Let's talk about a RLC series circuit. R, L and C are in series with a battery and a switch. The switch is open. L and C are discharged.

At t=0, the switch is closed and the battery (V1) feeds the circuit.

To find the equation for the voltage across the capacitor, I apply KVL to the circuit and get this:

\$ V_1(t) = Ri + L\frac{di}{dt} + V_C(t) \$

this later turns in to this:

\$ \frac{V_1}{LC} = \frac{d^2V_C(t)}{dt^2} + \frac{R}{L} \frac{dV_C(t)}{dt} + \frac{1}{LC}V_C(t) \$

After a few hours or brain melting, I get the final result for the equation of voltage across the capacitor


\$ V_C(t) = (At + B) \thinspace e^{-\alpha t} \$


\$ V_C(t) = Ae^{m_1t} + Be^{m_2t} \$


\$ V_C(t) = e^{- \alpha t}[K_1 \thinspace Cos(\omega_d t) + K_2 \thinspace Sin(\omega_d t) ] \$

The question now is: how do I find the current equations for the three cases?

The only thing I can see is that the current on a capacitor is equal to

\$ i_c = C \frac{dv}{dt} \$

If R, L and C are in series, the current is the same for the three.

So, all I have to do is to take the derivative of the voltage equations, that will give me, if there is no error in my math, the current equations...


\$ i(t) = e^{-\alpha t}(A -\alpha At -\alpha B) \$


\$ i(t) = m_1Ae^{m_1t} + m_2Be^{m_2t} \$


\$ i(t) = -\alpha e^{- \alpha t}(K1 Cos(\omega_d t) + K_2 Sin(\omega_d t) ] + \omega_d e^{- \alpha t}(K_2 Cos(\omega_d t) - K_1 Sin(\omega_d t) ] \$


  1. is this how you find the current equations for that kind of circuit, if not, please point me in the right direction.
  2. are these equations for the current correct?


EDIT: I have found this page that gives the same equations for current that I have found for voltage (??)

  • \$\begingroup\$ \$V_C(t)= \frac{1}{C}\int i dt\$, then differentiate the whole of your first equation. \$\endgroup\$
    – Chu
    Nov 14, 2018 at 0:42
  • \$\begingroup\$ what equation are you talking about? \$\endgroup\$
    – Duck
    Nov 14, 2018 at 0:52
  • \$\begingroup\$ See answer .... \$\endgroup\$
    – Chu
    Nov 14, 2018 at 1:00
  • \$\begingroup\$ Your differential equation looks right, but it appears you're lost in actually solving the math. Do you know how to determine if the circuit is under- critically, or over-damped? Do you know how to find the paremeters (\$a\$, \$\alpha\$, etc.)? That part is plain old math, not circuits any more. \$\endgroup\$
    – TimWescott
    Nov 14, 2018 at 1:55
  • \$\begingroup\$ Instead of finding the voltage across C, find the voltage equation across R then divide by R to find I. \$\endgroup\$
    – Andy aka
    Nov 14, 2018 at 12:58

1 Answer 1


In your first equation, write \$\small V_c= \frac{1}{C}\large \int \small i\:dt \$, then differentiate the whole of this equation (including the constant source voltage).

This gives:

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


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

Then the auxiliary equation is: \$\small m^2+\frac{R}{L}m+\frac{1}{LC}=0 \$, etc...

  • \$\begingroup\$ Looks nice until the rearranging, but then how is a second derivative equal to the square of the first derivative? \$\frac{d^2i}{dt^2} \ne (\frac{di}{dt})^2 \$ \$\endgroup\$
    – U.L.
    Nov 14, 2018 at 9:25
  • \$\begingroup\$ If I solve this equation, as you say, I end with the same equations for current and voltage. Is that correct? \$\endgroup\$
    – Duck
    Nov 14, 2018 at 14:27
  • \$\begingroup\$ It depends which component you're taking the voltage across - the current and voltage are the same shape for the resistor, but not for the capacitor or inductor. \$\endgroup\$
    – Chu
    Nov 15, 2018 at 0:34
  • \$\begingroup\$ @U.L. Read about the solution of 2nd order ODE's - the answer to your question is too long to give here, but suffice to say that \$\small m \ne \frac{di}{dt}\$ \$\endgroup\$
    – Chu
    Nov 15, 2018 at 0:38

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