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:
- is this how you find the current equations for that kind of circuit, if not, please point me in the right direction.
- 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 (??)