sentence is the key. Asking how to handle the discontinuity is what makes this into a conceptual question. $\endgroup$
    – David Z
    Feb 12, 2014 at 3:01
  • 3
    $\begingroup$ For the love of everything good in this world please don't use $i$ and $j$ to indicate current in a circuit. Those both usually mean $\sqrt{-1}$. In physics we usually use the $i$ and in engineering they use $j$ (because they use $i$ for current). Please use $I_1$ and $I_2$ or something for currents to avoid massive confusion. $\endgroup$
    – DanielSank
    May 29, 2015 at 3:48

2 Answers 2


This is how ideal and impossible circuits elements behave, but it's a starting point for a simple analysis: At discontinuous changes in circuits,

1) inductors have the same current immediately before and after the discontinuity, but can have discontinuous voltage changes. The current will then change exponentially/sinusoidally/both.

2) capacitors have the same voltage immediately before and after the discontinuity, but can have discontinuous current changes. The capacitor voltage will then change exponentially/sinusoidally/both.

3) the current and voltage associated with resistors can both change discontinuously, following $V_R = I_R R$.

4) You must be meticulous with sign conventions on these relationships.

At $t=0^-$, the current through the inductor is constant, so the voltage across the inductor is zero. That means $i$ (through the resistor) is also zero and the voltage across the capacitor is $V(t)$ with the rightmost plate at the higher potential, if $V(t)>0$. There is no current flowing into the capacitor because it is fully charged, so no current is flowing through the inductor.

At $t=0^+$, the voltage is turned off. Technically, there are two ways to interpret this: $V(t)$ is replaced by a straight wire (a short, which is what EEs do when they kill a voltage source) or, $V(t)$ is totally removed and an open takes its place (which would be like having a switch in series with the source). The behaviors will be different, but the starting conditions of the inductor current and capacitor voltage are the same.

The inductor current will initially be zero, and the voltage across the capacitor is $V(0^-)$.


If the objective is to determine parameters at steady state it could be simpler to approach this situation with these 2 assumptions in mind :

At $t=0$ : capacitor behaves like zero resistance , inductor like infinite resistance At $t \rightarrow $\infty$ : capacitor behaves like infinite resistance , inductor like zero resistance

Based on this at $t=0$ current $i= V/R$ current $j=0$ : first derivative of $Q(0)$

At $t \rightarrow \infty$ the circuit gets broken so $i=j=0$

  • 1
    $\begingroup$ Are you aware that you can use TeX formatting on this site? I edited the answer to use it. Please hit the edit button so you can see how it works. Also, I would encourage you to revisit the answer and add punctuation. $\endgroup$
    – DanielSank
    May 29, 2015 at 3:49

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