What are the initial conditions used to find the coefficients in the current equations?

by SpaceDog   Last Updated November 14, 2018 16:25 PM

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:


\$ 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) ] \$

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


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


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


\$ 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?

Related Questions

Updated November 14, 2018 00:25 AM

Updated August 28, 2018 11:25 AM

Updated August 28, 2018 23:25 PM

Updated April 20, 2018 21:25 PM

Updated February 18, 2016 03:11 AM