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Suppose I was dealing with the following voltage equation for the RLC series circuit:

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

To know the values of A and B I would do the two initial conditions:

  1. voltage at t=0 (I would solve the equation for t=0)
  2. dv/dt at t=0 (I would derivate the equation and solve for t=0)

Now suppose that I want to do the particular conditions for the current case to discover A and B.

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

What should I do? I suppose the conditions are the same, so, if I have the current equation, I have to integrate that to get voltage's but doing so, I will generate a constant of integration that will be a third unknown (I suppose it will be V0).

The whole thing is sounding a little strange.

In resume: I give you the i(t) equation of a series RLC circuit in the form

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

How do you get, for example, the values of A and B in the form of variables, I mean in terms of formula that can be used for any case?

How is that really solved?

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  • \$\begingroup\$ A is current at t = 0 and B is di/dt at t = 0. \$\endgroup\$ – Andy aka Aug 28 '18 at 11:39
  • \$\begingroup\$ thanks. What parameter is di/dt in terms of circuit? It must be something from the real world. I mean, dv/dt = i/c, but what about di/dt? \$\endgroup\$ – SpaceDog Aug 28 '18 at 11:41
  • \$\begingroup\$ ah, I see, di/dt = V/L, right? \$\endgroup\$ – SpaceDog Aug 28 '18 at 11:46
  • \$\begingroup\$ Yes because at t = 0 the inductor has the highest impedance and therefore it dictates the initial rate of change of current as per V = L di/dt because R, L and C are in series. \$\endgroup\$ – Andy aka Aug 28 '18 at 11:56
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    \$\begingroup\$ Did you really mean to have the same exponents (\$ e^{m_1t}\$)? If so, the A and B terms cannot be separated. \$\endgroup\$ – Chu Aug 28 '18 at 13:50
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Now suppose that I want to do the particular conditions for the current case to discover A and B. What should I do? I suppose the conditions are the same, so, if I have the current equation, I have to integrate that to get voltage's

If you have the current equation then A = current at t = 0 and B = di/dt at t = 0. Because all components are in series, the inductor defines the current and, di/dt = V/L as per the well-known equation for an inductor: -

$$V = L\dfrac{di}{dt}$$

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