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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

CRITICALLY DAMPED

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

OVERDAMPED

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

UNDERDAMPED

\$ 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...

CRITICALLY DAMPED

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

OVERDAMPED

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

UNDERDAMPED

\$ 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) ] \$

Questions:

  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?

thanks


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

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  • \$\begingroup\$ \$V_C(t)= \frac{1}{C}\int i dt\$, then differentiate the whole of your first equation. \$\endgroup\$ – Chu Nov 14 '18 at 0:42
  • \$\begingroup\$ what equation are you talking about? \$\endgroup\$ – SpaceDog Nov 14 '18 at 0:52
  • \$\begingroup\$ See answer .... \$\endgroup\$ – Chu Nov 14 '18 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 '18 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 '18 at 12:58
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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:

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

Rearranging:

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

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

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  • \$\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 '18 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\$ – SpaceDog Nov 14 '18 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 '18 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 '18 at 0:38

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