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Suppose 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.

I apply KVL to the circuit and find three equations:

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) ] \$

To find the coefficients of those equations I apply the two initial conditions:

  1. I solve the equations for t=0 and for the initial voltage across the capacitor.
  2. I take the derivative of the equation and solve for t=0.

Now lets talk about the current equations.

I never understood why but apparently the current equations are the same, or

CRITICALLY DAMPED

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

OVERDAMPED

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

UNDERDAMPED

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

What are the two conditions I must use for the current equations to find the coefficients?

In the voltage equations I used the initial voltage across the capacitor and the derivative of voltage (current).

Now I have the current equations.

One condition must be to solve the equations for t=0, but what about the second condition? How do I find the coefficients of the current equations?

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For a series RLC network, think about what the circuit looks like at DC steady state. After the switch closes, current will start to flow through all the components. However, eventually the capacitor will charge fully.

At this point, what does the capacitor look like? What does this say about the current in the circuit? How do these descriptions relate to your initial and final conditions?

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  • \$\begingroup\$ The inductor will prevent the current from flowing, as soon as the switch is closed, isn't it? inductors hate sudden changes in current. When the capacitor is charged, current is zero. Sorry but what you say does not make any sense for me, in terms of what I have asked. \$\endgroup\$ – SpaceDog Nov 14 '18 at 17:01
  • \$\begingroup\$ @SpaceDog I've made a minor edit to my answer. You're correct that the current in the circuit is still zero at \$t = 0^{+}\$. However, current will start to flow, and eventually the inductor will look just like a wire, right? And correct, when the capacitor is fully charged, the current stops. But this time, the voltage across the capacitor is different than it was at \$t = 0\$. \$\endgroup\$ – Shamtam Nov 14 '18 at 17:24
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The current through the inductor cannot change abruptly. So, iL(0-) = iL(0+). You can use inductor current at time instant 0 i.e. iL(0) as an initial condition.

For the second initial condition you can use the following formula diL(0)/dt = vL(0)/L. To find vL(0), you can apply KVL at your circuit for the time instant 0.

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