# How do I calculate the unknowns of the current equation for a RLC series circuit?

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?

• A is current at t = 0 and B is di/dt at t = 0. – Andy aka Aug 28 '18 at 11:39
• 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? – SpaceDog Aug 28 '18 at 11:41
• ah, I see, di/dt = V/L, right? – SpaceDog Aug 28 '18 at 11:46
• 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. – Andy aka Aug 28 '18 at 11:56
• Did you really mean to have the same exponents ($e^{m_1t}$)? If so, the A and B terms cannot be separated. – Chu Aug 28 '18 at 13:50

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