# Currents/voltages in RLC circuit

#### wtrow

Joined Oct 21, 2009
18
Please excuse my poorly drawn circuit. It has been in its current position for a long time, and at t=0 the switch is moved. I have to find the voltage of the capacitor and inductor at t=0, and the current of the inductor and capacitor at t=0, the voltage on the capacitor at t=infinity, and current on inductor at t=infinity. In other words, i need to find Vc(0), Ic(0), VL(0), IL(0), Vc(infinity), and IL(infinity). I'm pretty sure Vc(0)= 12V (equal to voltage across 4ohm resistor at t=0- since voltage cant change instantly in capacitors) and IL(0)= 3A (18V/(2+4), skip 5 ohm because L is treated as short circuit) If someone could tell me if those are right and how to get the rest that would be great, thanks!

#### t_n_k

Joined Mar 6, 2009
5,455
Your assumptions so far are correct. For VL(0):

At the instant the switch opens, the inductor current will initially be unchanged - i.e. 3A. The capacitor voltage is 12V at this instant. The next step in reasoning becomes a little more complicated. If the 5 ohm were not there, the assumption would be that the inductor voltage would be zero at t=0. Since, the drop across the 2 ohm would be 2x3 = 6V, the capacitor voltage is 12V and V2ohm(0) + Vc(0) =18V - so VL(0) would be zero as V2ohm(0) + Vc(0) just balances the 18V source.
But the 5 ohm is there, so does it have any effect at t=0?
If a current I were flowing in the 5 ohm away from the source (left-to-right) at t=0, then the voltage drop across the 2 ohm would be greater than 6V which would be inconsistent with the assumption of a current flowing in that direction - VC(0) still being 12V.
Conversely, if a current was flowing the other way (right-to-left) in the 5 ohm, then there would also be an inconsistency with the voltage drop across the 2 ohm being less than 6V. The only conclusion possible then is that the current in the 5 ohm at t=0 is zero and VL(0) is also zero.

For Ic(0):

Since the current in L is 3A and this same current can only flow into C at t=0 (with the switch now being open), then Ic(0)=3A.

For t=∞:

Current will flow into C until steady state conditions are again reached. Irrespective of the circuit damping (which is unknown but >0), the steady state voltage across C at t=∞ must be 18V. From that you should be able to deduce IL(∞)

#### wtrow

Joined Oct 21, 2009
18
Thank you so much for your help! I was taught that capacitors are to be treated as open circuits when they are fully charged, so at t=infinity, IL would be zero, since the circuit is incomplete, correct?

#### t_n_k

Joined Mar 6, 2009
5,455
Thank you so much for your help! I was taught that capacitors are to be treated as open circuits when they are fully charged, so at t=infinity, IL would be zero, since the circuit is incomplete, correct?
Correct - IL(∞)=0

I would argue that the capacitor steady state voltage just balances the source voltage and no further charge transfer (or current flow) can take place.