How do I find the current equations for the RLC Series Circuit?

by SpaceDog   Last Updated November 14, 2018 00:25 AM

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


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


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


\$ 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...


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


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


\$ 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) ] \$


  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?


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